math-eval/TAL-SCQ5K huggingface dataset:
Problem: Two dice are thrown. What is the probability that the product of the two numbers is a multiple of $5 ?$ (2001 AMC 8 Problem, Question \#14) 
Options:
aoVal: A
content: $\frac{1}{36}$ 
aoVal: B
content: $\frac{1}{18}$ 
aoVal: C
content: $\frac{1}{6}$ 
aoVal: D
content: $\frac{11}{36}$ 
aoVal: E
content: $\frac{1}{3}$ 
Answer: D
Analysis: This is equivalent to asking for the probability that at least one of the numbers is a multiple of 5 , since if one of the numbers is a multiple of 5 , then the product with it and another integer is also a multiple of 5 , and if a number is a multiple of 5 , then since 5 is prime, one of the factors must also have a factor of 5 , and 5 is the only multiple of 5 on a die, so one of the numbers rolled must be a 5 . To find the probability of rolling at least one 5 , we can find the probability of not rolling a 5 and subtract that from 1, since you either roll a 5 or not roll a 5 . The probability of not rolling a 5 on either dice is $\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)=\frac{25}{36}$. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of 5 , is $1-\frac{25}{36}=\frac{11}{36}, \text{D}$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Louis chooses a whole number from $$1$$ to $$30$$ at random. What is the probability that the number he chooses is a prime number? 
Options:
aoVal: A
content: $$\frac{3}{10}$$. 
aoVal: B
content: $$\frac{2}{5}$$. 
aoVal: C
content: $$\frac{1}{6}$$. 
aoVal: D
content: $$\frac{13}{30}$$. 
aoVal: E
content: $$\frac{1}{3}$$ 
Answer: E
Analysis: Prime numbers from $$1$$ to $$30$$:  $$2$$, $$3$$, $$5$$, $$7$$, $$11$$, $$13$$, $$17$$, $$19$$, $$23$$, $$29$$  Thus, there are $$10$$ prime numbers from $$1$$ to $$30$$.  Therefore, the probability that the number she chooses is a prime number is $$\frac{10}{30}=\frac{1}{3}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: The product of two whole numbers is $$30$$. What is the least possible value of their sum? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$31$$ 
Answer: B
Analysis: The product of two whole numbers is $$30$$. If the numbers are $$5$$ and $$6$$, their sum is $$5+6=11$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: A box contains $5$ red balls and $3$ white balls that are identical in all aspects except color. One ball is drawn at random from the box and then replaced. The box is then thoroughly shaken so that the balls are arranged at random again and a second ball is drawn randomly from the box. What is the probability of drawing white ball for both time? 
Options:
aoVal: A
content: $$\frac{5}{8}$$ 
aoVal: B
content: $$\frac{3}{8}$$ 
aoVal: C
content: $$\frac{25}{64}$$ 
aoVal: D
content: $$\frac{9}{64}$$ 
Answer: D
Analysis: $$D$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: How many numbers are there between $$0 \sim 50$$ that do not consist of digit \textquotesingle$$6$$\textquotesingle?． 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$50$$ 
Answer: C
Analysis: $$6, 16, 26, 36, 46$$  $$50$$ numbers minus $$5$$ numbers -\/-\textgreater{} $$45$$ numbers 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Problem: How many $$4$$-digit numbers greater than $$1000$$ are there that use the four digits of $$2012$$? ($$2012$$ AMC $$8$$ Problems, Question \#$10$) 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$12$$ 
Answer: D
Analysis: There are $3$ numbers that can be formed between $1000$ and $1999$, and $3\times2\times1=6$ numbers that can be formed with $2$ in the first place.  So the answer is $3+6=9$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 Driver seat, 1 front passenger seat, and 2 back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there? (2003 AMC 8 Problem, Question \#16) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$24$$ 
Answer: D
Analysis: There are only 2 people who can go in the driver\textquotesingle s seat-Bonnie and Carlo. Any of the 3 remaining people can go in the front passenger seat. There are 2 people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are $2 \cdot 3 \cdot 2$ or 12 ways. The answer is then (D) 12 . 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: The sum of the digits of $$1993$$ is $$1+9+9+3$$, or $$22$$. At some time in the future, the sum of the digits of a year will be $$33$$. This will \emph{first} occur in the century. 
Options:
aoVal: A
content: $$21\text{st}$$ 
aoVal: B
content: $$60\text{th}$$ 
aoVal: C
content: $$70\text{th}$$ 
aoVal: D
content: $$80\text{th}$$ 
Answer: C
Analysis: The sum of the digits of $$1993$$ is $$1+9+9+3$$, or $$22$$. At some time in the future, the sum of the digits of a year will be $$33$$. Since $$9+9+9 = 27$$, this will \emph{first} occur in $$6999$$, the $$70\text{th}$$ century. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Splitting Whole Numbers
Problem: Harriet tells Topaz that she is thinking of three positive integers, not necessarily all different. She tells her that the product of her three integers is $$36$$. She also tells her the sum of her three integers. However, Topaz still cannot work out what the three integers are. What is the sum of Harriet\textquotesingle s three integers? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$16$$ 
Answer: C
Analysis: The possible groups of three integers with product $$36$$ are $$(1,1,36)$$, $$(1,2,18)$$, $$(1,3,12)$$, $$(1,4,9)$$, $$(1,6,6)$$, $$(2,2,9)$$, $$(2,3,6)$$ and $$(3,3,4)$$ with sums $$38$$, $$21$$, $$16$$, $$14$$, $$13$$, $$13$$, $$11$$ and $$10$$ respectively. The only value for the sum that occurs twice is $$13$$. Hence, since Topaz does not know what the three integers chosen are, the sum of Harriet\textquotesingle s three integers is $$13$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: There are some two-digit numbers. The sum of the digits in the tens place and the ones place is 7. How many such numbers are there? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$4$$ 
Answer: A
Analysis: 16,25,34,43,52,61,70 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: There is a two-digit number. The product of its digits is $$18$$. The sum of the digits is $$11$$. The difference of the digits of this number is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$17$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$9$$ 
Answer: D
Analysis: $3\times6=18$, $3+6=9$  $2\times9=18$, $9+2=11$  Thus, the two digits of the number is $9$ and $2$.  The difference is $9-2=7$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: The clock is showing $$11\text{am}$$ now.  When the hour hand has turned through $$165{}^{}\circ$$, what time will it be? 
Options:
aoVal: A
content: $$11.30\text{pm}$$ 
aoVal: B
content: $$12.30\text{pm}$$ 
aoVal: C
content: $$4.30\text{pm}$$ 
aoVal: D
content: $$5.30\text{pm}$$ 
aoVal: E
content: $$11.30\text{pm}$$ 
Answer: C
Analysis: Each hour, the hour hand turns $$30{}^{}\circ$$, so it will take $$5\frac{1}{2}$$ hours to turn $$165{}^{}\circ$$; the time will therefore be $$4.30\text{pm}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: In a particular game, each of 4 players rolls a standard 6 -sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo\textquotesingle s first roll was a 5 , given that he won the game? 
Options:
aoVal: A
content: $\frac{61}{216}$ 
aoVal: B
content: $\frac{367}{1296}$ 
aoVal: C
content: $\frac{41}{144}$ 
aoVal: D
content: $\frac{185}{648}$ 
aoVal: E
content: $\frac{11}{36}$ 
Answer: C
Analysis: Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a 5 is just the probability that the highest roll in the first round is 5 . Let $P(x)$ indicate the probability that event $x$ occurs. We find that $P($ No one rolls a 6$)-P($ No one rolls a 5 or 6$)=P($ The highest roll is a 5$)$, so $$ \begin{gathered} P(\text { No one rolls a } 6)=\left(\frac{5}{6}\right)^{4}, \textbackslash\textbackslash{} P(\text { No one rolls a } 5 \text { or } 6)=\left(\frac{2}{3}\right)^{4}, \textbackslash\textbackslash{} P(\text { The highest roll is a } 5)=\left(\frac{5}{6}\right)^{4}-\left(\frac{4}{6}\right)^{4}=\frac{5^{4}-4^{4}}{6^{4}}=\frac{369}{1296}=(\text { C }) \frac{41}{144} \text {. } \end{gathered} $$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Miley had some keychains at first. She gave $$\frac{6}{7}$$ of them to her friends. Her mother then gave her another $$8$$ keychains. She had $$15$$ keychains in the end. How many keychains did Miley give to her friends in total? 
Options:
aoVal: A
content: $$42$$ 
aoVal: B
content: $$48$$ 
aoVal: C
content: $$49$$ 
aoVal: D
content: $$90$$ 
Answer: A
Analysis: $\frac{1}{7}$ of her keychains $=15-8=7$  $\textasciitilde$  Given to friends  $=\frac{6}{7}$ of her keychains  $=7\times6$  $=42$ 
Knowledge points: Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Word Problems Involving Fractions and Percentages->Finding a Whole Given a Part and the Percentage
Overseas Competition->Knowledge Point->Counting Modules
Problem: Martin uses the five digits $$0$$, $$1$$, $$3$$, $$7$$, $$9$$ to make some numbers (each digit can only be used once).  How many different three-digit numbers can be made? 
Options:
aoVal: A
content: $$60$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$40$$ 
Answer: C
Analysis: $4\times4\times3=48$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Ann sleeps just $$8$$ hrs. each day, so in $$10$$ days, she\textquotesingle s \emph{awake~}~\uline{~~~~~~~~~~}~hrs. 
Options:
aoVal: A
content: $$10\times 16$$ 
aoVal: B
content: $$8\times 10$$ 
aoVal: C
content: $$8\times 24$$ 
aoVal: D
content: $$16\times 24$$ 
Answer: A
Analysis: Ann\textquotesingle s awake $$16$$ hours each day. In $$10$$ days, that\textquotesingle s $$(10\times 16)$$ hours. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: A bag contains 4 red chips, 2 blue chips, and 3 white chips. If one chip is drawn at random, what is the probability that the chip will not be red? 
Options:
aoVal: A
content: $\dfrac29$ 
aoVal: B
content: $\dfrac49$ 
aoVal: C
content: $\dfrac59$ 
aoVal: D
content: $\dfrac69$ 
aoVal: E
content: $\dfrac79$ 
Answer: C
Analysis: $1-\dfrac49=\dfrac59$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: In how many ways can the letters in $CPCCKBY$ be rearranged so that two or more $C$s are not adjacent to each other? 
Options:
aoVal: A
content: $$240$$ 
aoVal: B
content: $$180$$ 
aoVal: C
content: $$64$$ 
aoVal: D
content: $$96$$ 
aoVal: E
content: $$140$$ 
Answer: A
Analysis: There are $3$ $C$s in total now with other $4$ letters remaining.  There are $\_4P\_4$ ways for us to arrange the remaining $4$ letters\textquotesingle{} positions.  So the answer is $\_4P\_4\times \_5C\_3 =240$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: In a particular game, each of $4$ players rolls a standard $6$-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo\textquotesingle s first roll was a $5$, given that he won the game? 
Options:
aoVal: A
content: $\frac{61}{216}$ 
aoVal: B
content: $\frac{367}{1296}$ 
aoVal: C
content: $\frac{41}{144}$ 
aoVal: D
content: $\frac{185}{648}$ 
aoVal: E
content: $\frac{11}{36}$ 
Answer: C
Analysis: Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a $5$ is just the probability that the highest roll in the first round is $5$. Let $P(x)$ indicate the probability that event $x$ occurs. We find that $P$ (No one rolls a $6$)-$P$ (No one rolls a $5$ or $6$)$=P$ (The highest roll is a $5$), so  $$ \begin{gathered} P(\text {No one rolls a } 6)=\left(\frac{5}{6}\right)^{4}, \textbackslash\textbackslash{} P(\text {No one rolls a } 5 \text { or } 6)=\left(\frac{2}{3}\right)^{4}, \textbackslash\textbackslash{} P(\text {The highest roll is a } 5)=\left(\frac{5}{6}\right)^{4}-\left(\frac{4}{6}\right)^{4}=\frac{5^{4}-4^{4}}{6^{4}}=\frac{369}{1296}=(\text {C}) \frac{41}{144} \text {. } \end{gathered} $$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: How many two-digit numbers are there where the tens digit is greater than the ones digit? 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$26$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$36$$ 
aoVal: E
content: $$45$$ 
Answer: E
Analysis: $$1+2+3+4+5+6+7+8+9=45$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: How many three-digit numbers are there that have the sum of their digits equal to $5$? (For example, $122$ is such a number, because $1+2+2=5$.)． 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$25$$ 
aoVal: E
content: $$30$$ 
Answer: B
Analysis: Method 1:  $104, 113, 122, 131, 140$;  $203, 212, 221, 230$;  $302, 311, 320$;  $401, 410$;  $500$  $5+4+3+2+1=15$  Method 2:$5=0+0+5=0+1+4=0+2+3=1+1+3=1+2+2$  When the three-digit number is formed by $0$, $0$, $5$, it can only be $500$, therefore $1$ number;  when the three-digit number is formed by $0$, $1$, $4$, it can be $104$, $140$, $401$ or $410$, therefore $4$ numbers;  when the three-digit number is formed by $0$, $2$, $3$, it can be $203$, $230$, $302$, or $320$, therefore $4$ numbers；  when the three-digit number is formed by $1$, $1$, $3$, it can be $113$, $131$, or $311$, therefore $3$ numbers;  when the three-digit number is formed by $1$, $2$, $2$, it can be $122$, $212$, or $221$, therefore $3$ numbers;  in total there are $1+4+4+3+3=15$ numbers that meet the requirement of the question. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Splitting Whole Numbers
Problem: $97+101+102+104+105=$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$500$$ 
aoVal: B
content: $$505$$ 
aoVal: C
content: $$507$$ 
aoVal: D
content: $$509$$ 
Answer: D
Analysis: $97+101+102+104+105=100\times5-3+1+2+4+5=509$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: Four students asked their teacher, Mr. Carter, to line up with them to take a picture.  ①If Mr. Carter does not want to stand on an end, how many different ways can they line up for the picture?  ②If Mr. Carter insists on standing on an end, how many different ways can they line up for the picture? 
Options:
aoVal: A
content: $72$ , $24$ 
aoVal: B
content: $96$ , $24$ 
aoVal: C
content: $72$ , $48$ 
aoVal: D
content: $96$ , $48$ 
Answer: C
Analysis: ①$$3\times 4\times 3\times 2\times 1=72$$ ,  ②$$2\times 4\times 3\times 2\times 1=48$$ . 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Adele has $1$ blue and $1$ red marble in her hand. Bobby has $1$ blue, $1$ yellow, and $2$ red marbles in his hand. Each of them randomly picks a marble to show the other. What is the probability that the colors are the same? 
Options:
aoVal: A
content: $\dfrac{1}{4}$ 
aoVal: B
content: $\dfrac{1}{3}$ 
aoVal: C
content: $\dfrac{3}{8}$ 
aoVal: D
content: $\dfrac{1}{2}$ 
aoVal: E
content: $\dfrac{2}{3}$ 
Answer: C
Analysis: The probability that both show a blue marble is $\dfrac{1}{2}\times \dfrac{1}{4}=\dfrac{1}{8}$. The probability that both show a red marble is $\dfrac{1}{2}\times \dfrac{2}{4}=\dfrac{1}{4}$. Therefore, the probability is $\dfrac{1}{4}+\dfrac{1}{8}=\dfrac{3}{8}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Which of the following statements is wrong  ? 
Options:
aoVal: A
content: The probability of a certain event to happen is $$1$$. 
aoVal: B
content: The probability of an impossible event to happen is $$0$$. 
aoVal: C
content: The probability of an indefinite event to happen is between $$0$$ and $$1$$. 
aoVal: D
content: Indefinite events include impossible events. 
Answer: D
Analysis: Impossible events are definite events.  So $$\text{D}$$ is wrong.  $$\text{A}$$, $$\text{B}$$, and $$\text{C}$$ are right.  So the answer is $$\text{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Abby, Bret, Carl, and Dana are seated in a row of four seats numbered \#$$1$$ to \#$$4$$. Joe looks at them and says:  ``Bret is next to Carl."  "Abby is between Bret and Carl."  However, each one of Joe's statements is false. Bret is actually sitting in seat \#$$3$$. Who is sitting in seat \#$$2$$? ． 
Options:
aoVal: A
content: Abby 
aoVal: B
content: Bret 
aoVal: C
content: Carl 
aoVal: D
content: Dana 
Answer: D
Analysis: We know that Carl does not sit next to Bret, so he must sit in seat \#$$1$$. Since Abby is not between Bret and Carl, she must sit in seat \#$$4$$. Finally, Dana has to take the last seat available, which is \#$$2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Bruno Mars is packing $3$ shirts for a trip. He just randomly grabs $3$ shirts from his closet without observing them. The closet contains $10$ shirts: $5$ striped, $3$ plaid, and $2$ solid-colored ones. What is the probability that he chooses $2$ striped shirts and $1$ solid-colored shirt? 
Options:
aoVal: A
content: $\frac12$ 
aoVal: B
content: $\frac13$ 
aoVal: C
content: $\frac1{12}$ 
aoVal: D
content: $\dfrac{1}{6}$ 
Answer: D
Analysis: $\dfrac{\_5C\_2 \times~~\_2C\_1}{\_{10}C\_3}=\dfrac{10\times2}{120}=\dfrac{2}{12}=\dfrac{1}{6}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Eddie is ordering lunch at a fast food restaurant with the following menu.  Mains: Sandwich, Burger, Pizza  Sides: Chicken Wings, Salad  Drinks: Coffee, Tea, Coke  If Eddie chooses to buy one main, one side, and one drink, how many different ways can he order lunch? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$18$$ 
Answer: D
Analysis: $$3\times2\times 3=18$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Problem: A conductor wanted to make a trio consisting of a violinist, a pianist, and a drummer. He had to choose one of two violinists, one of two pianists, and one of two drummers. How many possible combinations did he have? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$25$$ 
Answer: C
Analysis: $$2\times2\times2=8$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: The room numbers of a hotel are all three-digit numbers. The first digit represents the floor and the last two digits represent the room number. The hotel has rooms on five floors, numbered $$1$$ to $$5$$. It has $$35$$ rooms on each floor, numbered $$n01$$ to $$n35$$ where $$n$$ is the number of the floor. In numbering all the rooms, how many times will the digit $$2$$ be used? 
Options:
aoVal: A
content: $$60$$ 
aoVal: B
content: $$65$$ 
aoVal: C
content: $$95$$ 
aoVal: D
content: $$100$$ 
aoVal: E
content: $$105$$ 
Answer: E
Analysis: Each floor has $$35$$ rooms. On every floor except floor $$2$$, the digit $$2$$ will be used for rooms \textquotesingle$$n02$$\textquotesingle, \textquotesingle$$n12$$\textquotesingle, \textquotesingle$$n20$$\textquotesingle{} to \textquotesingle$$n29$$\textquotesingle{} (including \textquotesingle$$n22$$\textquotesingle) and \textquotesingle$$n32$$\textquotesingle. Hence the digit $$2$$ will be used $$14$$ times on each floor except floor $$2$$. On floor $$2$$, the digit $$2$$ will be used an extra $$35$$ times as the first digit of the room number. Therefore the total number of times the digit $$2$$ will be used is $$5\times14+35 = 105$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Problem: What is the angle between the hour hand and the minute hand at seven o\textquotesingle clock?~ ~ ．(Only consider angles less than 180°) 
Options:
aoVal: A
content: $50^{}\circ $ 
aoVal: B
content: $120^{}\circ $ 
aoVal: C
content: $135^{}\circ $ 
aoVal: D
content: $150^{}\circ $ 
aoVal: E
content: $165^{}\circ $ 
Answer: D
Analysis: The smaller angle is $\frac 5{12}$ of a full circle.  A full circle has $360$ degrees, so the angle is $\frac 5{12}\times 360^{}\circ =150^{}\circ $.  So, the answer is $\rm D$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Reading the Clock
Problem: How many numbers are there between $$0 \sim 50$$ that do not consist of digit \textquotesingle$$6$$\textquotesingle?． 
Options:
aoVal: A
content: $$42$$ 
aoVal: B
content: $$43$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$46$$ 
aoVal: E
content: $$50$$ 
Answer: D
Analysis: $$6, 16, 26, 36, 46$$  $$51$$ numbers minus $$5$$ numbers -\/-\textgreater{} $$45$$ numbers 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration