math-eval/TAL-SCQ5K huggingface dataset:
Problem: Which of the following numbers has the smallest prime factor? (2003 AMC 8 Problem, Question \#2) 
Options:
aoVal: A
content: $$55$$ 
aoVal: B
content: $$57$$ 
aoVal: C
content: $$58$$ 
aoVal: D
content: $$59$$ 
aoVal: E
content: $$61$$ 
Answer: C
Analysis: The smallest prime factor is $2$, and $58$ is the only multiple of $2$ among these five numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: Given that $$6$$ and $$9$$ are multiples of $$3$$, which of the followings might not be a multiple of $$3$$? 
Options:
aoVal: A
content: $$9 + 6$$ 
aoVal: B
content: $$9 - 6$$ 
aoVal: C
content: $$2 \times 9$$ 
aoVal: D
content: $$2 \times 9 + 1$$ 
Answer: D
Analysis: $$\text{A}$$: corresponding to $$\text{a}+\text{b}$$; $$\text{B}$$: corresponding to $$\text{a}-\text{b}$$; $$\text{C}$$: $$\text{na}$$, where $$\text{n}$$\emph{\emph{~}}is an integer. We choose $$\text{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: How many square numbers and cube numbers are there in the list below?  $18$, $27$, $64$, $72$, $81$, $162$, $196$, $324$, $343$, $496$, $529$, $675$, $729$, $784$, $841$, $951$, $1000$ 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$12$$ 
Answer: C
Analysis: Square numbers: $64$, $81$, $196$, $324$, $529$, $729$, $784$, $841$  Cube numbers: $27$, $64$, $343$, $729$, $1000$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers
Source: 其它
Problem: Which of the following is a composite number? 
Options:
aoVal: A
content: $$59$$ 
aoVal: B
content: $$61$$ 
aoVal: C
content: $$63$$ 
aoVal: D
content: $$67$$ 
Answer: C
Analysis: $63$ is a composite number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: How many numbers of the following are divisible by $9$?  $$\textasciitilde$$  $452 \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} 387\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1057\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~108\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~496\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1233$  $\textasciitilde$  $\textasciitilde$ 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: B
Analysis: $387, 108$, and $1233$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: What is the product of the least common multiple of $$6$$ and $$18$$ and the greatest common factor of $$6$$ and $$18$$? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$54$$ 
aoVal: D
content: $$108$$ 
Answer: D
Analysis: The least common multiple of $$6$$ and $$18$$ is $$18$$. The greatest common factor of $$6$$ and $$18$$ is $$6$$. Finally, $$6\times18 = 108$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: Which of the following numbers are prime numbers?  $137$~ ~ ~ ~ ~ ~ ~ ~ ~$151$~ ~ ~ ~ ~ ~ ~ ~ ~$237$~ ~ ~ ~ ~ ~ ~ ~ ~$301$ 
Options:
aoVal: A
content: $137$ and $237$ 
aoVal: B
content: $137$ and $301$ 
aoVal: C
content: $237$ and $301$ 
aoVal: D
content: $137$ and $151$ 
Answer: D
Analysis: $3$ is the factor of $237$ because $3 \times79 = 237$; $7$ is the factor of $301$ because $7 \times 43 = 301$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Problem: The expression $$1\times 2\times 3\times \cdots \times n$$ has exactly $$128$$ consecutive zeros at the end of its result, then the maximum value of $$n$$ is~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$405$$ 
aoVal: B
content: $$109$$ 
aoVal: C
content: $$500$$ 
aoVal: D
content: $$524$$ 
aoVal: E
content: $$539$$ 
Answer: D
Analysis: There are $$128$$ zeros at the end of the result, and the estimated answer is close to $$500$$.  $$\left[ \frac{500}{5} \right]+\left[ \frac{500}{25} \right]+\left[ \frac{500 }{125} \right]=100+20+4=124$$,  Just add $$4$$ numbers including $$5$$: $$505$$, $$510$$, $$515$$, $$520$$, so the maximum value of $$n$$ that satisfies the condition is $$524$$ . 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: Think Lab buys $24$ experimental apparatus. But two of the digits of the total price on the invoice are not clear, which only show $\overline{\square 8\square 2}$ dollars. The purchasing agent says the two unclear numbers on the invoice are the same. Given that the single price of this kind of apparatus is an integer, how much is it? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$328$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: C
Analysis: The number of total price can be divided by $3$ and $8$. It can be divided by $8$ which means the tens digit must be $3$ or $7$. And when the tens digit is $7$, the number can be divided by $3$.  Thus, the total price is $7872$ dollars, and each of them is $328$ dollars. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
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 $$8$$.  ($$2010$$ AMC $$8$$ Problem, Question \#$$22$$) 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: I think of a number.  When I divide it by $$2$$, the remainder is $$1$$.  When I divide it by $$3$$, the remainder is $$2$$.  What is the smallest possible value of the number?~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
Answer: A
Analysis: $$2\times3-1=5$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Source: 其它
Problem: 375+753+537+357+573+735= ． 
Options:
aoVal: A
content: $$3330$$ 
aoVal: B
content: $$1000$$ 
aoVal: C
content: $$3333$$ 
aoVal: D
content: $$1333$$ 
Answer: A
Analysis: $$3+5+7=15$$，  $$15+15=30$$，  $$30+300+3000=3330$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: Jam has some pieces of candy. He wants to share with some kids. If he shares the candy among $8$ kids equally, there will be $2$ pieces left. If he shares the candy among $9$ kids equally, there will be $3$ pieces left. If he shares the candy among $10$ kids equally, there will be $4$ pieces left. How many pieces of candy are there? 
Options:
aoVal: A
content: $321$ 
aoVal: B
content: $354$ 
aoVal: C
content: $720$ 
aoVal: D
content: $360$ 
aoVal: E
content: $240$ 
Answer: B
Analysis: The LCF of $8, 9,$ and $10$ is $8\times9\times10\div2=360$.  $360-6=354$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Source: 其它
Problem: Which of the following number pairs do not contain any divisible relationship? 
Options:
aoVal: A
content: $(12+78), 2$ 
aoVal: B
content: $(39+61), 3$ 
aoVal: C
content: $(44+82), 4$ 
aoVal: D
content: $(25+5100), 5$ 
Answer: C
Analysis: If both numbers in the parentheses are multiple of the smaller number, the number in the parentheses should be divisible by the smaller number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Becky is separating 53 cherries into bag A, bag B and bag C. It is known that there are even number of cherries in both bag A and bag B. Could you tell whether the number of cherries in bag C is even or odd number? 
Options:
aoVal: A
content: Odd 
aoVal: B
content: Even 
Answer: A
Analysis: Number of cherries in bag C: 53 (odd number) - even - even  Number of odd character = 1  Therefore, the number of cherries in bag C is odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: Think Lab buys $24$ experimental apparatus. But two of the digits of the total price on the invoice are not clear, which only show $\overline{\square 8\square 2}$ dollars. The purchasing agent says the two unclear digits on the invoice are the same. Given that the single price of this kind of apparatus is an integer, how much is it? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$328$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: C
Analysis: The number of total price can be divided by $3$ and $8$. It can be divided by $8$ which means the tens digit must be $3$ or $7$. And when the tens digit is $7$, the number can be divided by $3$.  Thus, the total price is $7872$ dollars, and each of them is $328$ dollars. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: The product of $$4$$ odd numbers is always． 
Options:
aoVal: A
content: even 
aoVal: B
content: odd 
aoVal: C
content: less than $$20$$ 
aoVal: D
content: greater than $$20$$ 
Answer: B
Analysis: The product of odd numbers is always odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Problem: There are three whole numbers $A$, $B$ and $C$ ($B\neq1$). $A\times B=21$, $B\times C=57$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$23$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$29$$ 
Answer: D
Analysis: $21=3\times 7$  $57=3\times 19$  Because $B$ is the factor both number contains, $B=3$  Thus, $A=7$, $C=19$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: Micky and Marcy want to make dumplings. They start with different speeds. There are two kinds of boxes. The smaller one of them could contain $9$ dumplings, and the bigger one could contain $17$ dumplings. Micky works with the smaller one, and Marcy works with the bigger one. What is the smallest number of dumplings that Micky needs to make until they start with a new box respectively at the same time? 
Options:
aoVal: A
content: $$143$$ 
aoVal: B
content: $$150$$ 
aoVal: C
content: $$153$$ 
aoVal: D
content: $$163$$ 
aoVal: E
content: $$173$$ 
Answer: C
Analysis: The least common multiple of $9$ and $17$: $9 \times 17 = 153$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Common Multiples and Least Common Multiples->Least Common Multiple of Two Numbers
Source: 其它
Problem: The $5$-digit number $\overline{2018U}$ is divisible by $9$. What is the remainder when this number is divided by $8$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: B
Analysis: We use the property that the digits of a number must sum to a multiple of $9$ if it is divisible by $9$. This means $2+0+1+8+U$ must be divisible by $9$. The only possible value for U then must be $7$. Since we are looking for the remainder when divided by $8$, we can ignore the thousands. The remainder when $187$ is divided by $8$ is $(\rm B)$ $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: $$\sqrt{41^{2} - 9^{2}} + \sqrt{8^{2} + 15^{2}}=$$． 
Options:
aoVal: A
content: $$56$$ 
aoVal: B
content: $$57$$ 
aoVal: C
content: $$58$$ 
aoVal: D
content: $$58$$ 
Answer: B
Analysis: $$\sqrt{41^{2} - 9^{2}} + \sqrt{8^{2} + 15^{2}}=\sqrt{1600} + \sqrt{289} = 40 + 17 = 57$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: How many prime numbers are less than $$10$$? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: C
Analysis: The prime numbers less than $$10$$ are $$2$$, $$3$$, $$5$$, and $$7$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)
Problem: A two-digit prime number is still prime when the digits of its first and tenth digits are exchanged. There are  such prime numbers. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: D
Analysis: $$11$$，$$13$$，$$17$$，$$31$$，$$37$$，$$71$$，$$73$$，$$79$$，$$97$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: In a Fibonacci-like sequence $$1,3,4,7,11,18\cdots $$(where each term is the sum of the two previous terms, starting from the third term), what is the remainder when the $$5555^{}\text{th}$$ term is divided by $$5$$？ 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: E
Analysis: Nil 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Problem: The $5$-digit number $\overline{2018U}$ is divisible by $9$. What is the remainder when this number is divided by $8$? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: A
Analysis: We use the property that the digits of a number must sum to a multiple of $9$ if it is divisible by $9$. This means $2+0+1+8+U$ must be divisible by $9$. The only possible value for U then must be $7$. Since we are looking for the remainder when divided by $8$, we can ignore the thousands. The remainder when $187$ is divided by $8$ is $(\rm B)$ $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: If the product of an even number and an odd number is $$840$$, what is the largest possible value of this odd number? 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$105$$ 
aoVal: D
content: $$420$$ 
Answer: C
Analysis: If the product of an even and an odd number is $$840={2^{3}}\times105$$, then the largest possible value of the odd number is $$105$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Problem: Modify a digit in $$675479$$ so that this six-digit number is divisible by $$25$$. What is the modified six-digit number? 
Options:
aoVal: A
content: $$675480$$ 
aoVal: B
content: $$675475$$ 
aoVal: C
content: $$675470$$ 
aoVal: D
content: $$625479$$ 
Answer: B
Analysis: change the number on ones digit from $$9$$to$$5$$，$$675475$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders
Problem: Which of the following is a factor of $$380$$?　 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$10$$ 
Answer: D
Analysis: Since $$380 =10\times38$$, $$10$$ is a factor of $$380$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: $$18$$ apples are equally given to $$9$$ kids. How many apples can every kid get? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: $$12$$ 
Answer: D
Analysis: $$18\div9=2$$  $$18=2+2+2+2+2+2+2+2+2$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders
Problem: Which of the following numbers is not prime? 
Options:
aoVal: A
content: $$13 $$ 
aoVal: B
content: $$19 $$ 
aoVal: C
content: $$89$$ 
aoVal: D
content: $$93$$ 
Answer: D
Analysis: $$93=31\times 3$$ , its factors are $1$，$31$，$3$，$93$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: If $n$ and $m$ are integers and $n^{2}+m^{2}$ is even, which of the following is impossible? (2014 AMC 8 Problem, Question \#13) 
Options:
aoVal: A
content: $n$ and $m$ are even 
aoVal: B
content: $n$ and $m$ are odd 
aoVal: C
content: $n+m$ is even 
aoVal: D
content: $n+m$ is odd 
aoVal: E
content: none of these are impossible 
Answer: D
Analysis: The question asks which one is impossible, all we need to do is find one possible way that the others are possible. After trying, when $n$ and $m$ are both even or odd, the calculation works, so $D$ is not correct. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Which of the following is not a factor of $48$? 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$16$$ 
Answer: C
Analysis: $14$ can not divide $48$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: Find the value for lcm $$\left[ 12,18\right]$$. 
Options:
aoVal: A
content: $$36$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$72$$ 
aoVal: D
content: $$48$$ 
Answer: A
Analysis: List the prime factorization for $$12$$ and $$18$$ first. $$12=2^{2}\times3$$ and $$18=3^{2}\times2$$. The largest exponent for $$2$$ is $$2$$, and the larger exponent for $$3$$ is $$2$$, thus the least common multiple for $$12$$ and $$18$$ is $$2^{2}\times3^{2}=36$$. We choose $$\text{A}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: If $$a=1$$, $$b=2$$, and $$c=3$$, then $$\overline{abc}=123$$. Find the value of $$m$$ according to this rule.  $$\overline{m21}=8\times \overline{m9}+3m$$. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$9$$ 
Answer: A
Analysis: $$\begin{eqnarray}\overline{m21}\&=\&8\times \overline{m9}+3m\textbackslash\textbackslash{} 100m+21\&=\&80m+72+3m\textbackslash\textbackslash{} 17m\&=\&51\textbackslash\textbackslash{} m\&=\&3.\end{eqnarray}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: Answer the question below:  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: A kind of water plant grows so fast that it doubles every day. If one plant is put into the pond on the first day, it will turn to two plants on the second day, and on the $26$\textsuperscript{th} days, they can fill the pond. If $8$ water plants are put into the pond on the first day, how many days will it take to fill the pond? 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$25$$ 
Answer: C
Analysis: It takes three days for one plant to turn to $8$ plants. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: A number has the same digit in its tens place and its hundredth place. How many times greater is the value of the digit in the tens place than the value of the digit in the hundredth place? 
Options:
aoVal: A
content: $0.1$ 
aoVal: B
content: $100$ 
aoVal: C
content: $1000$ 
aoVal: D
content: $10,000$ 
Answer: C
Analysis: Let the digit be $d$  $10d\div 0.01d=1000$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases
Source: 其它
Problem: If an equilateral triangle has in-teger sides, its perimeter \emph{cannot} be． 
Options:
aoVal: A
content: $$915$$ 
aoVal: B
content: $$615$$ 
aoVal: C
content: $$315$$ 
aoVal: D
content: $$115$$ 
Answer: D
Analysis: The perimeter of an equilateral $$\triangle $$ with integer sides is divisible by $$3$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: How many different primes are in the prime factorisation of $$2016$$? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
Answer: A
Analysis: $$2016=2\times2\times2\times2\times2\times3\times3\times7$$; there are $$3$$ different primes. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)
Problem: If add $$1994$$ 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: prime　 
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