From: cliff@watson.ibm.com (cliff)
Subject: Cliff Puzzle 7: 3x3 recursion
Date: Thu, 15 Oct 1992 22:50:47 GMT
Disclaimer: This posting represents the poster's views, not necessarily those of IBM
Organization: A

Title: Cliff Puzzle 7: 3x3 Recursion                                            
From: cliff@watson.ibm.com                                                      
                                                                                
If you respond to this puzzle, if possible please send me your name,            
address, affiliation, iand e-mail address
so I can credit you in the future if needed.
If you like, tell me a little bit        
about yourself so I can cite you appropriately if you provide unique            
information.  PLEASE ALSO directly mail me a copy of your response in           
addition to any responding you do in the newsgroup.  I will assume it is        
OK to describe your answer in any article or publication I may write in         
the future, with attribution to you, unless you state otherwise.                
Thanks, Cliff Pickover                                                          
                                                                                
* * *                                                                           
                                                                                
Consider the 3x3 array below.  All nine digits are used exactly once.           
                                                                                
1 9 2                                                                           
3 8 4                                                                           
5 7 6                                                                           
                                                                                
Notice that "384" is twice the number in the first row, and that                
"576" is three times the number in the first row.                               
                                                                                
Questions:                                                                      
1.  Are there other ways of arranging the numbers to produce the same           
result using each digit only once and the same rules?                           
Remember, the second row must be twice the first.  The third row                
must be 3 times the first row.                                                  
                                                                                
2.  Start with the number in the last row (e.g "576" or any other               
solution you may find) and continue to form another 3x3 matrix using the        
same rules with the new starting number.  In other words, 
the last number becomes the first, and the number in         
the new second row must be twice the first.  The third row must be three  
times the first.  (For this problem you may truncate any digits in the          
beginning.  For example, 1384 would become 384.)                                
                                                                                
Keep going.  How many matrices can you create before it is impossible           
to continue?  Again, each digit must be used only once                          
in each matrix.                                                                 
Newsgroups: rec.puzzles