math-eval/TAL-SCQ5K huggingface dataset:
Problem: Which of the followings is not a multiple of $3$? 
Options:
aoVal: A
content: $213$ 
aoVal: B
content: $214$ 
aoVal: C
content: $216$ 
aoVal: D
content: $219$ 
Answer: B
Analysis: $$\text{A}$$. $$2+1+3=6=2\times3$$;  $$\text{B}$$. $$2+1+4=7$$, and $$7$$ is not a multiple of $$3$$;  $$\text{C}$$. $$2+1+6=9=3\times3$$;  $$\text{D}$$. $$2+1+9=12=4\times3$$.  We choose $$\text{B}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: Which of the following products is an odd number? 
Options:
aoVal: A
content: $$23\times24$$ 
aoVal: B
content: $$24\times35$$ 
aoVal: C
content: $$42\times53$$ 
aoVal: D
content: $$53\times45$$ 
Answer: D
Analysis: The product is odd if and only if every factor is odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Problem: How many positive factors of $$36$$ are also multiples of $$4$$ ? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: B
Analysis: $$36={{2}^{2}}\times {{3}^{2}}$$, so the number of its factors would be $$\left( 2+1 \right)\times \left( 2+1 \right)=9$$, Among them, there are $$\left( 2+1 \right)\times 1=3$$ factors which has $${{2}^{2}}$$ as its factors. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? (2006 AMC 8 Problem, Question \#23) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: A
Analysis: The counting numbers that leave a remainder of 4 when divided by 6 are $4,10,16,22,28,34, \cdots$ The counting numbers that leave a remainder of 3 when divided by 5 are $3,8,13,18,23,28,33, \cdots$ So 28 is the smallest possible number of coins that meets both conditions. Because $4 \cdot 7=28$, there are (A) 0 coins left when they are divided among seven people. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Among numbers like $5$, $55$, $555$, $5555$, $$\cdots$$, how many of them are perfect squares? 
Options:
aoVal: A
content: $0$ 
aoVal: B
content: $1$ 
aoVal: C
content: $2$ 
aoVal: D
content: Countless 
Answer: B
Analysis: Only $5$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers
Source: 其它
Problem: Dividing a certain two$$-$$digit number by $$7$$ leaves a remainder of $$5$$; dividing it by $$11$$ leaves a remainder of $$9$$. What is the smallest possible value of this number? 
Options:
aoVal: A
content: $$40$$ 
aoVal: B
content: $$54$$ 
aoVal: C
content: $$75$$ 
aoVal: D
content: $$152$$ 
Answer: C
Analysis: The number when added by $$2$$ is divisible by $$7$$ and by $$11$$. Hence, the smallest value is $$7\times 11-2=75$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders
Problem: If a four-digit number $$\overline{5ab4}$$ is a perfect square number, then $$a+b=$$． 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
Answer: B
Analysis: First of all $${{70}^{2}}=4900$$, So the number is between $$70$$ and $$80$$.  According to the last digit, the number should be $$72 $$ or $$78 $$.  $$\because {{72}^{2}}=5184$$，$${{78}^{2}}=6084$$．  $$\therefore a+b=1+8=9$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers
Problem: The natural numbers from $$1$$ to $$99$$ inclusive are divided into n groups such that the following conditions hold:  Each number belongs to exactly one group.  Each group contains at least two numbers.  If two numbers belong to the same group, then their sum is not divisible by $$3$$.  What is the smallest number $$n$$ which satisfies the above conditions? ． 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$33$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$66$$ 
Answer: B
Analysis: We can divide the $$99$$ numbers into $$3$$ groups according to their remainder when divided by $$3$$. $$1$$, $$4$$, $$7\_\cdots$$ they belong to the group $$A$$, since they leave a remainder of $$1$$ when divided by $$3$$. $$2$$, $$5$$, $$8\_\cdots$$ they belong to the group $$B$$, since they leave a remainder of $$2$$ when divided by $$3$$. $$3$$, $$6$$, $$9\_\cdots$$ they belong to the group $$C$$, since they leave a remainder of $$0$$ when divided by $$3$$. As we know, the members in group $$C$$ cannot get together. Also, when the member of group $$A$$ and the member of group $$B$$ get together, they will create the sum which is a multiple of $$3$$. So the $$33$$ members of group $$C$$ will be in $$33$$ groups. Group $$A$$ and group $$B$$ can join in the $$33$$ groups however they want. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: If the four-digit number $$28X8$$ is divisible by $$3$$, how many possible values are there for $$X$$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: D
Analysis: $$2+8+8=18$$, $$18$$ is multiple of $$3$$  Thus, $$X$$ itself must be a multiple of $$3$$, or $$0$$.  We have $$3, 6, 9$$ and $$0$$. Four possible value. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: How many three-digit numbers have an odd number of factors? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$21$$ 
aoVal: E
content: $$22$$ 
Answer: E
Analysis: It can be shown that a positive integer has an odd number of factors if and only if it is square. ~The smallest three-digit square number is $${{10}^{2}}=100$$ and the largest is $${{31}^{2}}=961$$. Hence there are $$31-9=22$$ three-digit numbers which have an odd number of factors. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: ♥ $$\times$$ ☺ $$=$$ ♦  ☺ is an even number.  Which of the following gives an odd answer? 
Options:
aoVal: A
content: ♦ $$-\textasciitilde7$$ 
aoVal: B
content: ☺ $$+$$ ♦ 
aoVal: C
content: ☺ $$\times$$ ☺ 
aoVal: D
content: ♦ $$\times$$~♦ 
Answer: A
Analysis: ♥ $$\times$$ ☺ $$=$$ ♦  Since ☺ is an even number,~♦ must also be an even number.  ♦ $$-\textasciitilde7$$ is the only option to given an odd answer because even $$-$$ odd $$=$$ odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Problem: Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textbackslash$ 1 A 2$. 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: We know that a number is divisible by $11$ if the odd numbers added together minus the even numbers added together(or vice versa) is a multiple of $11$ . So, we have $1+2-A=$ a multiple of $11$ . The only multiple that works here is $0$ , as $11 \cdot 0=0$. Thus, $A=(D)$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Which of the following is not a factor of $$2016$$? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$7$$ 
Answer: A
Analysis: $$2016=7\times288=8 \times252=9\times224$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)
Problem: Jam has some pieces of candy. He wants to share with some kids. If he shares the candy among $8$ kids on average, there will be $2$ pieces left. If he shares the candy among $9$ kids on average, there will be $3$ pieces left. If he shares the candy among $10$ kids on average, 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: $8\times9\times10\div2=360$  $360-6=354$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Source: 其它
Problem: Chloe is working on this equation: $$475+17\times 58+990-19\times 32+33\times 111$$.  Her answer is $$5681$$.  Without calculating, do you think Chloe\textquotesingle s answer is correct or wrong? Explain why. 
Options:
aoVal: A
content: Correct 
aoVal: B
content: Wrong 
Answer: B
Analysis: Her calculation was wrong. The answer is supposed to be an even number.  $$117\times32$$，$$19\times12$$ are even numbers and $$133\times11$$ is an odd number.  So in that case, we can think of the equation as:  odd + even + even - even + odd = even  Hence, the answer should be an even number! However, Chloe\textquotesingle s answer is an odd number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Problem: If the four-digit number $3P78$ is divisible by $3$, how many possible values are there for $P$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: use divisibility rule.  3+ 7+8+R = divisible by 3.  18 + R = divisible by 3.  smallest possible number for R is 0, max possible amount for R is 9  3x6=18 - 18 =0 (can be divisible by 3)  3x7=21 -18 = 3  3x 8= 24 -18 =6  3x 9= 27 - 18 = 9  Total = 4 ways. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: John loves collecting stamps! If he divides the number of stamps he has by $$32$$, then he will have $$30$$ remaining stamps; if he divides the number of stamps he has by $$9$$, he will have $$7$$ remaining stamps; if he divides the number of stamps he has by $$7$$, he will have $$5$$ remaining stamps. How many stamps, at least, does John have? 
Options:
aoVal: A
content: $$2014$$ 
aoVal: B
content: $$2015$$ 
aoVal: C
content: $$2016$$ 
aoVal: D
content: $$2017$$ 
Answer: A
Analysis: The total number of stamps 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: Prime factorise $$24\times 105$$． 
Options:
aoVal: A
content: $${{2}^{3}}\times {{3}^{2}}\times 5\times 7$$ 
aoVal: B
content: $${{2}^{4}}\times {{3}^{2}}\times 5$$ 
aoVal: C
content: $${{2}^{4}}\times {{3}^{2}}\times 5\times 7$$ 
Answer: A
Analysis: $$24\times 105={{2}^{3}}\times {{3}^{2}}\times 5\times 7$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Problem: Which of the following is the correct expression of quinary (base-$5$ numeral system)? 
Options:
aoVal: A
content: $\left (8231\right )\_5$ 
aoVal: B
content: $\left (2001\right )\_5$ 
aoVal: C
content: $\left (4341\right )\_7$ 
aoVal: D
content: $\left (2345\right )\_5$ 
Answer: B
Analysis: In quinary, the base number must be $5$, and all the digits in the parentheses must be less than $5$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Properties and Applications of Number Bases
Problem: $$2010$$ is divided by $$N$$ and gets a remainder of $$15$$. There are~\uline{~~~~~~~~~~}~possible values of $$N$$. 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$16$$ 
Answer: D
Analysis: $$2010-15=1995=3\times 5\times 7\times 19$$  $$\left( 1+1 \right)\times \left( 1+1 \right)\times \left( 1+1 \right)\times \left( 1+1 \right)=16$$ factors.  Remove $$1$$, $$3$$, $$5$$, $$7$$, and $$15$$ from them, we can get $$16-5=11$$ possible values. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number
Problem: Timothy writes down the number 24. He reverses the digits to make the number 42. He then works out that 42 is 18 more than his starting number, 24.  Nicole writes down a whole number between 10 and 99. She also reverses the digits of her number. She finds that this makes a number that is 72 more than her starting number.  What was the last digit of Nicole's starting number? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$9$$ 
Answer: E
Analysis: Let the original number be ab, and reverse it to be ba, calculated by the place value principle, ba-ab=72  10b+a-(10a+b)=72, 9b-9a=72, b-a=8, b=9 and a=1. So the original number is 19, last digit is 9. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: How many three-digit numbers are divisible by $13$ ? (2004 AMC 8 Problem, Question \#18) 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$67$$ 
aoVal: C
content: $$69$$ 
aoVal: D
content: $$76$$ 
aoVal: E
content: $$77$$ 
Answer: C
Analysis: Let $k$ be any positive integer so that $13 k$ is a multiple of 13 . For the smallest three-digit number, $13 k\textgreater100$ and $k\textgreater\frac{100}{13} \approx 7.7$. For the greatest three-digit number, $13 k\textless999$ and $k\textless\frac{999}{13} \approx 76.8$. The number $k$ can range from 8 to 76 so there are $(\mathbf{C}) 69$ threedigit numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Which of the following numbers are prime numbers?  $$\textasciitilde$$  $179$~ ~ ~ ~ ~ ~ ~ ~ ~$129$~ ~ ~ ~ ~ ~ ~ ~ ~$187$~ ~ ~ ~ ~ ~ ~ ~ ~$157$  $$\textasciitilde$$ 
Options:
aoVal: A
content: Only $179$ 
aoVal: B
content: $179$ and $$187$$ 
aoVal: C
content: $179$ and $157$ 
aoVal: D
content: $179$, $157$, and $187$ 
Answer: C
Analysis: $129=3\times43$. $187=11\times17$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Problem: The square root of the square root of $$16$$ is ． 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$64$$ 
Answer: A
Analysis: The square root of $$16$$ is $$4$$ and the square root of $$4$$ is $$2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Candy buys $$15$$ bottles of drinks, and they are placed in order on the counter in the following way: $$2$$ bottles of milk, a bottle of juice, a bottle of coke, $$2$$ bottles of milk, a bottle of juice, a bottle of coke$$\ldots $$ According to the pattern, the $$15$$\textsuperscript{th} bottle is． 
Options:
aoVal: A
content: $$\text{milk}$$ 
aoVal: B
content: $$\text{juice}$$ 
aoVal: C
content: $$\text{coke}$$ 
Answer: B
Analysis: $$15\div 4=3\text{R}3$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Problem: How many factors $2000$ that have more than $2$ factors are there? (As an example, $12$ has $6$ factors, namely $1$, $2$, $3$, $4$, $6$ and $12$. But $2$ and $3$ only two factors.) 
Options:
aoVal: A
content: $$13$$ 
aoVal: B
content: $$14$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$16$$ 
aoVal: E
content: $$17$$ 
Answer: E
Analysis: $2000=2^{3}\times5^{3}$, the number of its factors would be $(4+1)(3+1)=20$, but $1$, $2$, and $5$ do not meet the condition. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: Which of the following prime factorization is incorrect? 
Options:
aoVal: A
content: $$12=2\times2\times3$$ 
aoVal: B
content: $$51=3\times17$$ 
aoVal: C
content: $$8=2\times2\times2$$ 
aoVal: D
content: $$45=5\times9$$ 
Answer: D
Analysis: $$45=5\times3\times3$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Problem: In Walmart, apples are sold in pack of eight. How many apples will you get, if you buy five packs? 
Options:
aoVal: A
content: $$32$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$56$$ 
Answer: B
Analysis: $5\times 8=40$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: The sum of the first four square numbers is $$30$$.  What is the sum of the first five square numbers? 
Options:
aoVal: A
content: $$55$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$25$$ 
aoVal: D
content: $$45$$ 
aoVal: E
content: $$65$$ 
Answer: A
Analysis: $30+25=55$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: In the division expression $$28\div$$~\uline{~~~~~~~~~~}~$$=$$~\uline{~~~~~~~~~~}~$$\text{r}4$$, how many different pairs of number are there to fill the gaps? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: $$1$$ 
Answer: B
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: $$\sqrt{10^{2} - 6^{2}} + \sqrt{3^{2} + 4^{2}}=$$． 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$13$$ 
Answer: D
Analysis: $$\sqrt{10^{2} - 6^{2}} + \sqrt{3^{2} + 4^{2}}=\sqrt{64} + \sqrt{25} = 8 + 5 = 13$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: There are three whole numbers $A$, $B$ and $C$ ($B\neq 1$). $A\times B=45$, $B\times C=50$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$23$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$29$$ 
Answer: C
Analysis: $45=3\times 3\times 5$  $50=2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=9$, $C=10$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: $$539-142$$ divided by $$4$$ has a remainder of~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: B
Analysis: $$539\div 4$$ R $$3$$；$$142\div 4$$ R $$2$$；$$539-142$$ divided by $$4$$, the remainder is 3-2=1. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: Which of the followings is divisible by $$8$$ and $$7$$? 
Options:
aoVal: A
content: $$56$$ 
aoVal: B
content: $$49$$ 
aoVal: C
content: $$42$$ 
aoVal: D
content: $$16$$ 
Answer: A
Analysis: $$56 = 8 \times 7$$, $$56 = 7 \times 8$$, so we choose $$\rm A$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: Given $$a$$ is a factor of $37$, then ． 
Options:
aoVal: A
content: $a$ can only be $1$ 
aoVal: B
content: $a$ can only be $37$ 
aoVal: C
content: $a$ can be 1 or $37$ 
aoVal: D
content: $a$ can be any number 
Answer: C
Analysis: "$$a$$ is a factor of $37$" means $37$ is divisible by $a$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: How many $$2$$-digit whole numbers have no odd factor except $$1$$? 
Options:
aoVal: A
content: $$49$$ 
aoVal: B
content: $$45$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$0$$ 
Answer: C
Analysis: The $$3$$ such whole numbers are $$16$$, $$32$$, and $$64$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers
Problem: Of the integers from $$1$$ to $$1000$$, how many are multiples of $$3$$, $$4$$, \emph{and} $$5$$? 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$17$$ 
Answer: C
Analysis: A multiple of $$3$$, $$4$$, and $$5$$ is a multiple of $$60$$, and $$16\times60\textless{}1000 \textless{}17\times60$$. 
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 $C$ $(B\ne 1)$. $A\times B=15$, $B\times C=35$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$35$$ 
Answer: C
Analysis: $15=3\times 5$  $35=5\times 7$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=3$, $C=7$,  $A+B+C=15$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
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: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: C
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: The sum of two prime numbers is $$99$$. What is the difference between the two numbers? 
Options:
aoVal: A
content: $$89$$ 
aoVal: B
content: $$92$$ 
aoVal: C
content: $$95$$ 
aoVal: D
content: $$97$$ 
Answer: C
Analysis: Because $$99$$ is an odd number, one of these two prime numbers must be an even prime number. The only choice is $$2$$. So the other is $$99-2 = 97$$. So, the difference between the two numbers is $$95$$. Therefore, we choose $$\rm C$$ . 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers
Problem: The result of $$29+30+31+\cdots\cdots+87+88$$ is an~\uline{~~~~~~~~~~}~number. 
Options:
aoVal: A
content: odd 
aoVal: B
content: even 
Answer: B
Analysis: $88-29+1=60$  $60\div2=30$  Odd numbers all paired. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers