PATENT ABSTRACT
A choice generation method for a numerical multiple choice question comprises steps: establishing a numerical multiple choice question, and obtaining a correct numerical answer of the numerical multiple choice question; determining the number of numerical choices for the numerical multiple choice question; and generating a plurality of the numerical choices either greater or smaller than the correct numerical answer according to a choice-answer relationship. The present invention applies to generate numerical choices for a numerical multiple choice question, and the generated numerical choices are different from the numerical correct answer.

PATENT DESCRIPTION
BACKGROUND OF THE INVENTION 
       [0001]    1. Field of the Invention 
         [0002]    The present invention relates to a choice generation method for a multiple choice question, particularly to a choice generation method for a numerical multiple choice question, 
         [0003]    2. Description of the Related Art 
         [0004]    In an examination, multiple choice questions are more convenient for test scorers. However, the conventional numerical choices of a multiple choice question are usually of fixed values or order. Even though the numerical choices are arranged randomly, the order of the numerical choices is still the same after the numerical choices are rearranged according to their values. Thus, the respondents may select the answers not via solving the questions hut via memorizing the answers. An approach to solve the problem is to reform the questions and choices and add them to the test item database. However, such an approach will make the test item database bulky and hard to administrate. 
         [0005]    In order to solve the abovementioned problems, some test item databases provide fill-in-the-blank questions containing some dependent variables, such as n1to10 for example, a test question is “the length of a park trail is 40 m, and the jogging speed of a person is n1to10 m/sec; how much time will the person take to walk through the park trail?” The equation to solve the question is “time=length÷speed=40 /n1to10”. Suppose that n1to10=7. Thus is obtained the answer 40/7=5.7142 . . . by substituting 7 for n1to10 the equation. Therefore, the respondent should fill [0/7=5.7142 . . . ] the blank. 
         [0006]    However, how many digits behind the decimal point should be used is likely to be a controversial issue in the abovementioned fill-in-the blank questions. Therefore, some test item databases provide choice-type calculation questions to solve the abovementioned problem, wherein each test question has some dependent variables (such as n1 to 10). Further, the correct answer of a choice-type calculation question may be substituted into various equations to generate choices of the choice-type calculation question, and the choices are varied with the value of the dependent variable in the choice-type calculation question. For example, the choices of a choice-type calculation question may be stated as follows: a: (40/n1 to 10) seconds; b: (40/n1 to 10)*2 seconds; c: (40/n1 to 10)/2 seconds; d; (40/n1 to 10)/3 seconds, wherein a is a correct answer and b, c, d are incorrect answers. Suppose that n1 to 10=5. Thus, the choice-type calculation question is stated as follows: the length of a park trail is 40 m, and the jogging speed of a person is 5 m/see; how much time will the person take to walk through the park trail a, 8.00 seconds; b. 2.67 seconds; c, 16.0 seconds; d. 4.00 seconds. The choices arranged from large to small are 16, 8, 4, 2.67. and the correct answer is 8. Suppose that n1 to 10=7. The choices are a. 5.71 seconds; b. 11.4 seconds; c. 2.86 seconds; d. 1.9 seconds. The choices arranged from large to small are 11.4, 5.71, 2,86, 1.9, and the correct answer is 5.71. Consequently, no matter whether n1 to 10=5 or 7, the correct answer is always the second largest choice. Thus, the respondents are easy to see through the rule and able to guess the correct answers. Then, the calculation questions become memorization tests. 
         [0007]    Accordingly, the present invention proposes a choice generation method for a multiple choice question to solve the abovementioned problems. 
       SUMMARY OF THE INVENTION 
       [0008]    The primary objective of the present invention is to provide a choice generation method for a numerical multiple choice question, which applies to an item database to generate choices of a numerical multiple choice question, and which uses a random integer variable to vary the order of the correct answer appearing in the sequence of the numerical choices 
         [0009]    Another objective of the present invention is to provide a choice generation method for a numerical multiple choice question, which generates numerical choices in form of an arithmetic or geometric sequence including the correct answer. 
         [0010]    To achieve the abovementioned objectives, the present invention proposes a choice generation method for a numerical multiple choice question, which comprises steps: establishing a numerical multiple choice question; obtaining a correct numerical answer of the numerical multiple choice question; determining the number of the numerical choices of the numerical multiple choice question; and generating a plurality of numerical choices either greater or smaller than the correct numerical answer according to a choice-answer relationship. 
         [0011]    Below, embodiments are described in detail to make easily understood the objectives, technical contents, characteristic and accomplishments of the present invention. 
     
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS  
         [0012]    The FIG. shows a flowchart of a choice generation method for a numerical multiple choice question according to one embodiment of the present invention. 
       
    
    
     DETAILED DESCRIPTION OF THE INVENTION  
       [0013]    Refer to the FIG. showing a flowchart of a choice generation method for a numerical multiple choice question according to one embodiment of the present invention. In Step S 10 , establish a numerical multiple choice question. For example, the length of a park trail is 40 m, and the jogging speed of a person is 5 m/sec; how much time will the person take to walk through the park trail After calculation, the correct answer of the question is 8 seconds. Next, in Step S 12 , determine the number of the numerical choices of the numerical multiple choice question. Next, in Step S 14 , generate a plurality of numerical choices either greater or smaller than the correct answer according to a choice-answer relationship, if the user intends to present the numerical choices in form of an arithmetic sequence including the correct answer, he may adopt a choice-answer relationship expressed by Equation (1): 
         [0000]    
       
         
           
             
               
                 
                   
                     choice 
                      
                     
                       ( 
                       c 
                       ) 
                     
                   
                   = 
                   
                     a 
                     + 
                     
                       
                         ∑ 
                         
                           b 
                           = 
                           0 
                         
                         
                           
                             [ 
                             
                               
                                 log 
                                 2 
                               
                                
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   1 
                                 
                                 ) 
                               
                             
                             ] 
                           
                           - 
                           1 
                         
                       
                        
                       
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         · 
                         
                           c 
                           b 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   1 
                   ) 
                 
               
             
           
         
       
     
         [0000]    wherein choice(e) is the value of the cth numerical choice, α the correct answer, and c b  the bth bit of c in the binary form (it is either 1 or 0), and wherein pm(d) may be expressed by Equation (2): 
         [0000]        pm ( d )=− d+X*d* 2   (2)
 
         [0000]    wherein d is a common difference, and X an integer variable of 0 or 1. The auxiliary random integer variable X is used to vary the order of the correct answer appearing in the sequence of the numerical choices. 
         [0014]    The process of the method of the present invention has been described above. The deduction of the equations and the exemplifications will be stated below. Equations (1) and (2) are mathematically deduced as follows: 
         [0000]    
       
         
           
             
               chioce 
                
               
                 ( 
                 c 
                 ) 
               
             
             = 
             
               
                 a 
                 + 
                 
                   
                     pm 
                      
                     
                       ( 
                       
                         d 
                         · 
                         
                           2 
                           0 
                         
                       
                       ) 
                     
                   
                   · 
                   
                     c 
                     0 
                   
                 
                 + 
                 
                   
                     pm 
                      
                     
                       ( 
                       
                         d 
                         · 
                         
                           2 
                           1 
                         
                       
                       ) 
                     
                   
                   · 
                   
                     c 
                     1 
                   
                 
                 + 
                 
                   
                     pm 
                      
                     
                       ( 
                       
                         d 
                         · 
                         
                           2 
                           2 
                         
                       
                       ) 
                     
                   
                   · 
                   
                     c 
                     2 
                   
                 
                 + 
                 … 
                 + 
                 
                   
                     pm 
                      
                     
                       ( 
                       
                         d 
                         · 
                         
                           2 
                           
                             
                               ⌈ 
                               
                                 
                                   log 
                                   2 
                                 
                                  
                                 
                                   ( 
                                   
                                     c 
                                     + 
                                     1 
                                   
                                   ) 
                                 
                               
                               ⌉ 
                             
                             - 
                             1 
                           
                         
                       
                       ) 
                     
                   
                   · 
                   
                     c 
                     
                       
                         [ 
                         
                           
                             log 
                             2 
                           
                            
                           
                             ( 
                             
                               c 
                               + 
                               1 
                             
                             ) 
                           
                         
                         ] 
                       
                       - 
                       1 
                     
                   
                 
               
               = 
               
                 a 
                 + 
                 
                   
                     ∑ 
                     
                       b 
                       = 
                       0 
                     
                     
                       
                         ⌈ 
                         
                           
                             log 
                             2 
                           
                            
                           
                             ( 
                             
                               c 
                               + 
                               1 
                             
                             ) 
                           
                         
                         ⌉ 
                       
                       - 
                       1 
                     
                   
                    
                   
                     
                       pm 
                        
                       
                         ( 
                         
                           d 
                           · 
                           
                             2 
                             b 
                           
                         
                         ) 
                       
                     
                     · 
                     
                       c 
                       b 
                     
                   
                 
               
             
           
         
       
     
         [0015]    The numerical choices generated by the present invention features irregularity and unpredictability, which will be demonstrated in Theorem 1 and Theorem 2. Before proving the two theorems, we have to propose and prove three lemmas. 
         [0000]    
       
         
           
             
                 
             
              
             
               
                 
                   Lemma 
                    
                   
                       
                   
                    
                   I 
                    
                   
                     : 
                   
                    
                   
                       
                   
                    
                   
                     c 
                     b 
                   
                 
                 = 
                 0 
               
               , 
               
                 ∀ 
                 
                   b 
                   &gt; 
                   
                     
                       ⌈ 
                       
                         
                           log 
                           2 
                         
                          
                         
                           ( 
                           
                             c 
                             + 
                             1 
                           
                           ) 
                         
                       
                       ⌉ 
                     
                     - 
                     1 
                   
                 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               Proof 
                
               
                 : 
               
                
               
                   
               
                
               A 
                
               
                   
               
                
               proof 
                
               
                   
               
                
               by 
                
               
                   
               
                
               contradiction 
                
               
                   
               
                
               is 
                
               
                   
               
                
               used 
                
               
                   
               
                
               
                 herein 
                 . 
                 
                   
 
                 
                  
                 
                     
                 
                  
                 Suppose 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 ∃ 
                 
                   b 
                   &gt; 
                   
                     
                       ⌈ 
                       
                         
                           log 
                            
                           
                             ( 
                             
                               c 
                               + 
                               1 
                             
                             ) 
                           
                         
                         - 
                         11 
                       
                       ⌉ 
                     
                     - 
                     1 
                   
                 
               
               , 
               
                 
                   c 
                   b 
                 
                 = 
                 1 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   c 
                   b 
                 
                 = 
                 
                   1 
                   ⇒ 
                   
                     c 
                     ≥ 
                     
                       2 
                       b 
                     
                   
                 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 
                   c 
                   + 
                   1 
                 
                 ≥ 
                 
                   
                     2 
                     b 
                   
                   + 
                   1 
                 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 
                   
                     
                       log 
                       2 
                     
                      
                     
                       ( 
                       
                         c 
                         + 
                         1 
                       
                       ) 
                     
                   
                   ≥ 
                   
                     
                       
                         log 
                         2 
                       
                        
                       
                         ( 
                         
                           
                             2 
                             b 
                           
                           + 
                           1 
                         
                         ) 
                       
                     
                     · 
                     
                       ⌈ 
                       
                         
                           log 
                           2 
                         
                          
                         
                           ( 
                           
                             c 
                             + 
                             1 
                           
                           ) 
                         
                       
                       ⌉ 
                     
                   
                   ≥ 
                   
                     ⌈ 
                     
                       
                         log 
                         2 
                       
                        
                       
                         ( 
                         
                           
                             2 
                             b 
                           
                           + 
                           1 
                         
                         ) 
                       
                     
                     ⌉ 
                   
                 
                 = 
                 
                   b 
                   + 
                   1 
                 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   b 
                   + 
                   1 
                 
                 ≤ 
                 
                   ⌈ 
                   
                     
                       log 
                       2 
                     
                      
                     
                       ( 
                       
                         c 
                         + 
                         1 
                       
                       ) 
                     
                   
                   ⌉ 
                 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 b 
                 ≤ 
                 
                   
                     ⌈ 
                     
                       
                         log 
                         2 
                       
                        
                       
                         ( 
                         
                           c 
                           + 
                           1 
                         
                         ) 
                       
                     
                     ⌉ 
                   
                   - 
                   1 
                 
               
               , 
               
                 b 
                 ≤ 
                 
                   
                     ⌈ 
                     
                       
                         
                           log 
                           2 
                         
                          
                         
                           ( 
                           
                             c 
                             + 
                             1 
                           
                           ) 
                         
                       
                       - 
                       1 
                     
                     ⌉ 
                   
                    
                   
                       
                   
                    
                   contradict 
                    
                   
                       
                   
                    
                   the 
                    
                   
                       
                   
                    
                   premise 
                 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               b 
               &gt; 
               
                 
                   ⌈ 
                   
                     
                       
                         log 
                         2 
                       
                        
                       
                         ( 
                         
                           c 
                           + 
                           1 
                         
                         ) 
                       
                     
                     - 
                     1 
                   
                   ⌉ 
                 
                 - 
                 1. 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   Therefore 
                    
                   
                       
                   
                    
                   is 
                    
                   
                       
                   
                    
                   proved 
                    
                   
                       
                   
                    
                   Lemma 
                    
                   
                       
                   
                    
                   
                     I 
                     . 
                     
                       
 
                     
                      
                     
                         
                     
                      
                     Lemma 
                   
                    
                   
                       
                   
                    
                   II 
                    
                   
                     : 
                   
                    
                   
                       
                   
                    
                   If 
                    
                   
                       
                   
                    
                   
                     choice 
                      
                     
                       ( 
                       c 
                       ) 
                     
                   
                 
                 = 
                 
                   a 
                   + 
                   
                     
                       ∑ 
                       
                         b 
                         = 
                         0 
                       
                       
                         
                           ⌈ 
                           
                             
                               log 
                               2 
                             
                              
                             
                               ( 
                               
                                 c 
                                 + 
                                 1 
                               
                               ) 
                             
                           
                           ⌉ 
                         
                         - 
                         1 
                       
                     
                      
                     
                       
                         pm 
                          
                         
                           ( 
                           
                             d 
                             · 
                             
                               2 
                               b 
                             
                           
                           ) 
                         
                       
                       · 
                       
                         c 
                         b 
                       
                     
                   
                 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 
                   choice 
                    
                   
                     ( 
                     
                       c 
                       + 
                       
                         2 
                         n 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     choice 
                      
                     
                       ( 
                       c 
                       ) 
                     
                   
                   + 
                   
                     pm 
                      
                     
                       ( 
                       
                         d 
                         · 
                         
                           2 
                           n 
                         
                       
                       ) 
                     
                   
                 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 
                   ∀ 
                   c 
                 
                 = 
                 0 
               
               , 
               1 
               , 
               2 
               , 
               … 
                
               
                   
               
               , 
               
                 
                   2 
                   n 
                 
                 - 
                 1. 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               Proof 
                
               
                 : 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   choice 
                    
                   
                     ( 
                     
                       c 
                       + 
                       
                         2 
                         n 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       a 
                       + 
                       
                         
                           ∑ 
                           
                             b 
                             = 
                             0 
                           
                           
                             
                               ⌈ 
                               
                                 
                                   log 
                                   2 
                                 
                                  
                                 
                                   ( 
                                   
                                     c 
                                     + 
                                     
                                       2 
                                       n 
                                     
                                     + 
                                     1 
                                   
                                   ) 
                                 
                               
                               ⌉ 
                             
                             - 
                             1 
                           
                         
                          
                         
                           
                             
                               pm 
                                
                               
                                 ( 
                                 
                                   d 
                                   · 
                                   
                                     2 
                                     b 
                                   
                                 
                                 ) 
                               
                             
                             · 
                             
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   
                                     2 
                                     n 
                                   
                                 
                                 ) 
                               
                               b 
                             
                           
                            
                           
                             ( 
                             1 
                             ) 
                           
                         
                       
                     
                      
                     
                       
 
                     
                      
                     
                         
                     
                     ∵ 
                     
                       ⌈ 
                       
                         
                           log 
                           2 
                         
                          
                         
                           ( 
                           
                             c 
                             + 
                             
                               2 
                               n 
                             
                             + 
                             1 
                           
                           ) 
                         
                       
                       ⌉ 
                     
                   
                   = 
                   
                     n 
                     + 
                     1 
                   
                 
               
               , 
               
                 
                   ∀ 
                   c 
                 
                 = 
                 0 
               
               , 
               1 
               , 
               2 
               , 
               … 
                
               
                   
               
               , 
               
                 
                   
                     
                       2 
                       n 
                     
                     - 
                     1 
                   
                    
                   
                     
 
                   
                   ∴ 
                   
                     ( 
                     1 
                     ) 
                   
                 
                 = 
                 
                   
                     a 
                     + 
                     
                       
                         ∑ 
                         
                           b 
                           = 
                           0 
                         
                         
                           
                             n 
                             + 
                             1 
                             - 
                             1 
                           
                           = 
                           n 
                         
                       
                        
                       
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         · 
                         
                           
                             ( 
                             
                               c 
                               + 
                               
                                 2 
                                 n 
                               
                             
                             ) 
                           
                           b 
                         
                       
                     
                   
                   = 
                   
                     
                       
                         a 
                         + 
                         
                           
                             ∑ 
                             
                               b 
                               = 
                               0 
                             
                             
                               n 
                               - 
                               1 
                             
                           
                            
                           
                             
                               pm 
                                
                               
                                 ( 
                                 
                                   d 
                                   · 
                                   
                                     2 
                                     b 
                                   
                                 
                                 ) 
                               
                             
                             · 
                             
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   
                                     2 
                                     n 
                                   
                                 
                                 ) 
                               
                               b 
                             
                           
                         
                         + 
                         
                           
                             ∑ 
                             
                               b 
                               = 
                               n 
                             
                             n 
                           
                            
                           
                             
                               
                                 pm 
                                  
                                 
                                   ( 
                                   
                                     d 
                                     · 
                                     
                                       2 
                                       b 
                                     
                                   
                                   ) 
                                 
                               
                               · 
                               
                                 
                                   ( 
                                   
                                     c 
                                     + 
                                     
                                       2 
                                       n 
                                     
                                   
                                   ) 
                                 
                                 b 
                               
                             
                              
                             
                                 
                             
                              
                             
                               ( 
                               2 
                               ) 
                             
                           
                         
                       
                        
                       
                         
 
                       
                        
                       
                           
                       
                       ∵ 
                       
                         
                           ( 
                           
                             c 
                             + 
                             
                               2 
                               n 
                             
                           
                           ) 
                         
                         b 
                       
                     
                     = 
                     
                       c 
                       b 
                     
                   
                 
               
               , 
               
                 
                   ∀ 
                   b 
                 
                 = 
                 0 
               
               , 
               1 
               , 
               2 
               , 
               … 
                
               
                   
               
               , 
               
                 
                   n 
                   - 
                   
                     1 
                      
                     
                         
                     
                      
                     
                       
                         and 
                          
                         
                            
                         
                          
                         
                             
                         
                         ( 
                         
                           c 
                           + 
                           
                             2 
                             n 
                           
                         
                         ) 
                       
                       n 
                     
                   
                 
                 = 
                 1 
               
               , 
               
                 
                   ∀ 
                   c 
                 
                 = 
                 0 
               
               , 
               1 
               , 
               2 
               , 
               … 
                
               
                   
               
               , 
               
                 
                   
                     
                       2 
                       n 
                     
                     - 
                     1 
                   
                    
                   
                     
 
                   
                   ∴ 
                   
                     ( 
                     2 
                     ) 
                   
                 
                 = 
                 
                   
                     a 
                     + 
                     
                       
                         ∑ 
                         
                           b 
                           = 
                           0 
                         
                         
                           n 
                           - 
                           1 
                         
                       
                        
                       
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         · 
                         
                           c 
                           b 
                         
                       
                     
                     + 
                     
                       
                         ∑ 
                         
                           b 
                           = 
                           n 
                         
                         n 
                       
                        
                       
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         · 
                         1 
                       
                     
                   
                   = 
                   
                     a 
                     + 
                     
                       
                         ∑ 
                         
                           b 
                           = 
                           0 
                         
                         
                           n 
                           - 
                           1 
                         
                       
                        
                       
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         · 
                         
                           c 
                           b 
                         
                       
                     
                     + 
                     
                       
                         pm 
                          
                         
                           ( 
                           
                             d 
                             · 
                             
                               2 
                               n 
                             
                           
                           ) 
                         
                       
                        
                       
                           
                       
                        
                       
                         ( 
                         3 
                         ) 
                       
                     
                   
                 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   
                     ( 
                     a 
                     ) 
                   
                    
                   
                       
                   
                    
                   When 
                    
                   
                       
                   
                    
                   c 
                 
                 = 
                 
                   2 
                   
                     n 
                     - 
                     1 
                   
                 
               
               , 
               
                 
                   2 
                   
                     n 
                     - 
                     1 
                   
                 
                 + 
                 1 
               
               , 
               
                 
                   2 
                   
                     n 
                     - 
                     1 
                   
                 
                 + 
                 2 
               
               , 
               … 
                
               
                   
               
               , 
               
                 
                   2 
                   n 
                 
                 - 
                 1 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 
                   ∵ 
                   
                     
                       ⌈ 
                       
                         
                           log 
                           2 
                         
                          
                         
                           ( 
                           
                             c 
                             + 
                             1 
                           
                           ) 
                         
                       
                       ⌉ 
                     
                     - 
                     1 
                   
                 
                 = 
                 
                   n 
                   - 
                   1 
                 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 
                   ∴ 
                   
                     
                       ∑ 
                       
                         b 
                         = 
                         0 
                       
                       
                         n 
                         - 
                         1 
                       
                     
                      
                     
                       
                         pm 
                          
                         
                           ( 
                           
                             d 
                             · 
                             
                               2 
                               b 
                             
                           
                           ) 
                         
                       
                       · 
                       
                         c 
                         b 
                       
                     
                   
                 
                 = 
                 
                   
                     
                       ∑ 
                       
                         b 
                         = 
                         0 
                       
                       
                         
                           ⌈ 
                           
                             
                               log 
                               2 
                             
                              
                             
                               ( 
                               
                                 c 
                                 + 
                                 1 
                               
                               ) 
                             
                           
                           ⌉ 
                         
                         - 
                         1 
                       
                     
                      
                     
                       
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         · 
                         
                           
                             c 
                             b 
                           
                            
                           
                             
 
                           
                            
                           
                               
                           
                           ( 
                           b 
                           ) 
                         
                       
                        
                       
                           
                       
                        
                       When 
                        
                       
                           
                       
                        
                       c 
                     
                   
                   = 
                   0 
                 
               
               , 
               1 
               , 
               2 
               , 
               … 
                
               
                   
               
               , 
               
                 
                   2 
                   
                     n 
                     - 
                     1 
                   
                 
                 - 
                 1 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 ∵ 
                 
                   
                     
                       ⌈ 
                       
                         
                           log 
                           2 
                         
                          
                         
                           ( 
                           
                             c 
                             + 
                             1 
                           
                           ) 
                         
                       
                       ⌉ 
                     
                     - 
                     1 
                   
                   &lt; 
                   
                     n 
                     - 
                     
                       1 
                        
                       
                           
                       
                        
                       and 
                     
                   
                 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   Lemma 
                    
                   
                       
                   
                    
                   1 
                    
                   
                     : 
                   
                    
                   
                       
                   
                    
                   
                     c 
                     b 
                   
                 
                 = 
                 0 
               
               , 
               
                 
                   
                     ∀ 
                     
                       b 
                       &gt; 
                       
                         
                           ⌈ 
                           
                             
                               log 
                               2 
                             
                              
                             
                               ( 
                               
                                 c 
                                 + 
                                 1 
                               
                               ) 
                             
                           
                           ⌉ 
                         
                         - 
                         1 
                       
                     
                   
                    
                   
                     
 
                   
                   ∴ 
                   
                     
                       ∑ 
                       
                         b 
                         = 
                         0 
                       
                       
                         n 
                         - 
                         1 
                       
                     
                      
                     
                       
                         pm 
                          
                         
                           ( 
                           
                             d 
                             · 
                             
                               2 
                               b 
                             
                           
                           ) 
                         
                       
                       · 
                       
                         c 
                         b 
                       
                     
                   
                 
                 = 
                 
                   
                     
                       
                         ∑ 
                         
                           b 
                           = 
                           0 
                         
                         
                           
                             ⌈ 
                             
                               
                                 log 
                                 2 
                               
                                
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   1 
                                 
                                 ) 
                               
                             
                             ⌉ 
                           
                           - 
                           1 
                         
                       
                        
                       
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         · 
                         
                           c 
                           b 
                         
                       
                     
                     + 
                     
                       
                         ∑ 
                         
                           b 
                           = 
                           
                             ⌈ 
                             
                               
                                 log 
                                 2 
                               
                                
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   1 
                                 
                                 ) 
                               
                             
                             ⌉ 
                           
                         
                         
                           n 
                           - 
                           1 
                         
                       
                        
                       
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         · 
                         0 
                       
                     
                   
                   = 
                   
                     
                       ∑ 
                       
                         b 
                         = 
                         0 
                       
                       
                         
                           ⌈ 
                           
                             
                               log 
                               2 
                             
                              
                             
                               ( 
                               
                                 c 
                                 + 
                                 1 
                               
                               ) 
                             
                           
                           ⌉ 
                         
                         - 
                         1 
                       
                     
                      
                     
                       
                         pm 
                          
                         
                           ( 
                           
                             d 
                             · 
                             
                               2 
                               b 
                             
                           
                           ) 
                         
                       
                       · 
                       
                         
                           c 
                           b 
                         
                          
                         
                           
 
                         
                         ( 
                         a 
                         ) 
                       
                     
                   
                 
               
               , 
               
                 
                   
                     ( 
                     b 
                     ) 
                   
                   ⇒ 
                   
                     ( 
                     3 
                     ) 
                   
                 
                 = 
                 
                   
                     a 
                     + 
                     
                       
                         ∑ 
                         
                           b 
                           = 
                           0 
                         
                         
                           
                             ⌈ 
                             
                               
                                 log 
                                 2 
                               
                                
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   1 
                                 
                                 ) 
                               
                             
                             ⌉ 
                           
                           - 
                           1 
                         
                       
                        
                       
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         · 
                         
                           c 
                           b 
                         
                       
                     
                     + 
                     
                       pm 
                        
                       
                         ( 
                         
                           d 
                           · 
                           
                             2 
                             n 
                           
                         
                         ) 
                       
                     
                   
                   = 
                   
                     
                       choice 
                        
                       
                         ( 
                         c 
                         ) 
                       
                     
                     + 
                     
                       pm 
                        
                       
                         ( 
                         
                           d 
                           · 
                           
                             2 
                             n 
                           
                         
                         ) 
                       
                     
                   
                 
               
               , 
             
           
         
       
     
         [0016]    Therefore is proved Theorem II. 
       Theorem I: 
       [0017]    If 
         [0000]    
       
         
           
             
               
                 choice 
                  
                 
                   ( 
                   c 
                   ) 
                 
               
               = 
               
                 a 
                 + 
                 
                   
                     ∑ 
                     
                       b 
                       = 
                       0 
                     
                     
                       
                         ⌈ 
                         
                           
                             log 
                             2 
                           
                            
                           
                             ( 
                             
                               c 
                               + 
                               1 
                             
                             ) 
                           
                         
                         ⌉ 
                       
                       - 
                       1 
                     
                   
                    
                   
                     
                       pm 
                        
                       
                         ( 
                         
                           d 
                           · 
                           
                             2 
                             b 
                           
                         
                         ) 
                       
                     
                     · 
                     
                       c 
                       b 
                     
                   
                 
               
             
             , 
           
         
       
     
         [0000]    choice(0)˜choice(2 n −1), ∀nεN the feature that they can generate 2 n  sets of arithmetic sequences each having α, an identical common difference d, and a total number of terms 2 n . 
       Proof: 
       [0018]      
         [0000]    
       
         
               
             
               
               
               
             
               
             
               
               
               
             
           
               
                   
               
             
             
               
                 When n = 1, 
               
               
                 choice(0) = a, choice(1) = a + pm(d · 2 0 ) = a + pm(d · 1) = a + pm(d) 
               
               
                 When pm(d) = −d, 
               
             
          
           
               
                 choice(1) = 
                 choice(0) = a 
                   
               
               
                 a + pm(d) = a − d 
                   
                   
               
             
          
           
               
                 When pm(d) = +d, 
               
             
          
           
               
                   
                 choice(0) = a 
                 choice(1) = 
               
               
                   
                   
                 a + pm(d) = a + d 
               
               
                   
               
             
          
         
       
     
         [0019]    Therefore, there are 2 n 2 1 =2 sets of arithmetic sequences each having α, an 
         [0020]    identical common difference d, and a total number of terms 2 n =2 1 =2. 
         [0021]    When n=2, 
         [0022]    It is learned from Lemma II that choice(c+2 1 )=choice(c)+pm(d·2 1l ). 
         [0023]    When c=2, express 2 of choice(2) by the binary system. 
         [0024]    Thus, choice(2)=choice(0+2 1 )=choice(0)+pm(d·2 1 )=choice(0)÷pm(d·2). 
         [0025]    When c=3, express 3 of choice(3) by the binary system. 
         [0026]    Thus, choice(3)=choice(1+2 1 )=choice(1)÷pm(d·2 1 )=choice(1)÷pm(d·2). 
         [0000]    
       
         
               
             
               
               
               
               
               
             
               
             
               
               
               
               
               
               
             
           
               
                   
               
             
             
               
                 When pm(d · 2) = −d · 2, 
               
             
          
           
               
                 choice(3) = 
                 choice(2) = 
                 choice(1) = 
                 choice(0) = a 
                   
               
               
                 choice(1) − 
                 choice(0) − 
                 a − d · 1 
               
               
                 d · 2 = 
                 d · 2 = 
               
               
                 a − d − d · 2 = 
                 a − d · 2 
               
               
                 a − d · 3 
               
               
                   
                 choice(2) = 
                 choice(3) = 
                 choice(0) = a 
                 choice(1) = 
               
               
                   
                 choice(0) − 
                 choice(1) − 
                   
                 a + d · 1 
               
               
                   
                 d · 2 = 
                 d · 2 = 
               
               
                   
                 a − d · 2 
                 a + d − d = 
               
               
                   
                   
                 a − d · 1 
               
             
          
           
               
                 When pm(d · 2) = +d · 2, 
               
             
          
           
               
                   
                 choice(1) = 
                 choice(0) = a 
                 choice(3) = 
                 choice(2) = 
                   
               
               
                   
                 a − d · 1 
                   
                 choice(1) + 
                 choice(0) + 
               
               
                   
                   
                   
                 d · 2 = 
                 d · 2 = 
               
               
                   
                   
                   
                 a − d + d · 2 = 
                 a + d · 2 
               
               
                   
                   
                   
                 a + d · 1 
               
               
                   
                   
                 choice(0) = a 
                 choice(1) = 
                 choice(2) = 
                 choice(3) = 
               
               
                   
                   
                   
                 a + d · 1 
                 choice(0) + 
                 choice(1) + 
               
               
                   
                   
                   
                   
                 d · 2 = 
                 d · 2 = 
               
               
                   
                   
                   
                   
                 a + d · 2 
                 a + d + d · 2 = 
               
               
                   
                   
                   
                   
                   
                 a + d · 3 
               
               
                   
                   
               
             
          
         
       
     
         [0027]    As shown in abovementioned cases of pm(d·2)=−d·2 and pm(d·2)=+d·2, when n=2, there are totally 2 n =2 2 =4 sets of arithmetic sequences each having α, an identical common difference d, and a total number of terms 2 n =2 2 =4. 
         [0028]    Suppose that when n=k, Theorem I is true. In other words, when n=k, there are 2 k  sets of arithmetic, sequences each having α, an identical, common difference d, and a total number of terms 2 k : 
         [0000]    
       
         
               
               
               
               
               
               
               
               
               
             
           
               
                   
               
             
             
               
                 a − d · 
                 a − d · 
                 a − d · 
                 . . . 
                 a 
                   
                   
                   
                   
               
               
                 (2 k  − 1) 
                 (2 k  − 2) 
                 (2 k  − 3) 
               
               
                   
                 a − d · 
                 a − d · 
                 . . . 
                 a 
                 a + d · 1 
               
               
                   
                 (2 k  − 2) 
                 (2 k  − 3) 
               
               
                   
                   
                 a − d · 
                 . . . 
                 a 
                 a + d · 1 
                 a + 
               
               
                   
                   
                 (2 k  − 3) 
                   
                   
                   
                 d · 2 
               
               
                   
                   
                   
                 . . . 
                 . . . 
                 . . . 
                 . . . 
                 . . . 
               
               
                   
                   
                   
                   
                 a 
                 a + d · 1 
                 a + 
                 . . . 
                 a + d · 
               
               
                   
                   
                   
                   
                   
                   
                 d · 2 
                   
                 (2 k  − 1) 
               
               
                   
               
             
          
         
       
     
         [0029]    When n=k+1, it is learned from Lemma II that choice(c+2 k )=choice(c)+pm(d·2 k ), ∀c=0,1,2, . . . , 2−1. 
         [0030]    When pm(d·2 k )=−d·2 k , 2 k  rows are generated in the table. As the next term is formed via adding −d·2 k  to the current term, the next term is arranged at the left of the current term in the table. When pm(d·2 k )=+d·2 k , there are also 2 k  rows generated in the table. As the next term is formed via adding +d·2 k  to the current term, the next term is arranged at the right of the current term in the table. As shown in abovementioned cases of pm(d·2 k )=−d·2 k  and pm(d·2 k )=+d·2 k , when n=k+1, there are totally 2 k +2 k =2 k+1  sets of arithmetic sequences each having α, an identical common difference d, and a total number of terms 2 k+1 . Thus is proved Theorem I. 
         [0031]    Below, a two-choice multiple choice question and a four-choice multiple choice question are used as the embodiments to demonstrate how the present invention generates numerical choices in form of an arithmetic sequence for a numerical multiple choice question. Firstly, define a two-choice numerical multiple choice question. The test question is “the length of a park trail is 40 m, and the jogging speed of a person is n1 to 10 m/sec; how much time will the person take to walk through the park trail?” The equation for solving the question is time=length÷speed=40/n1 to 10. The correct, answer is (40/n1 to 10) seconds. Thus, Choice 1 is (40/n1 to 10)+pm(d·2 0 ) wherein n1 to 10 is a random integer number of from 1 to 10. In this embodiment, n1 to 10 is assigned to be 5, and the correct answer is 8seconds. In this embodiment, d of pm(d·2 0 ) is assigned to be 1. However, d may be a random positive or negative integer in the present invention. Further, X is assigned to be 0 or 1 so as to vary the order of the correct answer appearing in the sequence of the numerical choices. 
         [0032]    When X=0, the numerical choices are calculated as follows: 
         [0000]    
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 0 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     the 
                      
                     
                         
                     
                      
                     correct 
                      
                     
                         
                     
                      
                     answer 
                   
                   ) 
                 
               
               = 
               
                 
                   choice 
                    
                   
                     ( 
                     0 
                     ) 
                   
                 
                 = 
                 
                   a 
                   = 
                   
                     
                       ( 
                       
                         
                           40 
                           / 
                           n 
                         
                          
                         
                             
                         
                          
                         1 
                          
                         
                             
                         
                          
                         to 
                          
                         
                             
                         
                          
                         10 
                       
                       ) 
                     
                     = 
                     
                       
                         ( 
                         
                           40 
                           / 
                           5 
                         
                         ) 
                       
                       = 
                       8 
                     
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               Choice 
                
               
                   
               
                
               1 
             
             = 
             
               
                 choice 
                  
                 
                   ( 
                   1 
                   ) 
                 
               
               = 
               
                 
                   a 
                   + 
                   
                     pm 
                      
                     
                       ( 
                       
                         d 
                         · 
                         
                           2 
                           0 
                         
                       
                       ) 
                     
                   
                 
                 = 
                 
                   
                     1 
                     + 
                     
                       pm 
                        
                       
                         ( 
                         
                           d 
                           · 
                           1 
                         
                         ) 
                       
                     
                   
                   = 
                   
                     
                       a 
                       + 
                       
                         pm 
                          
                         
                           ( 
                           d 
                           ) 
                         
                       
                     
                     = 
                     
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                               
                                   
                               
                                
                               
                                   
                               
                             
                              
                             10 
                           
                           ) 
                         
                         + 
                         
                           ( 
                           
                             
                               - 
                               1 
                             
                             + 
                             
                               X 
                               * 
                               2 
                             
                           
                           ) 
                         
                       
                       = 
                       
                         
                           
                             ( 
                             
                               40 
                               / 
                               5 
                             
                             ) 
                           
                           + 
                           
                             ( 
                             
                               
                                 - 
                                 1 
                               
                               + 
                               
                                 0 
                                 * 
                                 2 
                               
                             
                             ) 
                           
                         
                         = 
                         
                           
                             8 
                             - 
                             1 
                           
                           = 
                           7. 
                         
                       
                     
                   
                 
               
             
           
         
       
     
         [0033]    The choices arranged from large to small are 8, 7. The correct answer is the largest one among the choices. 
         [0034]    When X=1, the numerical choices arc calculated as follows: 
         [0035]    Choke 0 (the correct answer)=choice(0)=a=(40/n1 to 10)=(40/5)=8; 
         [0036]    Choice 1=(40/n1 to 10)+(−1+X*2)=(40/5)+(−1+1*2)=8+1=9. 
         [0037]    The choices arranged from large to small are 9, 8. The correct answer is the second largest one among the choices. 
         [0038]    Next, a four-choice multiple choice question is used as the embodiment to demonstrate the present invention. Similarly, X is 0 or 1 in the embodiment. From Lemma II, it is learned that choice(c+2 1 )=choice(c)+pm(d·2 1 ), ∀c=0,2 1   −1=0,1. Four    
         [0039]    numerical choices of the multiple choice question are calculated as follows. 
         [0000]    
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 0 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     the 
                      
                     
                         
                     
                      
                     correct 
                      
                     
                         
                     
                      
                     answer 
                   
                   ) 
                 
               
               = 
               
                 
                   choice 
                    
                   
                     ( 
                     0 
                     ) 
                   
                 
                 = 
                 
                   a 
                   = 
                   
                     
                       ( 
                       
                         
                           40 
                           / 
                           n 
                         
                          
                         
                             
                         
                          
                         1 
                          
                         
                             
                         
                          
                         to 
                          
                         
                             
                         
                          
                         10 
                       
                       ) 
                     
                     = 
                     
                       
                         ( 
                         
                           40 
                           / 
                           5 
                         
                         ) 
                       
                       = 
                       8 
                     
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   choice 
                    
                   
                     ( 
                     1 
                     ) 
                   
                 
                 = 
                 
                   
                     a 
                     + 
                     
                       pm 
                        
                       
                         ( 
                         
                           d 
                           · 
                           
                             2 
                             0 
                           
                         
                         ) 
                       
                     
                   
                   = 
                   
                     
                       a 
                       + 
                       
                         pm 
                          
                         
                           ( 
                           
                             d 
                             · 
                             1 
                           
                           ) 
                         
                       
                     
                     = 
                     
                       
                         a 
                         + 
                         
                           pm 
                            
                           
                             ( 
                             d 
                             ) 
                           
                         
                       
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         + 
                         
                           ( 
                           
                             
                               - 
                               1 
                             
                             + 
                             
                               X 
                               * 
                               2 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   choice 
                    
                   
                     ( 
                     2 
                     ) 
                   
                 
                 = 
                 
                   
                     choice 
                      
                     
                       ( 
                       
                         0 
                         + 
                         
                           2 
                           1 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       
                         choice 
                          
                         
                           ( 
                           0 
                           ) 
                         
                       
                       + 
                       
                         pm 
                          
                         
                           ( 
                           
                             d 
                             · 
                             
                               2 
                               1 
                             
                           
                           ) 
                         
                       
                     
                     = 
                     
                       
                         
                           choice 
                            
                           
                             ( 
                             0 
                             ) 
                           
                         
                         + 
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               2 
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         + 
                         
                           ( 
                           
                             
                               - 
                               2 
                             
                             + 
                             
                               
                                 X 
                                 2 
                               
                               * 
                               4 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 3 
               
               = 
               
                 
                   choice 
                    
                   
                     ( 
                     3 
                     ) 
                   
                 
                 = 
                 
                   
                     choice 
                      
                     
                       ( 
                       
                         1 
                         + 
                         
                           2 
                           1 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       
                         choice 
                          
                         
                           ( 
                           1 
                           ) 
                         
                       
                       + 
                       
                         pm 
                          
                         
                           ( 
                           
                             d 
                             · 
                             
                               2 
                               1 
                             
                           
                           ) 
                         
                       
                     
                     = 
                     
                       
                         
                           choice 
                            
                           
                             ( 
                             1 
                             ) 
                           
                         
                         + 
                         
                           pm 
                            
                           
                             ( 
                             
                               d 
                               · 
                               2 
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         + 
                         
                           ( 
                           
                             
                               - 
                               1 
                             
                             + 
                             
                               X 
                               * 
                               2 
                             
                           
                           ) 
                         
                         + 
                         
                           ( 
                           
                             
                               - 
                               2 
                             
                             + 
                             
                               
                                 X 
                                 2 
                               
                               * 
                               4 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             , 
           
         
       
     
         [0000]    wherein n1 to 10 is a random integer number of from 1 to 10 and X is either 0 or 1. In this embodiment, n1 to 10 is assigned to be 5, and (X, X 2 ) have four combinations; (0, 0), (1, 0), (0, 1)and (1, 1). 
         [0000]    
       
         
           
             
                 
             
              
             
               
                 When 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     X 
                     , 
                     
                       X 
                       2 
                     
                   
                   ) 
                 
                  
                 
                     
                 
                  
                 is 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     0 
                     , 
                     0 
                   
                   ) 
                 
               
               , 
               
                 
 
               
                
               
                 
                   
                     Choice 
                      
                     
                         
                     
                      
                     0 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         the 
                          
                         
                             
                         
                          
                         correct 
                          
                         
                             
                         
                          
                         answer 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       choice 
                        
                       
                         ( 
                         0 
                         ) 
                       
                     
                     = 
                     
                       a 
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         = 
                         
                           
                             ( 
                             
                               40 
                               / 
                               5 
                             
                             ) 
                           
                           = 
                           8 
                         
                       
                     
                   
                 
                 ; 
               
             
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         1 
                       
                       + 
                       
                         X 
                         * 
                         2 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         
                           - 
                           1 
                         
                         + 
                         
                           0 
                           * 
                           2 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       + 
                       
                         ( 
                         
                           - 
                           1 
                         
                         ) 
                       
                     
                     = 
                     7 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         2 
                       
                       + 
                       
                         
                           X 
                           2 
                         
                         * 
                         2 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         
                           - 
                           2 
                         
                         + 
                         
                           0 
                           * 
                           4 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       + 
                       
                         ( 
                         
                           - 
                           2 
                         
                         ) 
                       
                     
                     = 
                     6 
                   
                 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               Choice 
                
               
                   
               
                
               3 
             
             = 
             
               
                 
                   ( 
                   
                     
                       40 
                       / 
                       n 
                     
                      
                     
                         
                     
                      
                     1 
                      
                     
                         
                     
                      
                     to 
                      
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
                 + 
                 
                   ( 
                   
                     
                       - 
                       1 
                     
                     + 
                     
                       X 
                       * 
                       2 
                     
                   
                   ) 
                 
                 + 
                 
                   ( 
                   
                     
                       - 
                       2 
                     
                     + 
                     
                       
                         X 
                         2 
                       
                       * 
                       4 
                     
                   
                   ) 
                 
               
               = 
               
                 
                   
                     ( 
                     
                       40 
                       / 
                       5 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         1 
                       
                       + 
                       
                         0 
                         * 
                         2 
                       
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         2 
                       
                       + 
                       
                         0 
                         * 
                         4 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     8 
                     + 
                     
                       ( 
                       
                         - 
                         1 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         - 
                         2 
                       
                       ) 
                     
                   
                   = 
                   5. 
                 
               
             
           
         
       
     
         [0040]    The choices arranged from large to small are 8, 7, 6, 5. The correct answer is the largest one among the choices. 
         [0000]    
       
         
           
             
                 
             
              
             
               
                 When 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     X 
                     , 
                     
                       X 
                       2 
                     
                   
                   ) 
                 
                  
                 
                     
                 
                  
                 is 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     1 
                     , 
                     0 
                   
                   ) 
                 
               
               , 
               
                 
 
               
                
               
                 
                   
                     Choice 
                      
                     
                         
                     
                      
                     0 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         the 
                          
                         
                             
                         
                          
                         correct 
                          
                         
                             
                         
                          
                         answer 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       choice 
                        
                       
                         ( 
                         0 
                         ) 
                       
                     
                     = 
                     
                       a 
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         = 
                         
                           
                             ( 
                             
                               40 
                               / 
                               5 
                             
                             ) 
                           
                           = 
                           8 
                         
                       
                     
                   
                 
                 ; 
               
             
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         1 
                       
                       + 
                       
                         X 
                         * 
                         2 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         
                           - 
                           1 
                         
                         + 
                         
                           1 
                           * 
                           2 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       + 
                       1 
                     
                     = 
                     9 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         2 
                       
                       + 
                       
                         
                           X 
                           2 
                         
                         * 
                         2 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         
                           - 
                           2 
                         
                         + 
                         
                           0 
                           * 
                           4 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       + 
                       
                         ( 
                         
                           - 
                           2 
                         
                         ) 
                       
                     
                     = 
                     6 
                   
                 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               Choice 
                
               
                   
               
                
               3 
             
             = 
             
               
                 
                   ( 
                   
                     
                       40 
                       / 
                       n 
                     
                      
                     
                         
                     
                      
                     1 
                      
                     
                         
                     
                      
                     to 
                      
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
                 + 
                 
                   ( 
                   
                     
                       - 
                       1 
                     
                     + 
                     
                       X 
                       * 
                       2 
                     
                   
                   ) 
                 
                 + 
                 
                   ( 
                   
                     
                       - 
                       2 
                     
                     + 
                     
                       
                         X 
                         2 
                       
                       * 
                       4 
                     
                   
                   ) 
                 
               
               = 
               
                 
                   
                     ( 
                     
                       40 
                       / 
                       5 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         1 
                       
                       + 
                       
                         1 
                         * 
                         2 
                       
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         2 
                       
                       + 
                       
                         0 
                         * 
                         4 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     8 
                     + 
                     1 
                     + 
                     
                       ( 
                       
                         - 
                         2 
                       
                       ) 
                     
                   
                   = 
                   7. 
                 
               
             
           
         
       
     
         [0041]    The choices arranged from large to small are 9, 8, 7, 6. The correct answer is the second largest one among the choices. 
         [0000]    
       
         
           
             
                 
             
              
             
               
                 When 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     X 
                     , 
                     
                       X 
                       2 
                     
                   
                   ) 
                 
                  
                 
                     
                 
                  
                 is 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     0 
                     , 
                     1 
                   
                   ) 
                 
               
               , 
               
                 
 
               
                
               
                 
                   
                     Choice 
                      
                     
                         
                     
                      
                     0 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         the 
                          
                         
                             
                         
                          
                         correct 
                          
                         
                             
                         
                          
                         answer 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       choice 
                        
                       
                         ( 
                         0 
                         ) 
                       
                     
                     = 
                     
                       a 
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         = 
                         
                           
                             ( 
                             
                               40 
                               / 
                               5 
                             
                             ) 
                           
                           = 
                           8 
                         
                       
                     
                   
                 
                 ; 
               
             
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         1 
                       
                       + 
                       
                         X 
                         * 
                         2 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         
                           - 
                           1 
                         
                         + 
                         
                           0 
                           * 
                           2 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       + 
                       
                         ( 
                         
                           - 
                           1 
                         
                         ) 
                       
                     
                     = 
                     7 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         2 
                       
                       + 
                       
                         
                           X 
                           2 
                         
                         * 
                         2 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         
                           - 
                           2 
                         
                         + 
                         
                           1 
                           * 
                           4 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       + 
                       2 
                     
                     = 
                     10 
                   
                 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               Choice 
                
               
                   
               
                
               3 
             
             = 
             
               
                 
                   ( 
                   
                     
                       40 
                       / 
                       n 
                     
                      
                     
                         
                     
                      
                     1 
                      
                     
                         
                     
                      
                     to 
                      
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
                 + 
                 
                   ( 
                   
                     
                       - 
                       1 
                     
                     + 
                     
                       X 
                       * 
                       2 
                     
                   
                   ) 
                 
                 + 
                 
                   ( 
                   
                     
                       - 
                       2 
                     
                     + 
                     
                       
                         X 
                         2 
                       
                       * 
                       4 
                     
                   
                   ) 
                 
               
               = 
               
                 
                   
                     ( 
                     
                       40 
                       / 
                       5 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         1 
                       
                       + 
                       
                         0 
                         * 
                         2 
                       
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         2 
                       
                       + 
                       
                         1 
                         * 
                         4 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     8 
                     + 
                     
                       ( 
                       
                         - 
                         1 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         - 
                         2 
                       
                       ) 
                     
                   
                   = 
                   9. 
                 
               
             
           
         
       
     
         [0042]    The choices arranged from large to small are 10, 9, 8, 7. The correct answer is the third largest one among the choices. 
         [0000]    
       
         
           
             
                 
             
              
             
               
                 When 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     X 
                     , 
                     
                       X 
                       2 
                     
                   
                   ) 
                 
                  
                 
                     
                 
                  
                 is 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     1 
                     , 
                     1 
                   
                   ) 
                 
               
               , 
               
                 
 
               
                
               
                 
                   
                     Choice 
                      
                     
                         
                     
                      
                     0 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         the 
                          
                         
                             
                         
                          
                         correct 
                          
                         
                             
                         
                          
                         answer 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       choice 
                        
                       
                         ( 
                         0 
                         ) 
                       
                     
                     = 
                     
                       a 
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         = 
                         
                           
                             ( 
                             
                               40 
                               / 
                               5 
                             
                             ) 
                           
                           = 
                           8 
                         
                       
                     
                   
                 
                 ; 
               
             
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         1 
                       
                       + 
                       
                         X 
                         * 
                         2 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         
                           - 
                           1 
                         
                         + 
                         
                           1 
                           * 
                           2 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       + 
                       1 
                     
                     = 
                     9 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         2 
                       
                       + 
                       
                         
                           X 
                           2 
                         
                         * 
                         2 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         
                           - 
                           2 
                         
                         + 
                         
                           1 
                           * 
                           4 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       + 
                       2 
                     
                     = 
                     10 
                   
                 
               
             
             ; 
             and 
           
         
       
       
         
           
             
               Choice 
                
               
                   
               
                
               3 
             
             = 
             
               
                 
                   ( 
                   
                     
                       40 
                       / 
                       n 
                     
                      
                     
                         
                     
                      
                     1 
                      
                     
                         
                     
                      
                     to 
                      
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
                 + 
                 
                   ( 
                   
                     
                       - 
                       1 
                     
                     + 
                     
                       X 
                       * 
                       2 
                     
                   
                   ) 
                 
                 + 
                 
                   ( 
                   
                     
                       - 
                       2 
                     
                     + 
                     
                       
                         X 
                         2 
                       
                       * 
                       4 
                     
                   
                   ) 
                 
               
               = 
               
                 
                   
                     ( 
                     
                       40 
                       / 
                       5 
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         1 
                       
                       + 
                       
                         1 
                         * 
                         2 
                       
                     
                     ) 
                   
                   + 
                   
                     ( 
                     
                       
                         - 
                         2 
                       
                       + 
                       
                         1 
                         * 
                         4 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     8 
                     + 
                     1 
                     + 
                     2 
                   
                   = 
                   11. 
                 
               
             
           
         
       
     
         [0043]    The choices arranged from large to small are 11, 10, 9, 8. The correct answer is the fourth largest one among the choices. From the abovementioned examples, it is known that the variable X can be used to vary the order of the correct answer in the numerical choices. 
         [0044]    When the numerical choices are intended to be in form of a geometric sequence, they can be expressed by Equation (3): 
         [0000]    
       
         
           
             
               
                 
                   
                     choice 
                      
                     
                       ( 
                       c 
                       ) 
                     
                   
                   = 
                   
                     a 
                     · 
                     
                       
                         ∏ 
                         
                           b 
                           = 
                           0 
                         
                         
                           
                             ⌈ 
                             
                               
                                 log 
                                 2 
                               
                                
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   1 
                                 
                                 ) 
                               
                             
                             ⌉ 
                           
                           - 
                           1 
                         
                       
                        
                       
                           
                       
                        
                       
                         
                           [ 
                           
                             md 
                              
                             
                               ( 
                               
                                 r 
                                 
                                   2 
                                   b 
                                 
                               
                               ) 
                             
                           
                           ] 
                         
                         
                           c 
                           b 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   3 
                   ) 
                 
               
             
           
         
       
     
         [0000]    wherein choice(c) is the value of the cth choice, α the correct answer, and c b  the bth bit of c expressed in the binary system (0 or 1), and wherein md(r) is expressed by 
         [0000]    
       
         
           
             
               
                 
                   
                     md 
                      
                     
                       ( 
                       r 
                       ) 
                     
                   
                   = 
                   
                     
                       1 
                       r 
                     
                     + 
                     
                       X 
                       * 
                       
                         ( 
                         
                           r 
                           - 
                           
                             1 
                             r 
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   ( 
                   4 
                   ) 
                 
               
             
           
         
       
     
         [0000]    wherein r is the common ratio, and wherein X is a random integer variable (0 or 1) used to vary the order of the correct answer among the numerical choices. 
         [0045]    The deduction of Equations (3) and (4) is mathematically deduced as follows: 
         [0000]    
       
         
           
             
               choice 
                
               
                 ( 
                 c 
                 ) 
               
             
             = 
             
               
                 a 
                 · 
                 
                   
                     [ 
                     
                       md 
                        
                       
                         ( 
                         
                           r 
                           
                             2 
                             0 
                           
                         
                         ) 
                       
                     
                     ] 
                   
                   
                     c 
                     0 
                   
                 
                 · 
                 
                   
                     [ 
                     
                       md 
                        
                       
                         ( 
                         
                           r 
                           
                             2 
                             1 
                           
                         
                         ) 
                       
                     
                     ] 
                   
                   
                     c 
                     1 
                   
                 
                 · 
                 
                   
                     [ 
                     
                       md 
                        
                       
                         ( 
                         
                           r 
                           
                             2 
                             2 
                           
                         
                         ) 
                       
                     
                     ] 
                   
                   
                     c 
                     2 
                   
                 
                 · 
                 
                     
                 
                  
                 … 
                  
                 
                     
                 
                 · 
                 
                   
                     [ 
                     
                       md 
                        
                       
                         ( 
                         
                           r 
                           
                             2 
                             
                               
                                 ⌈ 
                                 
                                   
                                     log 
                                     2 
                                   
                                    
                                   
                                     ( 
                                     
                                       c 
                                       + 
                                       1 
                                     
                                     ) 
                                   
                                 
                                 ⌉ 
                               
                               - 
                               1 
                             
                           
                         
                         ) 
                       
                     
                     ] 
                   
                   
                     c 
                     
                       
                         ⌈ 
                         
                           
                             log 
                             2 
                           
                            
                           
                             ( 
                             
                               c 
                               + 
                               1 
                             
                             ) 
                           
                         
                         ⌉ 
                       
                       - 
                       1 
                     
                   
                 
               
               = 
               
                 a 
                 · 
                 
                   
                     ∏ 
                     
                       b 
                       = 
                       0 
                     
                     
                       
                         ⌈ 
                         
                           
                             log 
                             2 
                           
                            
                           
                             ( 
                             
                               c 
                               + 
                               1 
                             
                             ) 
                           
                         
                         ⌉ 
                       
                       - 
                       1 
                     
                   
                    
                   
                       
                   
                    
                   
                     
                       [ 
                       
                         md 
                          
                         
                           ( 
                           
                             r 
                             
                               2 
                               b 
                             
                           
                           ) 
                         
                       
                       ] 
                     
                     
                       c 
                       b 
                     
                   
                 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   Lemma 
                    
                   
                       
                   
                    
                   III 
                    
                   
                     : 
                   
                    
                   
                       
                   
                    
                   If 
                    
                   
                       
                   
                    
                   
                     choice 
                     ( 
                     c 
                     ) 
                   
                 
                 = 
                 
                   a 
                   · 
                   
                     
                       ∏ 
                       
                         b 
                         = 
                         0 
                       
                       
                         
                           ⌈ 
                           
                             
                               log 
                               2 
                             
                              
                             
                               ( 
                               
                                 c 
                                 + 
                                 1 
                               
                               ) 
                             
                           
                           ⌉ 
                         
                         - 
                         1 
                       
                     
                      
                     
                       
                         [ 
                         
                           md 
                            
                           
                             ( 
                             
                               r 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         ] 
                       
                       
                         c 
                         b 
                       
                     
                   
                 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 
                   choice 
                    
                   
                     ( 
                     
                       c 
                       + 
                       
                         2 
                         n 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     choice 
                      
                     
                       ( 
                       c 
                       ) 
                     
                   
                   · 
                   
                     md 
                      
                     
                       ( 
                       
                         r 
                         
                           2 
                           0 
                         
                       
                       ) 
                     
                   
                 
               
               , 
               
                 
                   ∀ 
                   c 
                 
                 = 
                 0 
               
               , 
               1 
               , 
               2 
               , 
               … 
                
               
                   
               
               , 
               
                 
                   2 
                   n 
                 
                 - 
                 1 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               Proof 
                
               
                 : 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   choice 
                    
                   
                     ( 
                     
                       c 
                       + 
                       
                         2 
                         n 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       a 
                       · 
                       
                         
                           ∏ 
                           
                             b 
                             = 
                             0 
                           
                           
                             
                               ⌈ 
                               
                                 
                                   log 
                                   2 
                                 
                                  
                                 
                                   ( 
                                   
                                     c 
                                     + 
                                     1 
                                   
                                   ) 
                                 
                               
                               ⌉ 
                             
                             - 
                             1 
                           
                         
                          
                         
                           
                             
                               [ 
                               
                                 md 
                                  
                                 
                                   ( 
                                   
                                     r 
                                     
                                       2 
                                       b 
                                     
                                   
                                   ) 
                                 
                               
                               ] 
                             
                             
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   
                                     2 
                                     b 
                                   
                                 
                                 ) 
                               
                               b 
                             
                           
                            
                           
                               
                           
                            
                           
                             ( 
                             4 
                             ) 
                           
                         
                       
                     
                      
                     
                       
 
                     
                      
                     
                         
                     
                     ∵ 
                     
                       ⌈ 
                       
                         
                           log 
                           2 
                         
                          
                         
                           ( 
                           
                             c 
                             + 
                             
                               2 
                               n 
                             
                             + 
                             1 
                           
                           ) 
                         
                       
                       ⌉ 
                     
                   
                   = 
                   
                     n 
                     + 
                     1 
                   
                 
               
               , 
               
                 
                   ∀ 
                   c 
                 
                 = 
                 0 
               
               , 
               1 
               , 
               2 
               , 
               … 
                
               
                   
               
               , 
               
                 
                   
                     
                       2 
                       n 
                     
                     - 
                     1 
                   
                    
                   
                     
 
                   
                   ∴ 
                   
                     ( 
                     4 
                     ) 
                   
                 
                 = 
                 
                   
                     a 
                     · 
                     
                       
                         ∏ 
                         
                           b 
                           = 
                           0 
                         
                         
                           
                             ⌈ 
                             
                               
                                 log 
                                 2 
                               
                                
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   
                                     2 
                                     b 
                                   
                                   + 
                                   1 
                                 
                                 ) 
                               
                             
                             ⌉ 
                           
                           - 
                           1 
                         
                       
                        
                       
                         
                           [ 
                           
                             md 
                              
                             
                               ( 
                               
                                 r 
                                 
                                   2 
                                   b 
                                 
                               
                               ) 
                             
                           
                           ] 
                         
                         
                           
                             ( 
                             
                               c 
                               + 
                               
                                 2 
                                 n 
                               
                             
                             ) 
                           
                           b 
                         
                       
                     
                   
                   = 
                   
                     
                       
                         a 
                         · 
                         
                           
                             ∏ 
                             
                               b 
                               = 
                               0 
                             
                             
                               n 
                               - 
                               1 
                             
                           
                            
                           
                               
                           
                            
                           
                             
                               
                                 [ 
                                 
                                   md 
                                    
                                   
                                     ( 
                                     
                                       r 
                                       
                                         2 
                                         b 
                                       
                                     
                                     ) 
                                   
                                 
                                 ] 
                               
                               
                                 
                                   ( 
                                   
                                     c 
                                     + 
                                     
                                       2 
                                       n 
                                     
                                   
                                   ) 
                                 
                                 b 
                               
                             
                             · 
                             
                               
                                 ∏ 
                                 
                                   b 
                                   = 
                                   n 
                                 
                                 n 
                               
                                
                               
                                   
                               
                                
                               
                                 
                                   
                                     [ 
                                     
                                       md 
                                        
                                       
                                         ( 
                                         
                                           r 
                                           
                                             2 
                                             b 
                                           
                                         
                                         ) 
                                       
                                     
                                     ] 
                                   
                                   
                                     
                                       ( 
                                       
                                         c 
                                         + 
                                         
                                           2 
                                           n 
                                         
                                       
                                       ) 
                                     
                                      
                                     b 
                                   
                                 
                                  
                                 
                                     
                                 
                                  
                                 
                                   ( 
                                   5 
                                   ) 
                                 
                               
                             
                           
                         
                       
                        
                       
                         
 
                       
                        
                       
                           
                       
                       ∵ 
                       
                         
                           ( 
                           
                             c 
                             + 
                             
                               2 
                               n 
                             
                           
                           ) 
                         
                         b 
                       
                     
                     = 
                     
                       c 
                       b 
                     
                   
                 
               
               , 
               
                 
                   ∀ 
                   b 
                 
                 = 
                 
                   
                     
                       [ 
                       
                         0 
                         , 
                         
                           n 
                           - 
                           1 
                         
                       
                       ] 
                     
                      
                     
                         
                     
                      
                     
                       
                         and 
                          
                         
                           
 
                         
                          
                         
                             
                         
                         ( 
                         
                           c 
                           + 
                           
                             2 
                             n 
                           
                         
                         ) 
                       
                       n 
                     
                   
                   = 
                   1 
                 
               
               , 
               
                 
                   ∀ 
                   c 
                 
                 = 
                 0 
               
               , 
               1 
               , 
               2 
               , 
               … 
                
               
                   
               
               , 
               
                 
                   
                     
                       2 
                       n 
                     
                     - 
                     1 
                   
                    
                   
                     
 
                   
                   ∴ 
                   
                     
                       ( 
                       5 
                       ) 
                     
                      
                     
                       a 
                       · 
                       
                         
                           ∏ 
                           
                             b 
                             = 
                             0 
                           
                           
                             n 
                             - 
                             1 
                           
                         
                          
                         
                             
                         
                          
                         
                           
                             
                               [ 
                               
                                 md 
                                  
                                 
                                   ( 
                                   
                                     r 
                                     
                                       2 
                                       b 
                                     
                                   
                                   ) 
                                 
                               
                               ] 
                             
                             
                               c 
                               b 
                             
                           
                           · 
                           
                             
                               ∏ 
                               
                                 
                                     
                                 
                                  
                                 
                                   b 
                                   = 
                                   n 
                                 
                               
                               n 
                             
                              
                             
                                 
                             
                              
                             
                               
                                 [ 
                                 
                                   md 
                                    
                                   
                                     ( 
                                     
                                       r 
                                       
                                         2 
                                         b 
                                       
                                     
                                     ) 
                                   
                                 
                                 ] 
                               
                               1 
                             
                           
                         
                       
                     
                   
                 
                 = 
                 
                   a 
                   · 
                   
                     
                       ∏ 
                       
                         b 
                         = 
                         0 
                       
                       
                         n 
                         - 
                         1 
                       
                     
                      
                     
                         
                     
                      
                     
                       
                         
                           
                             [ 
                             
                               md 
                                
                               
                                 ( 
                                 
                                   r 
                                   
                                     2 
                                     b 
                                   
                                 
                                 ) 
                               
                             
                             ] 
                           
                           
                             c 
                             b 
                           
                         
                         · 
                         
                           md 
                            
                           
                             ( 
                             
                               r 
                               n 
                             
                             ) 
                           
                         
                       
                        
                       
                           
                       
                        
                       
                         ( 
                         6 
                         ) 
                       
                     
                   
                 
               
             
           
         
       
       
         
           
             
                 
             
              
             
               
                 
                   
                     ( 
                     a 
                     ) 
                   
                    
                   
                       
                   
                    
                   When 
                    
                   
                       
                   
                    
                   c 
                 
                 = 
                 
                   2 
                   
                     n 
                     - 
                     1 
                   
                 
               
               , 
               
                 
                   2 
                   
                     n 
                     - 
                     1 
                   
                 
                 + 
                 1 
               
               , 
               
                 
                   2 
                   
                     n 
                     - 
                     1 
                   
                 
                 + 
                 2 
               
               , 
               … 
                
               
                   
               
               , 
               
                 
                   2 
                   n 
                 
                 - 
                 1 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 
                   ∵ 
                   
                     
                       ⌈ 
                       
                         
                           log 
                           2 
                         
                          
                         
                           ( 
                           
                             c 
                             + 
                             1 
                           
                           ) 
                         
                       
                       ⌉ 
                     
                     - 
                     1 
                   
                 
                 = 
                 
                   n 
                   - 
                   1 
                 
               
               , 
               
                 
 
               
                
               
                   
               
                
               
                 
                   ∴ 
                   
                     
                       ∏ 
                       
                         b 
                         = 
                         0 
                       
                       
                         n 
                         - 
                         1 
                       
                     
                      
                     
                         
                     
                      
                     
                       
                         [ 
                         
                           md 
                            
                           
                             ( 
                             
                               r 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         ] 
                       
                       
                         c 
                         b 
                       
                     
                   
                 
                 = 
                 
                   
                     
                       ∏ 
                       
                         b 
                         = 
                         0 
                       
                       
                         
                           ⌈ 
                           
                             
                               log 
                               2 
                             
                              
                             
                               ( 
                               
                                 c 
                                 + 
                                 1 
                               
                               ) 
                             
                           
                           ⌉ 
                         
                         - 
                         1 
                       
                     
                      
                     
                         
                     
                      
                     
                       
                         [ 
                         
                           md 
                            
                           
                             ( 
                             
                               r 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         ] 
                       
                        
                       
                         
                           c 
                           b 
                         
                          
                         
                           
 
                         
                          
                         
                             
                         
                         ( 
                         b 
                         ) 
                       
                        
                       
                           
                       
                        
                       When 
                        
                       
                           
                       
                        
                       c 
                     
                   
                   = 
                   0 
                 
               
               , 
               1 
               , 
               2 
               , 
               … 
                
               
                   
               
               , 
               
                 
                   2 
                   
                     n 
                     - 
                     1 
                   
                 
                 - 
                 1 
               
               , 
               
                 
 
               
                
               
                 
                   ∵ 
                   
                     
                       
                         ⌈ 
                         
                           
                             log 
                             2 
                           
                            
                           
                             ( 
                             
                               c 
                               + 
                               1 
                             
                             ) 
                           
                         
                         ⌉ 
                       
                       - 
                       1 
                     
                     &lt; 
                     
                       n 
                       - 
                       
                         1 
                          
                         
                             
                         
                          
                         and 
                          
                         
                             
                         
                          
                         Lemma 
                          
                         
                             
                         
                          
                         1 
                          
                         
                           : 
                         
                          
                         
                             
                         
                          
                         
                           c 
                           b 
                         
                       
                     
                   
                 
                 = 
                 0 
               
               , 
               
                 
                   
                     ∀ 
                     
                       b 
                       &gt; 
                       
                         
                           ⌈ 
                           
                             
                               log 
                               2 
                             
                              
                             
                               ( 
                               
                                 c 
                                 + 
                                 1 
                               
                               ) 
                             
                           
                           ⌉ 
                         
                         - 
                         1 
                       
                     
                   
                    
                   
                     
 
                   
                   ∴ 
                   
                     
                       ∏ 
                       
                         b 
                         = 
                         0 
                       
                       
                         n 
                         - 
                         1 
                       
                     
                      
                     
                         
                     
                      
                     
                       
                         [ 
                         
                           md 
                            
                           
                             ( 
                             
                               r 
                               
                                 2 
                                 b 
                               
                             
                             ) 
                           
                         
                         ] 
                       
                       
                         c 
                         b 
                       
                     
                   
                 
                 = 
                 
                   
                     
                       ∏ 
                       
                         b 
                         = 
                         0 
                       
                       
                         
                           ⌈ 
                           
                             
                               log 
                               2 
                             
                              
                             
                               ( 
                               
                                 c 
                                 + 
                                 1 
                               
                               ) 
                             
                           
                           ⌉ 
                         
                         - 
                         1 
                       
                     
                      
                     
                       
                         
                           [ 
                           
                             md 
                              
                             
                               ( 
                               
                                 r 
                                 
                                   2 
                                   b 
                                 
                               
                               ) 
                             
                           
                           ] 
                         
                         
                           c 
                           b 
                         
                       
                       · 
                       
                         
                           ∏ 
                           
                             b 
                             = 
                             
                               ⌈ 
                               
                                 
                                   log 
                                   2 
                                 
                                  
                                 
                                   ( 
                                   
                                     c 
                                     + 
                                     1 
                                   
                                   ) 
                                 
                               
                               ⌉ 
                             
                           
                           
                             n 
                             - 
                             1 
                           
                         
                          
                         
                             
                         
                          
                         
                           
                             [ 
                             
                               md 
                                
                               
                                 ( 
                                 
                                   r 
                                   
                                     2 
                                     b 
                                   
                                 
                                 ) 
                               
                             
                             ] 
                           
                           
                             c 
                             b 
                           
                         
                       
                     
                   
                   = 
                   
                     
                       
                         ∏ 
                         
                           b 
                           = 
                           0 
                         
                         
                           
                             ⌈ 
                             
                               
                                 log 
                                 2 
                               
                                
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   1 
                                 
                                 ) 
                               
                             
                             ⌉ 
                           
                           - 
                           1 
                         
                       
                        
                       
                         
                           
                             [ 
                             
                               md 
                                
                               
                                 ( 
                                 
                                   r 
                                   
                                     2 
                                     b 
                                   
                                 
                                 ) 
                               
                             
                             ] 
                           
                           
                             c 
                             b 
                           
                         
                         · 
                         1 
                       
                     
                     = 
                     
                       
                         ∏ 
                         
                           b 
                           = 
                           0 
                         
                         
                           
                             ⌈ 
                             
                               
                                 log 
                                 2 
                               
                                
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   1 
                                 
                                 ) 
                               
                             
                             ⌉ 
                           
                           - 
                           1 
                         
                       
                        
                       
                         
                           
                             [ 
                             
                               md 
                                
                               
                                 ( 
                                 
                                   r 
                                   
                                     2 
                                     b 
                                   
                                 
                                 ) 
                               
                             
                             ] 
                           
                           
                             c 
                             b 
                           
                         
                          
                         
                           
 
                         
                          
                         
                           ( 
                           a 
                           ) 
                         
                       
                     
                   
                 
               
               , 
               
                 
                   
                     ( 
                     b 
                     ) 
                   
                   ⇒ 
                   
                     ( 
                     6 
                     ) 
                   
                 
                 = 
                 
                   
                     a 
                     · 
                     
                       
                         ∏ 
                         
                           b 
                           = 
                           0 
                         
                         
                           
                             ⌈ 
                             
                               
                                 log 
                                 2 
                               
                                
                               
                                 ( 
                                 
                                   c 
                                   + 
                                   1 
                                 
                                 ) 
                               
                             
                             ⌉ 
                           
                           - 
                           1 
                         
                       
                        
                       
                         
                           
                             [ 
                             
                               md 
                                
                               
                                 ( 
                                 
                                   r 
                                   
                                     2 
                                     b 
                                   
                                 
                                 ) 
                               
                             
                             ] 
                           
                           
                             c 
                             b 
                           
                         
                         · 
                         
                           md 
                            
                           
                             ( 
                             
                               r 
                               b 
                             
                             ) 
                           
                         
                       
                     
                   
                   = 
                   
                     
                       choice 
                        
                       
                         ( 
                         c 
                         ) 
                       
                     
                     · 
                     
                       md 
                        
                       
                         ( 
                         
                           r 
                           n 
                         
                         ) 
                       
                     
                   
                 
               
             
           
         
       
     
         [0046]    Thus is proved Lemma III. 
         [0047]    Theorem II: If 
         [0000]    
       
         
           
             
               
                 choice 
                  
                 
                   ( 
                   c 
                   ) 
                 
               
               = 
               
                 a 
                 · 
                 
                   
                     ∏ 
                     
                       b 
                       = 
                       0 
                     
                     
                       
                         ⌈ 
                         
                           
                             log 
                             2 
                           
                            
                           
                             ( 
                             
                               c 
                               + 
                               1 
                             
                             ) 
                           
                         
                         ⌉ 
                       
                       - 
                       1 
                     
                   
                    
                   
                       
                   
                    
                   
                     
                       [ 
                       
                         md 
                          
                         
                           ( 
                           
                             r 
                             
                               2 
                               b 
                             
                           
                           ) 
                         
                       
                       ] 
                     
                     
                       c 
                       b 
                     
                   
                 
               
             
             , 
           
         
       
     
         [0000]    choice(0)˜choices(2 n −1), ∀nεN feature that they can generate 2 n  sets of geometric sequences each having α, an identical common ratio r, and a total number of terms 2 n . 
       Proof: 
       [0048]      
         [0000]    
       
         
               
               
             
               
               
               
               
             
           
               
                   
                   
               
             
             
               
                   
                 When n = 1, choice(0) = a, and  
               
               
                   
                 choice(1) = a · md(r 2     a   ) = a · md(r 1 ) = a · md(r). 
               
               
                   
                 
                   
                     
                       
                         
                           
                             When 
                              
                             
                                 
                             
                              
                             
                               md 
                                
                               
                                 ( 
                                 r 
                                 ) 
                               
                             
                           
                           = 
                           
                             1 
                             r 
                           
                         
                         , 
                       
                     
                   
                 
               
             
          
           
               
                   
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   md 
                                    
                                   
                                     ( 
                                     r 
                                     ) 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     1 
                                     r 
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 choice(0) = a 
                   
               
               
                   
                 When md(r) = r, 
                   
                   
               
               
                   
                   
                 choice(0) = a 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   md 
                                    
                                   
                                     ( 
                                     r 
                                     ) 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   r 
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
               
               
                   
                   
               
             
          
         
       
     
         [0049]    There are 2 n =2 1 =2 sets of geometric sequences each having α, an identical common ratio r, and a total number of terms 2 n =1 1 =2. 
         [0050]    When n=2, 
         [0051]    it is learned from Lemma III that choice(c+2 1 )=choice(c)·md(r 2     1   ), ∀c=0,2 1 −1=0,1. 
         [0052]    Express 2 of choice(2) in the binary system. 
         [0053]    Thus, choice(2)=choice(0+2 1 )=choice(0)·md(r 2     1   )=choice(0)·md(r 2 ). 
         [0054]    Express 3 of choice(3) in the binary system. 
         [0055]    Thus, choice(3)=choice(1+2 1 )=choice(1)+md(r 2     l   )=choice(1)+md(r 2 ). 
         [0000]    
       
         
               
             
               
               
               
               
               
               
               
             
               
             
               
               
               
               
               
               
               
             
           
               
                   
               
             
             
               
                 
                   
                     
                       
                         
                           
                             When 
                              
                             
                                 
                             
                              
                             
                               md 
                                
                               
                                 ( 
                                 
                                   r 
                                   2 
                                 
                                 ) 
                               
                             
                           
                           = 
                           
                             1 
                             
                               r 
                               2 
                             
                           
                         
                         , 
                       
                     
                   
                 
               
             
          
           
               
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     3 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 choice 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                   · 
                                   
                                     1 
                                     
                                       r 
                                       2 
                                     
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   
                                     a 
                                     · 
                                     
                                       1 
                                       r 
                                     
                                   
                                    
                                   
                                     1 
                                     
                                       r 
                                       2 
                                     
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     1 
                                     
                                       r 
                                       2 
                                     
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     2 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 choice 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   
                                     ( 
                                     0 
                                     ) 
                                   
                                   · 
                                   
                                     1 
                                     
                                       r 
                                       2 
                                     
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     1 
                                     
                                       r 
                                       2 
                                     
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   md 
                                    
                                   
                                     ( 
                                     r 
                                     ) 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     1 
                                     r 
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 choice (0) = a 
                   
                   
                   
               
               
                   
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     2 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 choice 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   
                                     ( 
                                     0 
                                     ) 
                                   
                                   · 
                                   
                                     1 
                                     
                                       r 
                                       2 
                                     
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     1 
                                     
                                       r 
                                       2 
                                     
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     3 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 choice 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                   · 
                                   
                                     1 
                                     
                                       r 
                                       2 
                                     
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   r 
                                   · 
                                   
                                     1 
                                     
                                       r 
                                       2 
                                     
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     1 
                                     r 
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 choice (0) = a 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   md 
                                    
                                   
                                     ( 
                                     r 
                                     ) 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   r 
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                   
                   
               
             
          
           
               
                 When md(r 2 ) = r 2 , 
               
             
          
           
               
                   
                   
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   md 
                                    
                                   
                                     ( 
                                     r 
                                     ) 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     1 
                                     r 
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 choice (0) = a 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     3 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 choice 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                   · 
                                   
                                     r 
                                     2 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     1 
                                     r 
                                   
                                   · 
                                   
                                     r 
                                     2 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   r 
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     2 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 choice 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   
                                     ( 
                                     0 
                                     ) 
                                   
                                   · 
                                   
                                     r 
                                     2 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     r 
                                     2 
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                   
               
               
                   
                   
                   
                 choice (0) = a 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   md 
                                    
                                   
                                     ( 
                                     r 
                                     ) 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   r 
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     2 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 choice 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   
                                     ( 
                                     0 
                                     ) 
                                   
                                   · 
                                   
                                     r 
                                     2 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     r 
                                     2 
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
                 
                   
                     
                       
                         
                           
                             
                               
                                 
                                   choice 
                                    
                                   
                                     ( 
                                     3 
                                     ) 
                                   
                                 
                                 = 
                                   
                                  
                                 choice 
                               
                             
                           
                           
                             
                               
                                   
                                  
                                 
                                   
                                     ( 
                                     1 
                                     ) 
                                   
                                   · 
                                   
                                     r 
                                     2 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   r 
                                   · 
                                   
                                     r 
                                     2 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 = 
                                   
                                  
                                 
                                   a 
                                   · 
                                   
                                     r 
                                     3 
                                   
                                 
                               
                             
                           
                         
                           
                       
                     
                   
                 
               
               
                   
               
             
          
         
       
     
         [0056]    As shown in abovementioned cases of 
         [0000]    
       
         
           
             
               
                 md 
                  
                 
                   ( 
                   
                     r 
                     2 
                   
                   ) 
                 
               
               = 
               
                 
                   
                     1 
                     
                       r 
                       2 
                     
                   
                    
                   
                       
                   
                    
                   and 
                    
                   
                       
                   
                    
                   
                     md 
                      
                     
                       ( 
                       
                         r 
                         2 
                       
                       ) 
                     
                   
                 
                 = 
                 
                   r 
                   2 
                 
               
             
             , 
           
         
       
     
         [0000]    when n=2, there are 2 n =2 2 =4 sets of geometric sequences each having α, an identical common ratio r, and a total number of terms 2 n 2 2 =4. 
         [0057]    Suppose that when n=k, Theorem I is true, i.e. there are 2 k  sets of arithmetic sequences each having α, an identical common ratio r, and a total number of terms 2 k : 
         [0000]    
       
         
               
               
               
               
               
               
               
               
               
             
           
               
                   
               
             
             
               
                 
                   
                     
                       
                         a 
                         · 
                         
                           1 
                           
                             r 
                             
                               
                                 2 
                                 k 
                               
                               - 
                               1 
                             
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         a 
                         · 
                         
                           1 
                           
                             r 
                             
                               
                                 2 
                                 k 
                               
                               - 
                               2 
                             
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         a 
                         · 
                         
                           1 
                           
                             r 
                             
                               
                                 2 
                                 k 
                               
                               - 
                               3 
                             
                           
                         
                       
                     
                   
                 
                 . . .  
                 a 
                   
                   
                   
                   
               
               
                   
                 
                   
                     
                       
                         a 
                         · 
                         
                           1 
                           
                             r 
                             
                               
                                 2 
                                 k 
                               
                               - 
                               2 
                             
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         a 
                         · 
                         
                           1 
                           
                             r 
                             
                               
                                 2 
                                 k 
                               
                               - 
                               3 
                             
                           
                         
                       
                     
                   
                 
                 . . .  
                 a 
                 a · r 
                   
                   
                   
               
               
                   
                   
                 
                   
                     
                       
                         a 
                         · 
                         
                           1 
                           
                             r 
                             
                               
                                 2 
                                 k 
                               
                               - 
                               3 
                             
                           
                         
                       
                     
                   
                 
                 . . .  
                 a 
                 a · r 
                 a · r 2   
                   
                   
               
               
                   
                   
                   
                 . . .  
                 . . .  
                 . . .  
                 . . .  
                 . . .  
                   
               
               
                   
                   
                   
                   
                 a 
                 a · r 
                 a · r 2   
                 . . .  
                 a · r 2     k     −1   
               
               
                   
               
             
          
         
       
     
         [0058]    When n=k+1, 
         [0059]    It is learned from Lemma III that choice(c+2 k )=choice(c)·md(r 2     n   ), ∀c=0,1,2, . . . , 2 n −1. 
         [0060]    When 
         [0000]    
       
         
           
             
               
                 md 
                 ( 
                 
                   r 
                   
                     2 
                     k 
                   
                 
                 ) 
               
               = 
               
                 1 
                 
                   r 
                   
                     2 
                     k 
                   
                 
               
             
             , 
           
         
       
     
         [0000]    2 k  rows are generated in the table. As the next term is formed via multiplying the current term and 
         [0000]    
       
         
           
             
               1 
               
                 r 
                 
                   2 
                   k 
                 
               
             
             , 
           
         
       
     
         [0000]    the next term is arranged at the left of the current term in the table. When md(r 2     k   )=r 2     k   , there are also 2 k  rows. As the next 
         [0061]    term is formed via multiplying the current term and r 2     k   , the next term is arranged at the right of the current terra in the table. As shown in abovementioned eases of 
         [0000]    
       
         
           
             
               
                 md 
                 ( 
                 
                   r 
                   
                     2 
                     k 
                   
                 
                 ) 
               
               = 
               
                 
                   
                     1 
                     
                       r 
                       
                         2 
                         k 
                       
                     
                   
                    
                   
                       
                   
                    
                   and 
                    
                   
                       
                   
                    
                   
                     md 
                     ( 
                     
                       r 
                       
                         2 
                         k 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   r 
                   
                     2 
                     k 
                   
                 
               
             
             , 
           
         
       
     
         [0000]    when n=k+1, there are totally 2 k +2 k =2 k+1  sets of geometric sequences each having α, an identical common ratio, and a total number of terms 2 k+1 . Thus is proved Theorem II. 
         [0062]    Below, a two-choice multiple choice question and a four-choice multiple choice question are used as the embodiments to demonstrate how the present invention generates numerical choices in form of a geometric sequence for a numerical multiple choice question. Firstly, define a numerical multiple choice question. The test question is “the length of a park trail is 40 m, and the jogging speed of a person is n1 to 10 m/sec; how much time will the person take to walk through the park trail?” The equation for solving the question is time=length÷speed40/n1 to 10. The correct answer is 40/n1 to 10 seconds. In the embodiment of a two-choice multiple choice question, the correct answer is 40/n1 to 10 seconds, and Choice 1 is (40/n1 to 10)*md(r) seconds, wherein n1 to 10 is a random integer number of from 1 to 10. In this embodiment, n1 to 10 assigned to be 5, and the correct answer is 8 seconds. In this embodiment, r of md(r) is assigned to be 1. However, r may be a random positive or negative integer in the present invention. Further, X is assigned to be 0 or 1 so as to vary the order of die correct answer appearing in the sequence of the numerical choices. 
         [0063]    When X=0, the numerical choices are calculated as follows: 
         [0064]    Choice 0 (the correct answer)=(40/n1 to 10)=(40/5)=8; 
         [0065]    Choice 1=(40/n1 to 10)·(0.5+X*1.5)=(40/5)·(0.5÷0*1.5)=(40/5)·0.5=4. 
         [0066]    The choices arranged from large to small are 8, 4. The correct answer is the largest one among the choices. 
         [0067]    When X=1, the numerical choices are calculated as follows: 
         [0068]    Choice 0 (the correct answer)=(40/n1 to 10)=(40/5)=8; 
         [0069]    Choice 1=(40/n1 to 10) (0.5+X*1.5)=(40/5)·(0.5+1*1.5)=(40/5)·2=16. 
         [0070]    The choices arranged from large to small are 16, 8. The correct answer is the second largest one among the choices. 
         [0071]    Next, a four-choice multiple choice question is used as the embodiment to demonstrate die present invention. Similarly, X is 0 or 1 in the embodiment. From Lemma III, it is learned that choice(c+ 2   1 )=choice(c)·md(r 2     1   ). Four numerical choices of the multiple choice question are calculated as follows: 
         [0000]    
       
         
           
             
                 
             
              
             
               
                 
                   Choice 
                    
                   
                       
                   
                    
                   0 
                    
                   
                       
                   
                    
                   
                     ( 
                     
                       the 
                        
                       
                           
                       
                        
                       correct 
                        
                       
                           
                       
                        
                       answer 
                     
                     ) 
                   
                 
                 = 
                 
                   
                     choice 
                      
                     
                       ( 
                       0 
                       ) 
                     
                   
                   = 
                   
                     a 
                     = 
                     
                       ( 
                       
                         
                           40 
                           / 
                           n 
                         
                          
                         
                             
                         
                          
                         1 
                          
                         
                             
                         
                          
                         to 
                          
                         
                             
                         
                          
                         10 
                       
                       ) 
                     
                   
                 
               
               ; 
             
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   choice 
                    
                   
                     ( 
                     1 
                     ) 
                   
                 
                 = 
                 
                   
                     a 
                     · 
                     
                       md 
                       ( 
                       
                         r 
                         
                           2 
                           0 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       a 
                       · 
                       
                         md 
                          
                         
                           ( 
                           
                             r 
                             1 
                           
                           ) 
                         
                       
                     
                     = 
                     
                       
                         a 
                         · 
                         
                           md 
                            
                           
                             ( 
                             r 
                             ) 
                           
                         
                       
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         * 
                         
                           ( 
                           
                             0.5 
                             + 
                             
                               X 
                               * 
                               1.5 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   choice 
                    
                   
                     ( 
                     2 
                     ) 
                   
                 
                 = 
                 
                   
                     choice 
                      
                     
                       ( 
                       
                         0 
                         + 
                         
                           2 
                           1 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       
                         choice 
                          
                         
                           ( 
                           0 
                           ) 
                         
                       
                       · 
                       
                         md 
                         ( 
                         
                           r 
                           
                             2 
                             1 
                           
                         
                         ) 
                       
                     
                     = 
                     
                       
                         
                           choice 
                            
                           
                             ( 
                             0 
                             ) 
                           
                         
                         · 
                         
                           md 
                            
                           
                             ( 
                             
                               r 
                               2 
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         * 
                         
                           ( 
                           
                             0.25 
                             + 
                             
                               
                                 X 
                                 2 
                               
                               * 
                               3.75 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
                 
             
              
             and 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 3 
               
               = 
               
                 
                   choice 
                    
                   
                     ( 
                     3 
                     ) 
                   
                 
                 = 
                 
                   
                     choice 
                      
                     
                       ( 
                       
                         1 
                         + 
                         
                           2 
                           1 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       
                         choice 
                          
                         
                           ( 
                           1 
                           ) 
                         
                       
                       + 
                       
                         md 
                         ( 
                         
                           r 
                           
                             2 
                             1 
                           
                         
                         ) 
                       
                     
                     = 
                     
                       
                         
                           choice 
                            
                           
                             ( 
                             1 
                             ) 
                           
                         
                         + 
                         
                           md 
                            
                           
                             ( 
                             
                               r 
                               2 
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           ( 
                           
                             
                               40 
                               / 
                               n 
                             
                              
                             
                                 
                             
                              
                             1 
                              
                             
                                 
                             
                              
                             to 
                              
                             
                                 
                             
                              
                             10 
                           
                           ) 
                         
                         * 
                         
                           ( 
                           
                             0.5 
                             + 
                             
                               X 
                               * 
                               1.5 
                             
                           
                           ) 
                         
                         * 
                         
                           ( 
                           
                             0.25 
                             + 
                             
                               
                                 X 
                                 2 
                               
                               * 
                               3.75 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             , 
           
         
       
     
         [0000]    wherein n1 to 10 is a random integer number of from 1 to 10, and wherein md(2)=(0.5+X*1.5) and md(4)=(0.25+X 2 *3.75), wherein either of X and X 2  is a random integer number selected from 0 and 1. In this embodiment, n1 to 10 is assigned to be 5, and (X, X 2 ) have four combinations: (0, 0), (1, 0), (0, 1) and (1, 1). 
         [0000]    
       
         
           
             
               When 
                
               
                   
               
                
               
                 ( 
                 
                   X 
                   , 
                   
                     X 
                     2 
                   
                 
                 ) 
               
                
               
                   
               
                
               is 
                
               
                   
               
                
               
                 ( 
                 
                   0 
                   , 
                   0 
                 
                 ) 
               
             
             , 
             
               
 
             
              
             
               
                 
                   Choice 
                    
                   
                       
                   
                    
                   0 
                    
                   
                       
                   
                    
                   
                     ( 
                     
                       the 
                        
                       
                           
                       
                        
                       correct 
                        
                       
                           
                       
                        
                       answer 
                     
                     ) 
                   
                 
                 = 
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   = 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     = 
                     8 
                   
                 
               
               ; 
             
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   * 
                   
                     ( 
                     
                       0.5 
                       + 
                       
                         X 
                         * 
                         1.5 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         0.5 
                         + 
                         
                           0 
                           * 
                           1.5 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       * 
                       0.5 
                     
                     = 
                     4 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   * 
                   
                     ( 
                     
                       0.25 
                       + 
                       
                         
                           X 
                           2 
                         
                         * 
                         3.75 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         0.25 
                         + 
                         
                           0 
                           * 
                           3.75 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       * 
                       0.25 
                     
                     = 
                     2 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
                 
             
              
             and 
           
         
       
       
         
           
             
               Choice 
                
               
                   
               
                
               3 
             
             = 
             
               
                 
                   ( 
                   
                     
                       40 
                       / 
                       n 
                     
                      
                     
                         
                     
                      
                     1 
                      
                     
                         
                     
                      
                     to 
                      
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
                 * 
                 
                   ( 
                   
                     0.5 
                     + 
                     
                       X 
                       * 
                       1.5 
                     
                   
                   ) 
                 
                 * 
                 
                   ( 
                   
                     0.25 
                     + 
                     
                       
                         X 
                         2 
                       
                       * 
                       3.75 
                     
                   
                   ) 
                 
               
               = 
               
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     * 
                     
                       ( 
                       
                         0.5 
                         + 
                         
                           0 
                           * 
                           1.5 
                         
                       
                       ) 
                     
                   
                   + 
                   
                     ( 
                     
                       0.25 
                       + 
                       
                         0 
                         * 
                         3.75 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     8 
                     * 
                     0.5 
                     * 
                     0.25 
                   
                   = 
                   1. 
                 
               
             
           
         
       
     
         [0072]    The choices arranged from large to small are 8, 4, 2, 1. The correct answer is the largest one among the choices. 
         [0000]    
       
         
           
             
                 
             
              
             
               
                 When 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     X 
                     , 
                     
                       X 
                       2 
                     
                   
                   ) 
                 
                  
                 
                     
                 
                  
                 is 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     1 
                     , 
                     0 
                   
                   ) 
                 
               
               , 
               
                 
 
               
                
               
                 
                   
                     Choice 
                      
                     
                         
                     
                      
                     0 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         the 
                          
                         
                             
                         
                          
                         correct 
                          
                         
                             
                         
                          
                         answer 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       ( 
                       
                         
                           40 
                           / 
                           n 
                         
                          
                         
                             
                         
                          
                         1 
                          
                         
                             
                         
                          
                         to 
                          
                         
                             
                         
                          
                         10 
                       
                       ) 
                     
                     = 
                     
                       
                         ( 
                         
                           40 
                           / 
                           5 
                         
                         ) 
                       
                       = 
                       8 
                     
                   
                 
                 ; 
               
             
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   * 
                   
                     ( 
                     
                       0.5 
                       + 
                       
                         X 
                         * 
                         1.5 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         0.5 
                         + 
                         
                           1 
                           * 
                           1.5 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       * 
                       2 
                     
                     = 
                     16 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   * 
                   
                     ( 
                     
                       0.25 
                       + 
                       
                         
                           X 
                           2 
                         
                         * 
                         3.75 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         0.25 
                         + 
                         
                           0 
                           * 
                           3.75 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       * 
                       0.25 
                     
                     = 
                     2 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
                 
             
              
             and 
           
         
       
       
         
           
             
               Choice 
                
               
                   
               
                
               3 
             
             = 
             
               
                 
                   ( 
                   
                     
                       40 
                       / 
                       n 
                     
                      
                     
                         
                     
                      
                     1 
                      
                     
                         
                     
                      
                     to 
                      
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
                 * 
                 
                   ( 
                   
                     0.5 
                     + 
                     
                       X 
                       * 
                       1.5 
                     
                   
                   ) 
                 
                 * 
                 
                   ( 
                   
                     0.25 
                     + 
                     
                       
                         X 
                         2 
                       
                       * 
                       3.75 
                     
                   
                   ) 
                 
               
               = 
               
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     * 
                     
                       ( 
                       
                         0.5 
                         + 
                         
                           1 
                           * 
                           1.5 
                         
                       
                       ) 
                     
                   
                   + 
                   
                     ( 
                     
                       0.25 
                       + 
                       
                         0 
                         * 
                         3.75 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     8 
                     * 
                     2 
                     * 
                     0.25 
                   
                   = 
                   4. 
                 
               
             
           
         
       
     
         [0073]    The choices arranged from large to small are 16, 8, 4, 2. The correct answer is the second largest one among the choices. 
         [0000]    
       
         
           
             
                 
             
              
             
               
                 When 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     X 
                     , 
                     
                       X 
                       2 
                     
                   
                   ) 
                 
                  
                 
                     
                 
                  
                 is 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     0 
                     , 
                     1 
                   
                   ) 
                 
               
               , 
               
                 
 
               
                
               
                 
                   
                     Choice 
                      
                     
                         
                     
                      
                     0 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         the 
                          
                         
                             
                         
                          
                         correct 
                          
                         
                             
                         
                          
                         answer 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       ( 
                       
                         
                           40 
                           / 
                           n 
                         
                          
                         
                             
                         
                          
                         1 
                          
                         
                             
                         
                          
                         to 
                          
                         
                             
                         
                          
                         10 
                       
                       ) 
                     
                     = 
                     
                       
                         ( 
                         
                           40 
                           / 
                           5 
                         
                         ) 
                       
                       = 
                       8 
                     
                   
                 
                 ; 
               
             
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   * 
                   
                     ( 
                     
                       0.5 
                       + 
                       
                         X 
                         * 
                         1.5 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         0.5 
                         + 
                         
                           0 
                           * 
                           1.5 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       * 
                       0.5 
                     
                     = 
                     4 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   * 
                   
                     ( 
                     
                       0.25 
                       + 
                       
                         
                           X 
                           2 
                         
                         * 
                         3.75 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         0.25 
                         + 
                         
                           1 
                           * 
                           3.75 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       * 
                       4 
                     
                     = 
                     32 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
                 
             
              
             and 
           
         
       
       
         
           
             
               Choice 
                
               
                   
               
                
               3 
             
             = 
             
               
                 
                   ( 
                   
                     
                       40 
                       / 
                       n 
                     
                      
                     
                         
                     
                      
                     1 
                      
                     
                         
                     
                      
                     to 
                      
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
                 * 
                 
                   ( 
                   
                     0.5 
                     + 
                     
                       X 
                       * 
                       1.5 
                     
                   
                   ) 
                 
                 * 
                 
                   ( 
                   
                     0.25 
                     + 
                     
                       
                         X 
                         2 
                       
                       * 
                       3.75 
                     
                   
                   ) 
                 
               
               = 
               
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     * 
                     
                       ( 
                       
                         0.5 
                         + 
                         
                           0 
                           * 
                           1.5 
                         
                       
                       ) 
                     
                   
                   + 
                   
                     ( 
                     
                       0.25 
                       + 
                       
                         1 
                         * 
                         3.75 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     8 
                     * 
                     0.5 
                     * 
                     4 
                   
                   = 
                   16. 
                 
               
             
           
         
       
     
         [0074]    The choices arranged from large to small are 32, 16, 8, 4. The correct answer is the third largest one among the choices. 
         [0000]    
       
         
           
             
                 
             
              
             
               
                 When 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     X 
                     , 
                     
                       X 
                       2 
                     
                   
                   ) 
                 
                  
                 
                     
                 
                  
                 is 
                  
                 
                     
                 
                  
                 
                   ( 
                   
                     1 
                     , 
                     1 
                   
                   ) 
                 
               
               , 
               
                 
 
               
                
               
                 
                   
                     Choice 
                      
                     
                         
                     
                      
                     0 
                      
                     
                         
                     
                      
                     
                       ( 
                       
                         the 
                          
                         
                             
                         
                          
                         correct 
                          
                         
                             
                         
                          
                         answer 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       ( 
                       
                         
                           40 
                           / 
                           n 
                         
                          
                         
                             
                         
                          
                         1 
                          
                         
                             
                         
                          
                         to 
                          
                         
                             
                         
                          
                         10 
                       
                       ) 
                     
                     = 
                     
                       
                         ( 
                         
                           40 
                           / 
                           5 
                         
                         ) 
                       
                       = 
                       8 
                     
                   
                 
                 ; 
               
             
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   * 
                   
                     ( 
                     
                       0.5 
                       + 
                       
                         X 
                         * 
                         1.5 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         0.5 
                         + 
                         
                           1 
                           * 
                           1.5 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       * 
                       2 
                     
                     = 
                     16 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
               
                 Choice 
                  
                 
                     
                 
                  
                 2 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         40 
                         / 
                         n 
                       
                        
                       
                           
                       
                        
                       1 
                        
                       
                           
                       
                        
                       to 
                        
                       
                           
                       
                        
                       10 
                     
                     ) 
                   
                   * 
                   
                     ( 
                     
                       0.25 
                       + 
                       
                         
                           X 
                           2 
                         
                         * 
                         3.75 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     + 
                     
                       ( 
                       
                         0.25 
                         + 
                         
                           1 
                           * 
                           3.75 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       8 
                       * 
                       4 
                     
                     = 
                     32 
                   
                 
               
             
             ; 
           
         
       
       
         
           
             
                 
             
              
             and 
           
         
       
       
         
           
             
               Choice 
                
               
                   
               
                
               3 
             
             = 
             
               
                 
                   ( 
                   
                     
                       40 
                       / 
                       n 
                     
                      
                     
                         
                     
                      
                     1 
                      
                     
                         
                     
                      
                     to 
                      
                     
                         
                     
                      
                     10 
                   
                   ) 
                 
                 * 
                 
                   ( 
                   
                     0.5 
                     + 
                     
                       X 
                       * 
                       1.5 
                     
                   
                   ) 
                 
                 * 
                 
                   ( 
                   
                     0.25 
                     + 
                     
                       
                         X 
                         2 
                       
                       * 
                       
                           
                       
                        
                       3.75 
                     
                   
                   ) 
                 
               
               = 
               
                 
                   
                     
                       ( 
                       
                         40 
                         / 
                         5 
                       
                       ) 
                     
                     * 
                     
                       ( 
                       
                         0.5 
                         + 
                         
                           1 
                           * 
                           1.5 
                         
                       
                       ) 
                     
                   
                   + 
                   
                     ( 
                     
                       0.25 
                       + 
                       
                         1 
                         * 
                         3.75 
                       
                     
                     ) 
                   
                 
                 = 
                 
                   
                     8 
                     * 
                     2 
                     * 
                     4 
                   
                   = 
                   64. 
                 
               
             
           
         
       
     
         [0075]    The choices arranged from large to small are 64, 32, 16, 8. The correct answer is the fourth largest one among the choices. From the abovementioned examples, it is known that the variable X can be used to vary the order of the correct answer in the numerical choices. 
         [0076]    In conclusion, the present invention proposes a choice generation method for a numerical multiple choice question, which applies to a numerical multiple choice question database, and which generates numerical choices in form of an arithmetic or geometric sequence, and which uses a random variable inside the equation to vary the order that the correct answer appears in the generated numeral choices. 
         [0077]    The embodiments described above are only to exemplify the present invention but not to limit the scope of the present invention. Any equivalent modification or variation according to the characteristic or spirit of the present invention is to be also included within the scope of the present invention.