math-eval/TAL-SCQ5K huggingface dataset:
Problem: If a four-digit number $$5A2A$$ can be divisible by $20$, the digit that $$A$$ represents is  . 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$8$$ 
Answer: A
Analysis: We consider the last two-digit to see if a number is divisible by $4$. We consider the last one digit to see if a number is divisible by $5$. If $$\overline{\textasciitilde5A2A\textasciitilde}$$ is divisible by $5$, the ones digit can only be $5$ or $0$. When the ones digit is $0$, the number that is formed by the last two digits, $20$, is divisible by $4$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: If $$1994$$ is added to any odd number, the sum will always be． 
Options:
aoVal: A
content: odd　 
aoVal: B
content: even　 
aoVal: C
content: $$1995$$ 
aoVal: D
content: $$0$$ 
Answer: A
Analysis: even number $$+$$ odd number $$=$$ odd 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 whole numbers are greater than $$9$$ and less than $$60$$?　 
Options:
aoVal: A
content: $$49$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$51$$ 
aoVal: D
content: $$59$$ 
Answer: B
Analysis: There are $$60$$ whole numbers from $$0$$ to $$59$$. That\textquotesingle s $$50$$ without $$0$$ to $$9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: If $$a$$, $$b$$ are prime numbers, and $$3a+7b=41$$, then $$a+b=$$． 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: B
Analysis: Based on the laws relating to the parity in addition and multiplication, either $$a$$ or $$b$$ must be $$2$$. If $$a = 2$$, then $$b = 5$$, and $$a + b = 7$$; if $$b = 2$$, then $$a = 9$$; $$9$$ is not a prime number, which doesn\textquotesingle t match the conditions in the question. Therefore, we choose B. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers
Problem: The hundreds digit of a three-digit number is $$2$$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$8$$ 
Answer: E
Analysis: $$\rm Method$$ $$1$$:  Let the hundreds, tens, and units digits of the original three-digit number be $$a$$, $$b$$, and $$c$$, respectively. We are given that $$a=c+2$$. The original three-digit number is equal to $$100a+10b+c=100(c+2)+10b+c=101c+10b+200$$. The hundreds, tens, and units digits of the reversed three-digit number are $$c$$, $$b$$, and $$a$$, respectively. This number is equal to $$100c+10b+a=100c+10b+(c+2)=101c+10b+2$$. Subtracting this expression from the expression for the original number, we get $$(101c+10b+200)-(101c+10b+2)=198$$ . Thus, the units digit in the final result is $$8$$.  $$\rm Method$$ $$2$$:  The result must hold for any three-digit number with its hundreds digit being $$2$$ more than the units digit. $$301$$ is such a number. Evaluating, we get $$301-103=198$$. Thus, the units digit in the final result is $$8$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: Cindy prepares burgers with two slices of beef each. A box of beef has $20$ slices. How many burgers can she prepare with all the three and a half box of beef? (Adapted from 2013 Math Kangaroo Problem, Level 3-4, Question \#7) 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$25$$ 
aoVal: E
content: $$30$$ 
Answer: B
Analysis: There are $3\times20+20\div2=70$ slices of beef. $70\div2=35$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders
Source: 其它
Problem: $$\overline{**45}$$, $$\overline{19*8}$$, $$\overline{23*1}$$, and $$\overline{3*49}$$ are four $4-$digit numbers with some unknown digits. Which number is possible to be a perfect square? 
Options:
aoVal: A
content: $$\overline{**45}$$ 
aoVal: B
content: $$\overline{19*8}$$ 
aoVal: C
content: $$\overline{23*1}$$ 
aoVal: D
content: $$\overline{3*49}$$ 
aoVal: E
content: None 
Answer: D
Analysis: According to the ones digit of perfect squares, we can eliminate choice $B$.  A perfect square ending with $5$ must be a multiple of $25$, so we can eliminate choice $A$.  $${{48}^{2}}=2304$$ and $${{49}^{2}}=2401$$ so we can eliminate choice $C$.  ${57}^{2}=3249$. We choose option $D$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers
Source: 其它
Problem: Today is $$1^{st}$$ January $$2023$$, which is a Sunday. Teacher Angel has $$5$$ candies in her bag. Every time she completed her Tuesday and Thursday class, she will award herself with $$2$$ candies. On which date would she have $$15$$ candies in her bag? 
Options:
aoVal: A
content: $$15^{th}$$ January $$2023$$ 
aoVal: B
content: $$16^{th}$$ January $$2023$$ 
aoVal: C
content: $$17^{th}$$ January $$2023$$ 
aoVal: D
content: $$18^{th}$$ January $$2023$$ 
aoVal: E
content: $$19^{th}$$ January $$2023$$ 
Answer: C
Analysis: $$16$$ days later. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Source: 其它
Problem: $$N$$ is a two$$-$$digit number. When $$N$$ is divided by $$5$$, the remainder is $$1$$. When $$N$$ is divided by $$11$$, the remainder is $$1$$. The smallest possible value of $N$ is~\uline{~~~~~~~~~~}~.(Adapted from $$2016$$ AMC $$8$$ Problem, Question \#$$5$$) 
Options:
aoVal: A
content: $$51$$ 
aoVal: B
content: $$54$$ 
aoVal: C
content: $$56$$ 
aoVal: D
content: $$61$$ 
aoVal: E
content: $$67$$ 
Answer: C
Analysis: Among $1, 56, 111$, \ldots~the smallest possible $$N$$ that satisfies the two conditions is $$56$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: Let $$N$$ be the greatest five$$-$$digit number whose digits have a product of $$120$$. What is the sum of the digits of $$N$$? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$20$$ 
Answer: D
Analysis: To make the largest possible five$$-$$digit $$N$$, you should make the number in the biggest digit as large as possible. $$120=8\times5\times3\times1\times1$$. So, the sum of digits is $$8+5+3+1+1=18$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization
Problem: The first $2018$ integers ($1$, $2$, $3$, $\cdots$, $2017$, $2018$) are written on the blackboard. What is the minimum number of integers that should be erased from the blackboard, so that the last digit of the product of the remaining integers is $2$? 
Options:
aoVal: A
content: $$402$$ 
aoVal: B
content: $$403$$ 
aoVal: C
content: $$404$$ 
aoVal: D
content: $$410$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: First, we need to remove all the integers which are multiples of $5$, otherwise the last digit of the products is $0$ or $5$.  Hence, $403$ integers need to be removed.  Next, note that the last digit of each of the products below is $6$.  $1\times2\times3\times4\times6\times7\times8\times9$,  $11\times12\times13\times14\times16\times17\times18\times19$,  $\cdots\cdots$  $2001\times2002\times2003\times2004\times2006\times2007\times2008\times2009$,  and the last digit of the product $2011\times2012\times2013\times2014\times2016\times2017\times2018$ is $4$.  Hence, we need to remove one more "$2$"and the last digit of the product will be $2$.  So the answer is $404$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders
Problem: Summer has some nuts and wants to divide them equally to $5$ kids. Everyone can get $7$ nuts at most. How many nuts does Summer have at most? 
Options:
aoVal: A
content: $$34$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$39$$ 
aoVal: E
content: $$40$$ 
Answer: D
Analysis: There are $5-1=4$ nuts left after dividing at most. Thus, Summer has $5\times7+4=39$ nuts at most. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Problem: The ones digit of $$9\times 8\times 7\times 6\times 5\times 4\times 3\times 3\times 4\times 5\times 6\times 7\times 8\times 9$$ is． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$9$$ 
Answer: A
Analysis: Since $$5\times 4 = 20$$, the ones digit of the given product must be $$0$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $$\overline{1A2}$$. What is the missing digit $A$ of this $3$-digit number? (2014 AMC 8 Problem, Question \#8) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: D
Analysis: $1+2-A=0$, $A=3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: How many 0s are there at the end of the product $$2\times3\times5\times7\times8\times12\times25$$. ． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: D
Analysis: 3 set of 2$\times$5. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Problem: What is the sum of the ten-thousands\textquotesingle{} digit and the millions\textquotesingle{} digit of $$1234567890$$? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
Answer: C
Analysis: The ten-thousands\textquotesingle digit plus the millions\textquotesingle~digit of $$1234567890$$ is $$6 + 4 = 10$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: Which of the following numbers is the odd one out? 
Options:
aoVal: A
content: $$1138$$ 
aoVal: B
content: $$1226$$ 
aoVal: C
content: $$1324$$ 
aoVal: D
content: $$1416$$ 
aoVal: E
content: $$1854$$ 
Answer: E
Analysis: $$1138 →1\times1\times3\times8=24$$  $$1226 →1\times2\times2\times6=24$$  $$1324 →1\times3\times2\times4=24$$  $$1416~→1\times4\times1\times6=24$$  $$1854 →1\times8\times5\times4=160$$  $$1854$$ is the odd one out as the product of its digits is not $$24$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers
Problem: The greatest odd factor of the product $$1\times 2\times 3\times 4\times 5\times 6$$ is． 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$75$$ 
Answer: C
Analysis: $$1\times 2\times 3\times 4\times 5\times (2\times 3)=2\times 4\times 2\times (1\times 3\times 5\times 3)=16\times 45$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: The product of any two multiples of $$3$$ must be a multiple of． 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$15$$ 
Answer: B
Analysis: Since $$3\times3 =9$$, the product of two multiples of 3 is divisible by $$9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: There are three whole numbers $A$, $B$ and $C$. $A\times B=35$, $B\times C=84$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$23$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$29$$ 
Answer: C
Analysis: $35=5\times 7$  $84=2\times 2\times 3\times 7$  Because $B$ is the factor both number contains, $B=7$  Thus, $A=5$, $C=12$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: Determine whether the following calculations give odd or even numbers.  (a) $$14327+21462-3583$$  (b) $$9377\times1525$$ 
Options:
aoVal: A
content: Both Odd 
aoVal: B
content: Both Even 
aoVal: C
content: Odd, Even 
aoVal: D
content: Even, Odd 
Answer: D
Analysis: Nil 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: When $$106$$ is divided by $$3$$, the remainder is． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: B
Analysis: $$ 3\times35 = 105$$; so $$106 \div 3 =35$$ with remainder $$106-105=1$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders
Problem: Find the value for lcm $$\left[ 4,6,8\right]$$. 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$48$$ 
Answer: C
Analysis: List the prime factorization for $$4$$, $$6$$ and $$8$$ first. $$4=2^{2}$$, $$6=2\times3$$, $$8=2^{3}$$. The largest exponent for $$2$$ is $$3$$,~~and the largest exponent for $$3$$ is $$1$$, thus the least common multiple for $$4$$, $$6$$, and $$8$$ is $$2^{3}\times3=24$$. We choose $$\rm C$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: John, Mike, Emily and Nick want to buy some apples from Walmart where the apples are sold in pack of six. How many packs should they buy to get the same amount of apples for each one of them? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: C
Analysis: For each one of them to get the same amount of apples, the total amount should be a multiple of $4$, because there are $4$ of them.  The apples are sold in pack of six. $3$ packs contain $18$ apples; $5$ packs contain $30$; $6$ packs contain $36$; $7$ packs contain $42$.  Among the number of $18, 30, 36, 42$, only $36$ can be divided by $4$. $36$ is a multiple of $4$. Therefore, they should buy $6$ packs. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: There is a book with 650 pages. Henry tears 31 paper from the book, each paper contains two pages. Is it possible that the sum of their page number equals to 2000? 
Options:
aoVal: A
content: $$Yes.$$ 
aoVal: B
content: $$No.$$ 
Answer: B
Analysis: If there are odd number of odd page number, the sum of page numbers is odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: $$1$$ thousand $$+9$$ hundreds $$+ 8$$ tens $$+ 18$$ ones $$=$$． 
Options:
aoVal: A
content: $$1918$$ 
aoVal: B
content: $$1988$$ 
aoVal: C
content: $$1998$$ 
aoVal: D
content: $$19818$$ 
Answer: C
Analysis: $$1000+900+80+18=1980+18=1998$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: The product of all $$4$$ sides of a square is $$1296$$. The sum of all $$4$$ sides of the square is． 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$36$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$72$$ 
Answer: A
Analysis: Let $x$ represent the length of each side of the square, then:  $x^{4} = 1296 \implies x^{2} = \sqrt{1296} = 36$.  Hence $x = \sqrt{36} = 6$ and $4x = 4\times 6 = 24$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Change a digit of the number $45879$ to make the new five-digit number be divisible by $125$. What is the new five-digit number? 
Options:
aoVal: A
content: $$45870$$ 
aoVal: B
content: $$45875$$ 
aoVal: C
content: $$45579$$ 
aoVal: D
content: $45875$ and $45870$ 
Answer: B
Analysis: The last three digits must be divisible by $125$, so it can only be $875$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: In this fictional "Old Island", all the numbers contain only odd digits. The order of the counting numbers is as follows:  1, 3, 5, 7, $\cdots $ , 19, 31, 33, $\cdots $  What is the 31st counting number in the island? 
Options:
aoVal: A
content: $$101$$ 
aoVal: B
content: $$111$$ 
aoVal: C
content: $$99$$ 
aoVal: D
content: $$113$$ 
aoVal: E
content: None of the above. 
Answer: B
Analysis: 1,3,5,7,9 -\/-\/-\/-\/-\/-\/-\/-\/-\/-⑤  11,13,15,17,19-\/-\/-\/-\/-\/-\/-⑤  31,33,35,37,39-\/-\/-\/-\/-⑤  51,53,55,57,59-\/-\/-\/-\/-⑤  71,73,75,77,79-\/-\/-\/-\/-⑤  91,93,95,97,99-\/-\/-\/-⑤  $$111$$  the 31st number is 111. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: Divide 5 numbers $$2$$、$$3$$、$$12$$、$$15$$ and $$30$$ into two groups to make the product of numbers in each group the same, so the two groups are ． 
Options:
aoVal: A
content: （$$2$$, $$3$$, $$15$$），（$$12$$, $$30$$） 
aoVal: B
content: （$$2$$, $$12$$, $$15$$），（$$3$$, $$30$$） 
aoVal: C
content: （$$2$$, $$3$$, $$30$$），（$$12$$, $$15$$） 
aoVal: D
content: （$$12$$, $$3$$, $$15$$），（$$2$$, $$30$$） 
Answer: C
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->Finding Factors Given the Product
Problem: How many prime numbers are there between $120$ and $140$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: A
Analysis: $127$, $131$, $137$, $139$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: Which of the following is not a prime number? 
Options:
aoVal: A
content: $$41$$ 
aoVal: B
content: $$51$$ 
aoVal: C
content: $$61$$ 
aoVal: D
content: $$71$$ 
Answer: B
Analysis: Since $$51=3\times17$$, $$51$$ is not a prime number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: Mom was cooking at home with the lights on when suddenly the power went out. Dad came home and pressed the switch four times, and then Mason came home and pressed the switch three times. When the power came back on, were the lights off or on? 
Options:
aoVal: A
content: On 
aoVal: B
content: Off 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Problem: Mike has $$17$$ apples and $$19$$ pears. He puts every $5$ fruits in each box. At least how many more fruits does he need to get $8$ boxes of fruits? (Adapted from 2001 Math Kangaroo Problem, Level 3-4, Question \#7) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: D
Analysis: $(17+19)\div5=7R1$  $5-1=4$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: A square has an area of $144\text{cm}^{2}$, what is the length of each side of the square? 
Options:
aoVal: A
content: $10\text{cm}$ 
aoVal: B
content: $11\text{cm}$ 
aoVal: C
content: $12\text{cm}$ 
aoVal: D
content: $13\text{cm}$ 
Answer: C
Analysis: The length of each side of the square is $\sqrt{144} = 12\text{cm}$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Vicky bought $26$ apples to divide to her $5$ cousins. If Vicky wants to give them all of the apples and also wants to make every cousin gets the same amount, at least how many more apples should Vicky buy? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$4$$ 
Answer: E
Analysis: $6\times5-26=4$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
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: If a natural number can be written as the sum of both two and three consecutive natural numbers, then we can call it a Think Number. What is the largest Think Number no larger than $5789$? 
Options:
aoVal: A
content: $$5786$$ 
aoVal: B
content: $$5787$$ 
aoVal: C
content: $$5788$$ 
aoVal: D
content: $$5789$$ 
aoVal: E
content: $$5784$$ 
Answer: B
Analysis: The number can be written as $$n+(n+1)=2n+1(n\geqslant 1)$$ and $$x+(x+1)+(x+2)=3x+3$$.  It must be a multiple of $3$ but leaves a remainder of $1$ when divided by $2$ but leaves a remainder of $1$.  $5790$ can be divisible by both $2$ and $3$, so it is $5790-3=5787$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: The sum of three $2-$digit consecutive numbers is the maximum 2-digit number. What is their product? 
Options:
aoVal: A
content: $$99$$ 
aoVal: B
content: $$25900$$ 
aoVal: C
content: $$35904$$ 
aoVal: D
content: $$34589$$ 
aoVal: E
content: $$39804$$ 
Answer: C
Analysis: $99\div3=33$  $32+33+34=99$  $32\times33\times34=35904$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: A student thinks of a natural number. She divides the number by $$9$$ and the remainder is $$7$$. What is the remainder when double that number is divided by $$9$$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: C
Analysis: The remainder of $$A$$ $$\div9$$ is $$7$$, and $$2A=A+A$$. Therefore the remainder of $$2A\div9$$ is $$7+7=14$$. $$14= 9+5$$, therefore the remainder is $$5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: Salah has collected more than $$20$$ football cards.  When he puts his cards in piles of four, he has three cards left over.  When he puts the cards in piles of five, he has four cards left over.  Which of the following could be the number of cards he has in total? 
Options:
aoVal: A
content: $$27$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$35$$ 
aoVal: D
content: $$39$$ 
aoVal: E
content: $$43$$ 
Answer: D
Analysis: The number of Salah\textquotesingle s cards is $$3$$ more than a multiple of $$4$$, and $$4$$ more than a multiple of $$5$$. Of the two-digit options given, the only one that satisfies both criteria is $$39(=9\times4+3 =7\times5+4)$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem