math-eval/TAL-SCQ5K huggingface dataset:
Problem: The square root of $$49$$ is ． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
Answer: A
Analysis: The square root of $$49$$ is $$7$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Danni wants to use the numbers $1$, $3$, $7$, and $9$ to make prime numbers less than $100$. How many different prime numbers can she make? (She can use the same number more than once.)  $$\textasciitilde$$ 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$14$$ 
Answer: C
Analysis: Using these $4$ numbers, we can make $20$ different numbers which are less than $100$. However, only these $12$ numbers are prime numbers. $3; 7; 11; 13; 17; 19; 31; 37; 71; 73; 79; 97$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: The numberis divisible by $$3\times3$$. 
Options:
aoVal: A
content: $$663$$ 
aoVal: B
content: $$603$$ 
aoVal: C
content: $$336$$ 
aoVal: D
content: $$303$$ 
Answer: B
Analysis: Add the digits to test: for $$603$$, $$6+0+3 =9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: Ivan is puting 13 cakes into box A, box B and box C. It is known that there are odd number of cakes in both box A and box B. Could you tell whether the number of cakes in box C is even or odd number? 
Options:
aoVal: A
content: Odd 
aoVal: B
content: Even 
Answer: A
Analysis: Number of cakes in box C: 13 (odd number) - odd - odd  Number of odd character = 3  Therefore, the number of cakes in box C is odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: The largest whole-number multiple of $$7$$ less than $$200$$ is ． 
Options:
aoVal: A
content: $$187$$ 
aoVal: B
content: $$189$$ 
aoVal: C
content: $$196$$ 
aoVal: D
content: $$197$$ 
Answer: C
Analysis: $$200\div7\textgreater28$$, so the largest such multiple is $$28\times7 =196$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: The smallest number greater than $2$ that leaves a remainder of $2$ when divided by $3,4,5$, or $6$ lies between what numbers? (2012 AMC 8 Problem, Question \# 15) 
Options:
aoVal: A
content: $40$ and $50$ 
aoVal: B
content: $51$ and $55$ 
aoVal: C
content: $56$ and $60$ 
aoVal: D
content: $61$ and $65$ 
aoVal: E
content: $66$ and $99$ 
Answer: D
Analysis: To find the answer to this problem, we need to find the least common multiple of $3,4,5,6$ and add $2$ to the result. The least common multiple of the four numbers is $60$ , and by adding $2$ , we find that that such number is $62$ . Now we need to find the only given range that contains $62$ . The only such range is answer (D), and so our final answer is $(\text{D}) 61$ and $65$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: What is the sum of all factors of $$24$$? 
Options:
aoVal: A
content: $$52$$ 
aoVal: B
content: $$60$$ 
aoVal: C
content: $$72$$ 
aoVal: D
content: $$84$$ 
Answer: B
Analysis: The sum of the factors is $$(3^{0}+3^{1})$$$$\times (2^{0}+2^{1}+2^{2}+2^{3})=60$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number
Problem: What is the greatest number of consecutive integers such that the sum of the digits of none of them is divisible by $$5$$? ． 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
Answer: D
Analysis: Five consecutive numbers can be: $$12$$, $$13$$, $$14$$, $$15$$, and $$16$$ (without carrying) or $$17$$, $$18$$, $$19$$, $$20$$, $$21$$ (with carrying in the tens place). Without carrying, among each of the $$5$$ consecutive integers, we can find one whose sum of digits is a multiple of $$5$$. So we need to carry. To make the number of integers the greatest, we can start from a number whose remainder is $$1$$ when divided by five. And the most important thing is, after we write the fourth number, the carry appears. For example: $$56$$, $$57$$, $$58$$, $$59$$, $$60$$, $$61$$, $$62$$, $$63$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: The number $$1$$ million is less than the number． 
Options:
aoVal: A
content: $$1$$ thousand 
aoVal: B
content: $$10$$ thousand 
aoVal: C
content: $$100$$ thousand 
aoVal: D
content: $$1$$ billion 
Answer: D
Analysis: The number $$1$$ billion is $$1$$ thousand times $$1$$ million. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: $\overline{392AB}$ is a multiple of $45$, and $\overline{B34}$ is a three-digit even number. What is the sum of $A$ and $B$? 
Options:
aoVal: A
content: $$13$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $4$ or $13$ 
Answer: A
Analysis: $45=5\times9$  $\overline{392AB}$ should be ended with $0$ or $5.$ But $0$ cannot be the first digit. Thus, $B$ can only be $5$ and $A$ should be $8.$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: There are three whole number $A$, $B$, $C$. $A\times B=21$, $B\times C=60$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$33$$ 
aoVal: C
content: $$35$$ 
aoVal: D
content: $$40$$ 
Answer: A
Analysis: $21=3\times 7$  $60=2\times 2\times 3\times 5$  Because $B$ is the factor both number contains, $B=3$  Thus, $A=7$, $C=20$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: Two whole numbers have a least common multiple of $60$.  -Each number is less than or equal to $12$.  -The greatest common factor of the two number is $2$.  What are the two numbers? 
Options:
aoVal: A
content: $6$ and $10$ 
aoVal: B
content: $5$ and $12$ 
aoVal: C
content: $10$ and $12$ 
aoVal: D
content: $12$ and $16$ 
Answer: C
Analysis: A: The least common multiple of $6$ and $10$ is $30$.  B: Two numbers are relatively prime.  C: The least common multiple is $60$, and the greatest common factor is $2$.  D: The least common multiple is $48$, and the greatest common factor is $4$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: There are some books in the central library. If Adam divides the number of the books in the library by $32$, there will be $30$ books remained; if he divides the number of books in the library by $9$, there will be $7$ books remained; if he divides the number of books in the library by $7$, there will be $5$ books remained. How many books, at least, are there in the central library? 
Options:
aoVal: A
content: $$2014$$ 
aoVal: B
content: $$2015$$ 
aoVal: C
content: $$2016$$ 
aoVal: D
content: $$2017$$ 
Answer: A
Analysis: The total number of books when divided by $$32$$, $$9$$, and $$7$$ leaves remainders of $$30$$, $$7$$, and $$5$$, respectively. In other words, the number when added by $$2$$ is divisible by $$32$$, $$9$$, and $$7$$. Hence, the smallest such number is $$32\times9\times7-2=2014$$. The answer is $$\rm A$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Problem: Order all prime numbers from least to greatest. What is the sum of the next two prime numbers after $43$? 
Options:
aoVal: A
content: $$89$$ 
aoVal: B
content: $$90$$ 
aoVal: C
content: $$100$$ 
aoVal: D
content: $$101$$ 
Answer: C
Analysis: $47+53=100$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: Pip swam $$3$$ back and forth in the lane for a total of $$156$$ meters, 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 $$156\div3=52$$meters  length of the lane$$52\div2=26$$meters 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Short Division
Source: 其它
Problem: How many positive factors of $$360$$ are also multiples of $$12$$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: E
Analysis: $$360={{2}^{3}}\times {{3}^{2}}\times5$$, so the number of its factors would be $$\left( 3+1 \right)\times \left( 2+1 \right)\times\left( 1+1 \right)=24$$. Among them, there are $$2\times 2\times(1+1)=8$$ factors which have $${{2}^{2}}\times 3$$ as its factors. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: What is the ones digit of the product $$80\times 70\times 60\times 50\times 40\times 30\times 20\times 10\times 5\times 2$$? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
Answer: A
Analysis: Since $$10$$ is a factor of this product, the ones digit is $$0$$. 
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: A recurring decimal can also be written by putting a bar over the digits that repeat. Thus, $$0.\overline {123}$$ means $$0.123123\cdots $$. The sum of $$0.\overline {234}$$, $$0.\overline {342}$$, and $$0.\overline {432}$$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$1.01$$ 
aoVal: C
content: $$1\frac1{111}$$ 
aoVal: D
content: $$1.009$$ 
Answer: C
Analysis: $$0.\overline{234} +0.\overline{342} +0.\overline{432} =1.\overline{009} $$$$=1\frac 1{111}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: Fill in the blank:~\uline{~~~~~~~~~~}~is $$4$$ tens $$5$$ ones greater than $$2$$ tens $$7$$ ones. 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$28$$ 
aoVal: C
content: $$62$$ 
aoVal: D
content: $$72$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: $$4$$ tens $$5$$ ones: $$45$$  $$2$$ tens $$7$$ ones: $$27$$  Greater than: $$45 + 27 = 72$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: If I multiply three different prime numbers, the product must have  positive divisors. 
Options:
aoVal: A
content: $$ 3 $$ 
aoVal: B
content: $$5 $$ 
aoVal: C
content: $$6 $$ 
aoVal: D
content: $$8$$ 
Answer: D
Analysis: Multiply the three smallest primes: $$2\times3 \times5=30$$, whose $$8$$ divisors are $$1$$, $$2$$, $$3$$, $$5$$, $$6$$, $$10$$, $$15$$, and $$30$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: Students\textquotesingle~seat numbers are from $$1$$ to $$35$$. How many times does digit $$3$$ appear? (adapted from $$2002$$ Math Kangaroo Problem, Level $$3$$ - $4$, Question \#$9$) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$16$$ 
aoVal: E
content: $$32$$ 
Answer: C
Analysis: The digit $3$ appears in the ones place: $3$, $13$, $23$,  The digit $3$ appears in the tens place: $30, 31, 32, 33, 34, 35$  In total, $3 + 6 = 9$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value->Questions Involving Applying Place Value Principle
Problem: An unknown number is an odd number greater than $$50$$ but less than $$100$$. The number is a multiple of $$3$$ and $$7$$. What is the value of the unknown number? 
Options:
aoVal: A
content: $$63$$ 
aoVal: B
content: $$77$$ 
aoVal: C
content: $$84$$ 
aoVal: D
content: $$91$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: A multiple of $$3$$ and $$7$$ also a multiple of $$21$$  $$21\times3=63$$  $$21\times4=even$$  $$21\times5=105$$, exceed  Thus, only $$63$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Source: 其它
Problem: Among the numbers below, how many numbers can be divisible by $4$?  $234$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $789$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $7756$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $8865$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $3728$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $8064$ 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$3$$ 
Answer: D
Analysis: $7756, 3728,$ and $8064$ can be divisible by $4$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: Della has a box of ping-pong balls. No matter 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: $$1^{st}$$ November $$2016$$ is a Tuesday, what day will it be $$30^{th}$$ November of the same year? 
Options:
aoVal: A
content: Monday 
aoVal: B
content: Wednesday 
aoVal: C
content: Thursday 
aoVal: D
content: Saturday 
aoVal: E
content: None of the above 
Answer: B
Analysis: $$30-1+1=30$$  $$30\div7=4R2$$  Tuesday -\/-\textgreater~Wednesday. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Source: 其它
Problem: is a factor of $$123123123$$. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: B
Analysis: $$123123123$$ is divisible by $$3$$, since the sum of its digits is $18$ which is divisible by $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: How many different prime numbers, when multiplied by $$11$$, have an even number as their product? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: B
Analysis: The product of two odd numbers is odd, so there is only one even product, $$2\times 11$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers
Problem: Which of the following statement is correct? 
Options:
aoVal: A
content: All prime numbers are odd numbers. 
aoVal: B
content: In every $3$ consecutive numbers, there must be a composite number. 
aoVal: C
content: All even numbers are composite numbers. 
aoVal: D
content: The sum of two different odd number must be a composite number. 
Answer: D
Analysis: A: Any non-$3$ multiple of 3 are composite  B: $1$, $2$, $3$ does not have composite number  C: $2$ is prime number  D: $1+$any odd number is multiple of $2$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: Let $a$ and $b$ be positive integers such that $a+a b=1443$ and $a b+b=1444$. Find $10 a+b$. 
Options:
aoVal: A
content: $$407$$ 
aoVal: B
content: $$408$$ 
aoVal: C
content: $$418$$ 
aoVal: D
content: $$419$$ 
aoVal: E
content: $$428$$ 
Answer: B
Analysis: $$ \left\textbackslash{\begin{array}{l} a(b+1)=1443=3 \times 13 \times 37=37 \times 39 \textbackslash\textbackslash{} b(a+1)=1444=38^{2} \end{array}\right. $$  Compare and $a=37, b=38 .$ So $10 a+b=370+38=408$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: When Ringo places his marbles into bags with $6$ marbles per bag, he has $4$ marbles left over. When Paul does the same with his marbles, he has $3$ marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with $6$ marbles per bag. How many marbles will be leftover? (2012 AMC 10B Problems, Question \#4) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: A
Analysis: In total, there were $3+4=7$ marbles left from both Ringo and Paul.We know that $7 \equiv 1(\bmod 6)$. This means that there would be $1$ marble leftover. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Which of the following is divisible by all of the integers from $$1$$ to $$10$$ inclusive? 
Options:
aoVal: A
content: $$23\times34$$ 
aoVal: B
content: $$34\times45$$ 
aoVal: C
content: $$45\times56$$ 
aoVal: D
content: $$56\times67$$ 
aoVal: E
content: $$67\times78$$ 
Answer: C
Analysis: Of the options given, $$23\times 34$$, $$56\times 67$$ and $$67\times 78$$ are all not divisible by $$5$$, so may be discounted. Also $$34$$ is not divisible by $$4$$ and $$45$$ is odd, so $$34\times 45$$ may also be discounted as it is not divisible by $$4$$. The only other option is $$45\times 56$$. As a product of prime factors, $$45\times 56=2^{3}\times3^{2}\times5\times7$$, so it is clear that it is divisible by all of the integers from $$1$$ to $$10$$ inclusive. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization
Problem: When two numbers are divided and the quotient is $$11$$ and remainder is $$5$$, the smallest value of the dividend is ． 
Options:
aoVal: A
content: $$55$$ 
aoVal: B
content: $$71$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$66$$ 
Answer: B
Analysis: $$6\times 11+5=71$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Problem: Leo prepares more than $400$ cupcakes for a party. Now he can divided all of them equally into $5$ piles. After his pet cat eats one of the cupcakes, he finds that now he can divide the remaining cupcakes equally into $6$ piles. Then, the naughty pet cat eat another one, and Leo divides the remaining cupcakes into $7$ piles. How many cupcakes did Leo make at least at the beginning? Find the sum of the three digits. 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$13$$ 
aoVal: E
content: $$14$$ 
Answer: A
Analysis: The number is a multiple of $5$. When it is divided by $6$, it will have a remainder of $1$; when it is divided by $7$, it will have a remainder of $2$.  Thus, if we add another $5$ to the number, it can be multiple of all the three numbers, which is $5\times6\times7=210$ at least. But the number is more than $400$, so we can get $210\times2-5=415.$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Source: 其它
Problem: There are exactly $$3$$ prime numbers between． 
Options:
aoVal: A
content: $$10$$ and $$20 $$ 
aoVal: B
content: $$20$$ and $$30$$ 
aoVal: C
content: $$30$$ and $$40$$ 
aoVal: D
content: $$40$$ and $$50$$ 
Answer: D
Analysis: The $$3$$ prime numbers between $$40$$ and $$50$$ are $$41$$, $$43$$, and $$47$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: The $4$-digit number $32B9$ is divisble by $3$. If $B$ is even, find the digit $B$. 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$8$$ 
Answer: C
Analysis: $$3+2+B+9$$  = 14+B is divisble by 3.  Max single digit possible is 9. Then 14+9 = 23. Range is between 15 to 23.  Multiple of 3 - 3 x 7=21, 3 x 6 =18, 3 x 5 = 15.  Since, it should be even number, 18-14=4. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
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? (2010 AMC 8 Problem, Question \#22) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$8$$ 
Answer: E
Analysis: The result must hold for any three-digit number with 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 $(\mathbf{E}) 8$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Which of the following is a factor of $30$ ? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: A
Analysis: $30=5 \times 6$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: A $$2-$$digit number is such that the product of the digits adding the sum of the digits is equal to the number itself. What is the ones digit of the number? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$9$$ 
Answer: E
Analysis: We can think of the number as $$10a+b$$, where $$a$$ and $$b$$ are digits. Since the number is equal to the product of the digits $$(a\cdot b)$$ plus the sum of the digits $$(a+b)$$, we can say that $$10a+b=a\cdot b+a+b$$. We can simplify this to $$10a=a\cdot b+a$$, and factor to $$(10)a=(b+1)a$$. Dividing by $$a$$, we have that $$b+1=10$$. Therefore, the units digit, $$b$$, is $$9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: Of the first $$100$$ whole numbers,~\uline{~~~~~~~~~~}~use the digit $$2$$ at least once. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$19$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$10$$ 
Answer: B
Analysis: $$2$$, $$12$$, $$20-29$$, $$32$$, $$42$$, $$52$$, $$62$$, $$72$$, $$82$$, and $$92$$ use a $$2$$; that\textquotesingle s $$19$$ numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers