math-eval/TAL-SCQ5K huggingface dataset:
Problem: If the sum of two prime numbers is $$39$$, the difference between these two prime numbers will be ． 
Options:
aoVal: A
content: $$29$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$33$$ 
aoVal: D
content: $$35$$ 
aoVal: E
content: $$37$$ 
Answer: D
Analysis: Because $$39$$ is an odd number, one of the two numbers must be an even prime number $$2$$. The other number is $$39-2=37$$, and their difference is $$35$$.  So the answer is $$\text{B}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: What is the sum of the distinct prime integer divisors of $2016 ?$ (2016 AMC 8 Problem, Question \#9) 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$49$$ 
aoVal: E
content: $$63$$ 
Answer: B
Analysis: The prime factorization is $2016=2^{5} \times 3^{2} \times 7$. Since the problem is only asking us for the distinct prime factors, we have $2,3,7$. Their desired sum is then (B) 12 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Which pair of numbers has a common factor greater than $$1$$? 
Options:
aoVal: A
content: $$4$$ and $$9$$ 
aoVal: B
content: $$6$$ and $$27$$ 
aoVal: C
content: $$27$$ and $$50$$ 
aoVal: D
content: $$33$$ and $$100$$ 
Answer: B
Analysis: The greatest common factors for the choices are $$1$$, $$3$$, $$1$$, $$1$$, respectively. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: Suppose it is now the month of February. What month will it be $$90$$ calendar months from now? 
Options:
aoVal: A
content: April 
aoVal: B
content: May 
aoVal: C
content: June 
aoVal: D
content: July 
aoVal: E
content: August 
Answer: E
Analysis: $$90\div12=7R6$$  $$6$$ months afterFebruary will be August 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Source: 其它
Problem: The greatest odd factor of $$30$$ is .　 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$21$$ 
Answer: C
Analysis: The smallest even factor is $$2$$; $$30\div2= 15$$, the greatest odd factor. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: There are two reading rooms at Think Town\textquotesingle s Library: Tulip\textquotesingle s and Lily\textquotesingle s. There are two lamps on every table in Tulip\textquotesingle s Reading Room. As for Lily\textquotesingle s Reading Room, there are three lamps on every table. Nini knows that the total number of lamps in the two reading rooms is an odd number, and the total number of tables in the two reading rooms is also an odd number. Which reading room has an odd number of tables? 
Options:
aoVal: A
content: Tulip\textquotesingle s Reading Room 
aoVal: B
content: Lily\textquotesingle s Reading Room 
aoVal: C
content: Both rooms 
aoVal: D
content: None of the rooms 
Answer: B
Analysis: There are two lamps on every table in Tulip\textquotesingle s Reading Room, so the number of lamps in it is an even number. The total number of lamps in the two rooms is odd, and thus the number of lamps in Lily\textquotesingle s Reading Room is an odd number. Since there are three lamps on every table in Lily\textquotesingle s Reading Room, the number of tables in it is also an odd number. However, the total number of tables in the two rooms is odd, so the number of tables in Tulip\textquotesingle s is even. Therefore, the answer is $$\text{B}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Odd and Even Applications
Problem: Any number that is divisible by both $$12$$ and $$5$$ must also be divisible by． 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$8$$ 
Answer: B
Analysis: A number divisible by $$12$$ \& $$5$$ is divisible by a product of their factors, such as $$3\times5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: Linda wrote down a natural number. When she divided the number by $$7$$, the remainder was $$5$$. What is the remainder when twice that number is divided by $$7$$? ． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: C
Analysis: The remainder of $$A$$ $$\div7$$ is $$5$$, and $$2A=A+A$$. Therefore the remainder of $$2A\div7$$ is $$5+5=10$$. $$10= 7+3$$, therefore the remainder is $$3$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: Which of the following cases can\textquotesingle t make an even number? 
Options:
aoVal: A
content: An odd number + An even number 
aoVal: B
content: An even number + An even number 
aoVal: C
content: An odd number + An odd number 
Answer: A
Analysis: The sum of an odd number and an even number is always an odd number. The sum of two numbers which are both even or both odd is always an even number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers
Problem: How many numbers listed below are positive?  $$23$$, $$32$$, $$-56$$, $$-98$$, $$7$$ 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: D
Analysis: $$23$$, $$32$$, and $$7$$ are positive; $$-56$$, $$-98$$ are negative. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: In the division expression $$28\div$$~\uline{~~~~~~~~~~}~$$=$$~\uline{~~~~~~~~~~}~$$\text{R}4$$, how many different combinations are there for the quotient and the divisor? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: D
Analysis: We can use the equation: divisor $$\times$$ quotient $$=$$ dividend $$-$$ remainder, so here we can get divisor $$\times$$ quotient $$=28-4=24$$. Therefore, the only possibilities are $$1$$ and $$24$$, $$2$$ and $$12$$, $$3$$ and $$8$$, and $$4$$ and $$6$$ for a total of four possible combinations. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Problem: Pip swam $$3$$ back and forth in the lane for a total of $$156$$ metres, how long is the lane in this pool? 
Options:
aoVal: A
content: $$52$$ 
aoVal: B
content: $$104$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$234$$ 
Answer: C
Analysis: one back and forth $$156\div3=52$$metres  length of the lane$$52\div2=26$$metres 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Short Division
Source: 其它
Problem: The sum of three different positive even numbers could be． 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$19$$ 
Answer: C
Analysis: The sum must be even, and it could be $$2+4+8=14$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers
Problem: Della has a box of ping pong balls. Each time she counts the balls $8$ by $8$, $10$ by $10$, or $12$ by $12$, there are always $3$ balls left. How many ping pong balls are there in the box at least? 
Options:
aoVal: A
content: $$60$$ 
aoVal: B
content: $$120$$ 
aoVal: C
content: $$123$$ 
aoVal: D
content: $$240$$ 
aoVal: E
content: $$243$$ 
Answer: C
Analysis: If the ping pong balls in the box are reduced by $3$, then there will be no extra balls when you count them $8$ by $8$, $10$ by $10$, or $12$ by $12$. It means that the number of the ping pong balls is a common multiple of $8$, $10$, and $12$ after being reduced by $3$. If you want to know how many ping pong balls there are at least, you can first find the least common multiple of $8$, $10$ and $12$, and then add $3$ to get the answer.  $$\begin{array}{l} {2\left\textbar{} \underline{\textasciitilde8\textasciitilde\textasciitilde10\textasciitilde\textasciitilde12}\right. }\textbackslash\textbackslash{\textasciitilde2\left\textbar{} \underline{4\textasciitilde\textasciitilde\textasciitilde5\textasciitilde\textasciitilde\textasciitilde6}\right. }\textbackslash\textbackslash{\textasciitilde\textasciitilde\textasciitilde\textasciitilde\textasciitilde2\textasciitilde\textasciitilde\textasciitilde5\textasciitilde\textasciitilde\textasciitilde3} \end{array}$$  $2\times2\times2\times5\times3=120$  $120+3=123$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Problem: There are some flowers along the corridor, arranged in the following order: $3$ red flowers, $2$ yellow flowers, $2$ pink flowers$\cdots$  If there are $100$ flowers altogether, how many red flowers are there altogether? ． 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$42$$ 
aoVal: D
content: $$44$$ 
Answer: D
Analysis: $$3+2+2=7$$;  $100\div7=14R2$;  $$14$$$\times$$$3$$+$$2$$=$$44$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Problem: There are three whole numbers $A$, $B$ and $C$ ($B\neq1$). $A\times B=55$, $B\times C=100$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$34$$ 
aoVal: D
content: $$36$$ 
Answer: D
Analysis: $55=5\times 11$  $100=2\times 2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=11$, $C=20$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: In how many ways can we represent the number $$2003$$ as a sum of two prime numbers? ($$2003$$ Math Kangaroo Problems, Level $$7-8$$, Question \#$$15$$) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: Such a representation is impossible. 
Answer: E
Analysis: If so, $2$ must be one of the prime numbers, and the other must be $2003-2=2001$, which is not prime. Thus, such a representation is impossible. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers->Knowing Prime and Composite Numbers
Problem: How many of the numbers $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, and $$9$$ can be written as a sum of exactly two of the other numbers? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
Answer: C
Analysis: I wrote $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, and $$9$$ on my page. Adding $$1$$ to $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, and $$8$$ results in $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, and $$9$$. We cannot get a sum of $$1$$ or $$2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: Which of the following is an even number? 
Options:
aoVal: A
content: $$321$$ 
aoVal: B
content: $$489$$ 
aoVal: C
content: $$644$$ 
aoVal: D
content: $$25$$ 
Answer: C
Analysis: even numbers end with 0,2,4,6,8 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: A three-digit number can be written as $\overline{6\square4}$~and this three-digit number is divisble by $4$. How many different digits can we fill in this square? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: E
Analysis: $0, 2, 4, 6,$ and $8$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: Jason has some baseball cards from the $$1920$$s. If he divides the number of cards he has by $$3$$, the remainder is $1$; if he divides the number of cards he has by $$5$$, the remainder is $3$; if he divides the number of cards he has by $$7$$, the remainder is $5$. Suppose Jason has at least $k$ cards, the sum of digits of $k$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$12$$ 
Answer: C
Analysis: The number of cards after adding $$2$$ is divisible by $$3$$, $$5$$, and $$7$$. Since the least common multiple of $$3$$, $$5$$, and $$7$$ is $$3\times5\times7=105$$, Jason has $$105-2=103$$ cards at least. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: If $$1\times2\times3\times4\times5\times6\times7\times8\times$$$$9\times10\times11\times12\times13\times$$$$14\times15 = 1307 674 368000$$,  how many times does the digit "$$0$$" appear in the product $$10\times20\times30\times40\times50\times60\times$$$$70\times80\times90\times100\times110\times120\times130\times140\times150$$? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$19$$ 
aoVal: D
content: $$20$$ 
Answer: C
Analysis: If $$1\times2\times3\times4\times5\times6\times7\times$$$$8\times9\times10\times11\times12\times13\times14\times15 = $$$$1307 674 368000$$, and we multiply each of these $$15$$ numbers by $$10$$, the new product will have an additional $$15$$ zeroes, and $$15+4=19$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization
Problem: Every positive integer is congruent modulo $9$ to the sum of its decimal digits.  Now, let $S(n)$ equal the sum of the digits of positive integer $n$. For example, $S(1507)=13$. For a particular positive integer $n, S(n)=1274$. Which of the following could be the value of $S(n+1)$? (Adapted From 2017 AMC 12A Problems, Question \#18) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$1239$$ 
aoVal: E
content: $$1265$$ 
Answer: D
Analysis: Note that $n \equiv S(n) \bmod 9$, so $S(n+1)-S(n) \equiv n+1-n=1 \bmod 9$. So, since $S(n)=1274 \equiv 5 \bmod 9 $, we have that $S(n+1) \equiv 6 \bmod 9$. Then, only one of the answer choices is congruent to $6 \bmod 9$, which is $(D)=1239$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Congruence
Source: 其它
Problem: How many different ways are there to form a $3-$digit number without repeating digits, using $1, 2, 3, 4,$ and $0$? 
Options:
aoVal: A
content: $$36$$ 
aoVal: B
content: $$48$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$125$$ 
Answer: B
Analysis: $4\times 4\times3=48$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Source: 其它
Problem: Arthur writes down three two-digit integers. One is square, one is prime and one is triangular.  He uses the digits $$1$$, $$2$$, $$3$$, $$4$$, $$5$$ and $$6$$ exactly once each.  Which largest prime does he write? ． 
Options:
aoVal: A
content: $$13 $$ 
aoVal: B
content: $$23 $$ 
aoVal: C
content: $$31 $$ 
aoVal: D
content: $$41 $$ 
aoVal: E
content: $$43$$ 
Answer: D
Analysis: First note that Arthur can write down three squares, namely $$16$$, $$25$$ and $$36$$. Also, he can write down four triangular numbers, namely $$15$$, $$21$$, $$36$$ and $$45$$. If he chooses $$16$$ and $$45$$ for the square and triangular number respectively, then the remaining digits are $$2$$ and $$3$$, the prime is $$23$$. If he chooses $$25$$ and $$36$$ then the remaining digits are $$1$$ and $$4$$, the prime is $$41$$. If he chooses $$36$$ for the square number, the remaining difits can be a prime. So the largest prime he write is $$41$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers
Problem: Each of the following products is an even number except． 
Options:
aoVal: A
content: $$11\times99$$ 
aoVal: B
content: $$44\times33$$ 
aoVal: C
content: $$55\times22$$ 
aoVal: D
content: $$88\times66$$ 
Answer: A
Analysis: The product is even $$except$$ when both numbers you multiply are odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Problem: Of the following, which has an odd quotient when divided by $$2$$? 
Options:
aoVal: A
content: $$456456456456456$$ 
aoVal: B
content: $$678 678678678678$$ 
aoVal: C
content: $$432432432432432$$ 
aoVal: D
content: $$876876876 876876$$ 
Answer: B
Analysis: To check for an odd quotient, divide only the last two digits by $$2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Problem: Three different numbers are chosen from the numbers 3, 5, 6 and 8.  They are then added together.  Which of these statements is/are correct?  \textbf{1} The total cannot be a multiple of 8.  \textbf{2} The total can be a multiple of 3.  \textbf{3} The total is always odd. 
Options:
aoVal: A
content: none of them 
aoVal: B
content: statement 1 only 
aoVal: C
content: statement 2 only 
aoVal: D
content: statement 3 only 
aoVal: E
content: statements 2 and 3 only 
Answer: A
Analysis: 3+5+8=16，$$3+5+6=14$$，$$5+6+8=19$$，$$16$$ is a multiple of 8, (1) wrong;  No multiples of 3, (2) wrong, 14,16 is even, 19 is odd. (3) Wrong 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: What is the angle between the hour hand and the minute hand at seven o\textquotesingle clock? ． 
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: Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris\textquotesingle s. What is the expected number of balls that occupy their original positions after these two successive transpositions? 
Options:
aoVal: A
content: $$1.6$$ 
aoVal: B
content: $$1.8$$ 
aoVal: C
content: $$2.0$$ 
aoVal: D
content: $$2.2$$ 
aoVal: E
content: $$2.4$$ 
Answer: D
Analysis: After the first swap, we do casework on the next swap. Case 1: Silva swaps the two balls that were just swapped There is only one way for Silva to do this, and it leaves 5 balls occupying their original position. Case 2: Silva swaps one ball that has just been swapped with one that hasn\textquotesingle t swapped There are two ways for Silva to do this, and it leaves 2 balls occupying their original positions. Case 3 : Silva swaps two balls that have not been swapped There are two ways for Silva to do this, and it leaves 1 ball occupying their original positions. Our answer is the average of all 5 possible swaps, so we get $$ \frac{5+2 \cdot 2+2 \cdot 1}{5}=\frac{11}{5}=(\text { D) } 2.2 $$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: A robot is facing south-east.  It makes 58 quarter-turns clockwise, then 93 quarter-turns anti-clockwise.  In which direction is the robot now facing? 
Options:
aoVal: A
content: north 
aoVal: B
content: north-east 
aoVal: C
content: north-west 
aoVal: D
content: south-east 
aoVal: E
content: south-west 
Answer: E
Analysis: 58 quarter-turn clockwise, 93 quarter-turn anti-clockwise, is the same thing as 35 and 1/4 quarter-turn counterclockwise, 35*1/4=8 turns more than 3/4, so it is facing southwest. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Source: 其它
Problem: $$2006$$ students participated in a survey. The survey stated that $$1500$$ of them participated in the Math Kangaroo contest, and $$1200$$ of them participated in an English Language contest. Out of the students who participated in the survey, how many participated in both contests if it is known that $$6$$ people did not take part in either of the competitions? ($$2006$$ Math Kangaroo Problems, Level $$7-8$$, Question \#$$5$$) 
Options:
aoVal: A
content: $$300$$ 
aoVal: B
content: $$500$$ 
aoVal: C
content: $$600$$ 
aoVal: D
content: $$700$$ 
aoVal: E
content: $$1000$$ 
Answer: D
Analysis: $(1200+1500)-(2006-6)=700$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: In how many ways can the letters in $BEEKBBPERPP$ be rearranged so that two or more $E$s do not appear together? 
Options:
aoVal: A
content: $$49200$$ 
aoVal: B
content: $$94080$$ 
aoVal: C
content: $$564480$$ 
aoVal: D
content: $$1800$$ 
aoVal: E
content: $$98400$$ 
Answer: B
Analysis: There are $3$ $E$s in total now with other $8$ letters remaining. But pay attention to $B$ and $P$: there are $3$ $B$s and $3$ $P$ here.  There are $\_8P\_5 \div \_3P\_3$ ways for us to arrange the $8$ letters\textquotesingle{} positions. Then, we can put the $3$ $E$s in the $9$ intervals.  So the answer is $\_8P\_5 \div \_3P\_3 \times \_9C\_3=94080$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: I have $$11$$ pieces of candy in each of three baskets. From each basket I take out one piece of candy in the following order: from the left, from the middle, from the right, from the middle, from the left, from the middle, from the right, and so on. What is the largest number of pieces of candy left in one of the baskets when the middle basket is empty? (2001 Math Kangaroo Problem, Level 3-4, Question \#23) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$11$$ 
Answer: D
Analysis: Start from either the left or the right and follow the steps until the basket in the middle has $$0$$ candy. And you can see that one side has $$5$$ pieces and the other side has $$6$$ pieces. So the largest number of pieces of candy is $$6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Operational Problem
Problem: A positive integer divisor of $12 !$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$?~ (2020 AMC 10A Problem, Question \#15) 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$23$$ 
Answer: E
Analysis: The prime factorization of $12 !$ is $2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7 \cdot 11$. This yields a total of $11 \cdot 6 \cdot 3 \cdot 2 \cdot 2$ divisors of $12 !$. In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization.  Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in $12 !$. Thus, there are $6 \cdot 3 \cdot 2$ perfect squares. (For $2$ , you can choose $0,2,4,6,8,$ or $10$, etc.  The probability that the divisor chosen is a perfect square is $$ \frac{6 \cdot 3 \cdot 2}{11 \cdot 6 \cdot 3 \cdot 2 \cdot 2}=\frac{1}{22} \Longrightarrow \frac{m}{n}=\frac{1}{22} \Longrightarrow m+n=1+22=\text { (E) } 23 $$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它