math-eval/TAL-SCQ5K huggingface dataset:
Problem: Abe has $1$ green and $1$ red jelly beans in his hand. Bob has $1$ green and $2$ yellow jelly beans in his hand. Each randomly picks a jelly bean to show to the other. What is the probability that the colours match? ． 
Options:
aoVal: A
content: $\dfrac{1}{2}$ 
aoVal: B
content: $\dfrac{1}{3}$ 
aoVal: C
content: $\dfrac{3}{4}$ 
aoVal: D
content: $\dfrac{1}{6}$ 
Answer: D
Analysis: The probability that both show a green bean is $\dfrac{1}{2}\cdot \dfrac{1}{3}=\dfrac{1}{6}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: If the sum of $$9$$ consecutive odd integers is $$1935$$, what is the sum of the next $$9$$ consecutive odd integers? 
Options:
aoVal: A
content: $$2015$$ 
aoVal: B
content: $$2017$$ 
aoVal: C
content: $$2097$$ 
aoVal: D
content: $$2099$$ 
Answer: C
Analysis: Each number in the second sum is $$18$$ greater than the corresponding number in the first sum. Thus the second sum is $$1935 +18 \times9 = 2097$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: A box contains $11$ cards, numbered from $1$ to $11$. One card is selected randomly from the box. What is the probability that the number on the selected card is greater than $7$? (adapted from 2017 AMC 8 Problem, Question \#10) 
Options:
aoVal: A
content: $\frac1{11}$ 
aoVal: B
content: $\frac4{11}$ 
aoVal: C
content: $\frac7{11}$ 
aoVal: D
content: $\frac{10}{11}$ 
aoVal: E
content: $\frac2{11}$ 
Answer: B
Analysis: There are $4$ numbers greater than $7$. Thus, the probability is $\frac4{11}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Three kids line up to play games. In how many different ways can they form the line? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: A
Analysis: $3\times 2\times 1=6$  There are six different ways for three kids to line up. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: A city has a bicycle hire scheme where it is possible to hire a bicycle for short journeys.  Last year I hired a bicycle $$60$$ times and rode for $$13$$ hours altogether.  For how long on average did I hire the bicycle on each ride? 
Options:
aoVal: A
content: $$13$$ minutes 
aoVal: B
content: $$23$$ minutes 
aoVal: C
content: $$39$$ minutes 
aoVal: D
content: $$47$$ minutes 
aoVal: E
content: $$73$$ minutes 
Answer: A
Analysis: The rider travels for $$13$$ hours over $$60$$ rides, which is anaverage time of $$\frac{13}{60}$$ of an hour per ride, hence $$13$$ minutes. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Using Formulas
Problem: A $$3$$-digit integer is called a \textquotesingle V-number\textquotesingle{} if the digits go \textquotesingle high-low-high\textquotesingle{} $$-$$ that is, if the tens digit is smaller than both the hundreds digit and the units (or \textquotesingle ones\textquotesingle) digit.  How many $$3$$-digit \textquotesingle V-numbers\textquotesingle{} are there? 
Options:
aoVal: A
content: $$120$$ 
aoVal: B
content: $$240$$ 
aoVal: C
content: $$285$$ 
aoVal: D
content: $$320$$ 
aoVal: E
content: $$400$$ 
Answer: C
Analysis: The smallest \textquotesingle V-number\textquotesingle{} is $$101$$ and the largest \textquotesingle V-mumber\textquotesingle{} is $$989$$.  Consider the tens digits. The smallest tens digit is $$0$$ and the largest tens digit is $$8$$.  If the tens digit is $$0$$, the hundreds digit can be $$1$$ to $$9$$, and the units digit can be $$1$$ to $$9$$, giving $$9 \times 9$$ possible \textquotesingle V-numbers\textquotesingle.  If the tens digit is $$1$$, then the hundreds digit can be $$2$$ to $$9$$ and the units digit can be $$2$$ to $$9$$, giving $$8 \times 8$$ possible \textquotesingle V-numbers\textquotesingle.  If the tens digit is $$d$$, where $$d$$ can be any digit from $$0$$ to $$8$$, the hundreds digit can be $$(d + 1)$$ to $$9$$ and the units digit can be $$(d + 1)$$ to $$9$$, giving $$(9-d)\times (9-d)$$ possible \textquotesingle V-numbers\textquotesingle.  The greatest value of $$d$$ is $$8$$. In this case, the hundreds digit can only be $$9$$ and the units digit can only be $$9$$, which gives just $$1 \times 1$$ possibilities.  This gives the total number of possible \textquotesingle V-numbers\textquotesingle{} to be $$9\times9+8\times8+\cdots +1\times1 = 285$$, which is the sum of the squares from $$1$$ to $$9$$ inclusive. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers
Problem: On $2021$ December $31$\textsuperscript{st}, Lucas, Jeremy, and Irene visited their grandpa together. Then Lucas visited him every $4$ days, Jeremy visited him every $5$ days, and Irene visited him every $6$ days. In the first three months of $2022$, how many days could the grandpa be visited by at least one person? 
Options:
aoVal: A
content: $$42$$ 
aoVal: B
content: $$52$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$46$$ 
aoVal: E
content: $$56$$ 
Answer: A
Analysis: In $31+28+31=90$ days, they could visit the grandpa in $[90\div4]+[90\div5]+[90\div6]-[90\div20]-[90\div12]-[90\div30]+[90\div60]=42$ days. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle
Source: 其它
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->Questions Involving Enumeration->Enumeration
Problem: Find the number of positive integers from $100$ to $300$ which is divisible by $6$, $8$ and $10$. 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$55$$ 
aoVal: C
content: $$65$$ 
aoVal: D
content: $$75$$ 
aoVal: E
content: None of the above 
Answer: E
Analysis: E 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle
Source: 其它
Problem: Jack, Sarah, and Jimmy participated in a maths competition.  Jack says: "I won the competition."  Sarah says: "I didn\textquotesingle t win the competition."  Jimmy says: "Jack didn\textquotesingle t win the competition."  Only one of them told the truth. Who won the maths competition? ． 
Options:
aoVal: A
content: $$$$Jack 
aoVal: B
content: Sarah 
aoVal: C
content: Jimmy 
Answer: B
Analysis: We can spot that Jack\textquotesingle s statement and Jimmy\textquotesingle s statement contradict each other, so one of them is telling the truth. Therefore, Sarah tells a lie. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning using Hypothesis
Problem: Manjari\textquotesingle s average score on six tests is $$82$$. Her average score on the $$7^{}\text{th}$$ and $$8^{}\text{th}$$ tests is $$98$$. What is her average score on all eight tests? ． 
Options:
aoVal: A
content: $$86$$ 
aoVal: B
content: $$88$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$94$$ 
Answer: A
Analysis: Total score on the first six tests is $$6\times82=492$$. Her total score on all eight tests is $$492+2\times98=688$$, and her average score is $$688\div8=86$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)
Problem: If three couples stand in a row for taking a photo, how many ways are there for only one couple to stand next to each other? 
Options:
aoVal: A
content: $$36$$ 
aoVal: B
content: $$72$$ 
aoVal: C
content: $$144$$ 
aoVal: D
content: $$240$$ 
aoVal: E
content: $$288$$ 
Answer: E
Analysis: $$\_{3}C\_1\times \_{2}C\_1\times \_{3}A\_3\times \_{2}C\_1=72$$,  $$\_{3}C\_1\times \_{2}C\_1\times \_{3}A\_3\times \_{2}C\_1\times \_{3}C\_1=216$$,  $$72+216=288$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: There is a ball in a box and three kids are guessing what colour it is.  \textbf{Val says: "The ball is white."}  \textbf{John says: "The ball is blue."}  \textbf{Elvis says: "I agree with Val."}  They open the box and find only one of them guessed right.  What colour is the ball? 
Options:
aoVal: A
content: white 
aoVal: B
content: blue 
aoVal: C
content: Uncertain 
Answer: B
Analysis: We can spot that Val\textquotesingle s guess and Elvis\textquotesingle{} guess are the same, so both of them must be wrong. Therefore, John guessed it right. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Ming, Fanny and other $$6$$ classmates sit at a row. Ming and Fanny cannot sit at $$2$$ ends at the same time. How many sitting arrangements are there? 
Options:
aoVal: A
content: $$30000$$ 
aoVal: B
content: $$32800$$ 
aoVal: C
content: $$34800$$ 
aoVal: D
content: $$38800$$ 
aoVal: E
content: $$40000$$ 
Answer: D
Analysis: $$(6\times7+2\times6)\times6!=54\times720=38800$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Problem: Professor Chang has ten different language books lined up on a bookshelf: three Arabic, three German, and four Spanish. How many ways are there to arrange the ten books on the shelf keeping the Arabic books together and keeping the Spanish books together? (Adapted from $2018$ AMC $8$ Problem, Question \#$16$) 
Options:
aoVal: A
content: $$1440$$ 
aoVal: B
content: $$2880$$ 
aoVal: C
content: $$5760$$ 
aoVal: D
content: $$17280$$ 
aoVal: E
content: $$34560$$ 
Answer: D
Analysis: Since the three Arabic books and four Spanish books have to be kept together, respectively, we can treat them both as just one book. That means we\textquotesingle re trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $\_5P\_5$. Now we multiply this product by $\_3P\_3\times \_4P\_4$~because there are $\_3P\_3$~ways to arrange just three Arabic books, and $\_4P\_4$~ways to arrange just four Spanish books. Multiplying all these together, we have the answer $D$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: Marcie is $4$ years old, and Sara is $6$ years old this year. $8$ years later, what is the sum of their ages? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$26$$ 
Answer: E
Analysis: $4+6+8+8=26$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Problem: Annie has four cards of different colors. She writes letter $B$ on the red card and blue card, and writes letter $O$ on the yellow card and green card. Now Annie puts the four cards in a box. Bob is going to draw three of them from the box randomly. How many different possible results can Bob get? Among them, how many can Bob get his name? 
Options:
aoVal: A
content: $6$; $3$ 
aoVal: B
content: $4$; $3$ 
aoVal: C
content: $6$; $2$ 
aoVal: D
content: $4$; $2$ 
aoVal: E
content: $6$; $1$ 
Answer: D
Analysis: There are $4$ ways to choose three from four. Two $B$s with one $O$ can form Bob\textquotesingle s name. Thus, $2$ of the $4$ ways can get Bob\textquotesingle s name. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: Among the natural numbers from $1$ to $600$, how many numbers are multiples of $3$ or $5$? 
Options:
aoVal: A
content: $$260$$ 
aoVal: B
content: $$280$$ 
aoVal: C
content: $$300$$ 
aoVal: D
content: $$320$$ 
Answer: B
Analysis: There are $200$ multiples of $3$ and there are $120$ multiples of of $5$. There are also $40$ multiples of $15$. By the Inclusion Exclusion Principle, there are $200+120-40=280$ numbers which are multiples of $3$ or $5$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: Eddie is ordering lunch at a fast food restaurant that has sandwiches and burgers on the lunch menu, along with coffee, milk, and tea as drink options. If Eddie chooses one food item and one drink item from the lunch menu, he hasdifferent ways to order lunch. 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: C
Analysis: $$2\times 3=6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Problem: There are $$17$$ balls in a bag. Each ball has a number from $$1$$ to $$17$$ on it. We randomly pick a ball from the bag. What is the smallest number of balls we have to pick in order to be sure that we have at least one pair of balls with a sum equal to $$18$$? ($$2005$$ Math Kangaroo Problem, Level $$9-10$$, Question $$ \textbackslash\# $$$$15$$ 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$17$$ 
Answer: C
Analysis: Among these numbers, there are $$8$$ pairs of numbers can get the sum of $$18$$($$1+17=2+16=3+15=4+14=5+13=6+12=7+11=$$$$8+10$$), and $$9$$ is useless. So in the worst case, after we choose $$9$$, we need $$8+1=9$$ more numbers to make sure a pair appears.  Thus, the answer is $$1+8+1=10$$.  Copyrighted material used with permission from Math Kangaroo in USA, NFP Inc. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Worst Case in Pigeonhole Principle Problems
Problem: $$14.81+19.28$$~\uline{~~~~~~~~~~}~$$19.82+14.21$$ 
Options:
aoVal: A
content: $$\textgreater$$ 
aoVal: B
content: $$\textless$$ 
aoVal: C
content: $$=$$ 
Answer: C
Analysis: Isolate the decimals and whole number parts 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Problem: A bus was supposed to arrive at Think Station at $$5:30$$ PM, but it arrived $45$ minutes earlier by changing another route because of a traffic accident. When did the bus reach Think Station? 
Options:
aoVal: A
content: $5:00$ PM 
aoVal: B
content: $4:30$ PM 
aoVal: C
content: $4:00$ PM 
aoVal: D
content: $5:30$ PM 
aoVal: E
content: $4:45$ PM 
Answer: E
Analysis: $5:30$-$45$ minutes=$4:45$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Source: 其它
Problem: Miruna had to multiply two $$2$$-digit numbers together, but she accidentally reversed the digits of both of them before multiplying and reached the answer $$209$$. Which of the following answers should she have obtained? 
Options:
aoVal: A
content: $$1001$$ 
aoVal: B
content: $$1003$$ 
aoVal: C
content: $$1005$$ 
aoVal: D
content: $$1007$$ 
aoVal: E
content: $$1009$$ 
Answer: A
Analysis: The prime factors of $$209$$ are $$11$$ and $$19$$, so these must have been the reversed numbers that Miruna multiplied. The correct multiplication was $$11 \times 91=1001$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: My average score on $$8$$ math tests is $$90$$. If my average score on the first $$5$$ tests was $$87$$, what was my average score on the last $$3$$ tests? 
Options:
aoVal: A
content: $$96$$ 
aoVal: B
content: $$95$$ 
aoVal: C
content: $$94$$ 
aoVal: D
content: $$93$$ 
Answer: B
Analysis: I scored a total of $$720$$ on all $$8$$ tests. The total of $$435$$ on the first $$5$$ tests leaves a total of $$285$$ for the last $$3$$ tests, so the average is $$285\div3 = 95$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Tiffany wants to pack $$9$$ shirts into several bags. There are at least $$2$$ shirts in each bag, and the number of shirts in each bag should be different. There are~\uline{~~~~~~~~~~}~ways to pack the shirts. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: B
Analysis: $$9=2+7=3+6=4+5$$  $$9=2+3+4$$  So there are $$3+1=4$$ ways 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: A fair $6$-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number? (2011 AMC 8 Problems, Question \#18) 
Options:
aoVal: A
content: $\dfrac{1}{6}$ 
aoVal: B
content: $\dfrac{5}{12}$ 
aoVal: C
content: $\dfrac{1}{2}$ 
aoVal: D
content: $\dfrac{7}{12}$ 
aoVal: E
content: $\dfrac{5}{6}$ 
Answer: D
Analysis: There are $6\cdot 6=36$ ways to roll the two dice, and $6$ of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are $\dfrac{30}{2}=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is $\dfrac{15+6}{36}=\dfrac{21}{36}=\boxed {\left (\text{D}\right )\dfrac{7}{12}}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Source: 其它
Problem: Sophia's average score on six tests is $$82$$. Her average scores on the $$7^{}\text{th}$$ and $$8^{}\text{th}$$ tests is $$98$$. What is her average score on all eight tests? ． 
Options:
aoVal: A
content: $$86$$ 
aoVal: B
content: $$88$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$94$$ 
Answer: A
Analysis: Sophia\textquotesingle s total score on the first six tests is $$6\times82=492$$. Her total score on all eight tests is $$492+2\times98=688$$, and her average score is $$688\div8=86$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Eddie finished reading a story book last week. He read an average of $19$ pages per day in the first six days of the week and $26$ pages in the last day. How many pages on average did Eddie read per day? 
Options:
aoVal: A
content: $$26$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: $(19\times6+26)\div7=140\div7=20$ pages. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)
Problem: Amy picks some number cards as shown below:  $13, 20, 14, 15, 19, 20, 20, 19, 19, 15, 19, 19, 20, 13, 15$.  What is the difference between their mode and median? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$19$$ 
aoVal: E
content: $$20$$ 
Answer: A
Analysis: She picks two $13$s, one $14$, three $15$s, five $19$s, and four $20$s.  The mode is $19.$  The median is $19.$  Thus, their difference should be $0.$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Maria arrived at the market at $$9:35$$. She spent $$1$$ hour $30$ minutes at the market, and it took her an hour to arrive home. When did she arrive home? 
Options:
aoVal: A
content: $11:05$ 
aoVal: B
content: $12:05$ 
aoVal: C
content: $11:45$ 
aoVal: D
content: $11:55$ 
aoVal: E
content: $12:00$ 
Answer: B
Analysis: $9:35+1:30+1=12:05$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Source: 其它
Problem: Granny has $$10$$ grandchildren. Alice is the oldest. One day, Granny notices that her grandchildren all have different ages. If the sum of her grandchildren\textquotesingle s ages is $$180$$, what is the youngest Alice can be? (2014 Math Kangaroo Problem, Level 5-6, Question \#30) 
Options:
aoVal: A
content: $$19$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$21$$ 
aoVal: D
content: $$22$$ 
aoVal: E
content: $$23$$ 
Answer: E
Analysis: Try to consider the problem of the youngest Alice in such way: make each grandchild has a similar age as possible, and the age of each grandchild should be different, that is, $$1 + 2 + 3 +\cdots 9 + 10 = 55$$; then $$180 - 55 = 125$$, $$125 \div 10 = 12R5$$, and $$5$$ is left. If every child\textquotesingle s age is added by $$12$$, then $$5$$ is left. Give the extra year to each of the $$5$$ oldest children. In doing so, the minimum age of Alice is $$10+12+1=23$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Calculate:  $$624\times14 =$$． 
Options:
aoVal: A
content: $$8866$$ 
aoVal: B
content: $$8976$$ 
aoVal: C
content: $$8736$$ 
aoVal: D
content: $$8636$$ 
Answer: C
Analysis: 624x14=624x(10+4)=624x10+624x4=6240+2496=8736 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Problem: Throw two dice of the same quality and size. The six sides of each die are marked with number of dots from $$1$$ to $$6$$, respectively. Among the following options,~\uline{~~~~~~~~~~}~is an impossible event. 
Options:
aoVal: A
content: The sum of dots is $$12$$. 
aoVal: B
content: The sum of dots is smaller than $$3$$. 
aoVal: C
content: The sum of dots is larger than $$4$$ but smaller than $$8$$. 
aoVal: D
content: The sum of dots is $$13$$. 
Answer: D
Analysis: The maximum sum is $$6+6=12$$, so "the sum of dots is $$13$$" is an impossible event.  So $$\text{D}$$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: There are  different ways for a librarian, lending six different books to three students, given that each student gets only a book. 
Options:
aoVal: A
content: $$120$$ 
aoVal: B
content: $$100$$ 
aoVal: C
content: $$96$$ 
aoVal: D
content: $$72$$ 
Answer: A
Analysis: The first student has six choices of books; the second has five; and the third has four. By the Rule of product, there is a total of $$6\times5\times4=120$$ways. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: What is the smallest possible sum of two positive integers whose product is $$240$$？ 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$34$$ 
aoVal: E
content: $$38$$ 
Answer: B
Analysis: Since the product of the two positive integers is $$240$$, the possible pairs of integers are $$\left( 1,240 \right)$$, $$\left( 2,120 \right)$$, $$\left( 3,80 \right)$$, $$\left( 4,60 \right)$$, $$\left( 5,48 \right)$$, $$\left( 6,40 \right)$$, $$\left( 8,30 \right)$$, $$\left( 10,24 \right)$$, $$\left( 12,20 \right)$$ and $$\left( 15,16 \right)$$. The respective sums of these pairs are $$241$$, $$122$$, $$83$$, $$64$$, $$53$$, $$46$$, $$38$$, $$34$$, $$32$$ and $$31$$. Of these, the smallest value is $$31$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations