math-eval/TAL-SCQ5K huggingface dataset:
Problem: Kate and Kerry are two dogs. Every $3$-hour, Kate goes to the dog park. Every $2$-hour, Kerry gos to the dog park. Today, Kate and Kerry first meet at $10:00\text{am}$. When will they meet each other again? 
Options:
aoVal: A
content: $12:00\text{pm}$ 
aoVal: B
content: $4:00\text{pm}$ 
aoVal: C
content: $6:00\text{pm}$ 
aoVal: D
content: $10:00\text{pm}$ 
Answer: B
Analysis: $LCM[2,3]=6$ 
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 characteristic of numbers that are divisible by $$4$$ is． 
Options:
aoVal: A
content: Last digit is divisible by $$4$$. 
aoVal: B
content: Last two digits are divisible by $$4$$. 
aoVal: C
content: Last three digits are divisible by $$4$$. 
aoVal: D
content: The sum of digits is divisible by $$4$$. 
Answer: B
Analysis: $$1$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders
Problem: $$\sqrt {2\times 4\times 8}\times \sqrt {8\times 8}=$$． 
Options:
aoVal: A
content: $$64$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$8$$ 
Answer: A
Analysis: $$\sqrt {2\times 4\times 8}\times \sqrt {8\times 8}=8\times 8=64$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Which of the following numbers is not the square of a whole number? 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$144$$ 
aoVal: C
content: $$196$$ 
aoVal: D
content: $$200$$ 
Answer: D
Analysis: All choices except $$200$$ are perfect squares since:  $$100=10^{2}$$  $$144=12^{2}$$  $$196=14^{2}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: How many factors does $36$ have? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$9$$ 
Answer: D
Analysis: $1, 2, 3, 4, 6, 9, 12, 18, 36$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: When from any three-digit number we subtract that number written backwards, the difference will always be a number that is divisible by:． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$9$$ 
Answer: D
Analysis: No matter how we write the three-digit number, the sum of its three digits is always the same. Therefore, the remainders of the two numbers divided by $$9$$ are also the same. The difference will always be a number that is divisible by $$9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Casting Out Nines
Problem: Which of the following has an odd number of whole-number factors? 
Options:
aoVal: A
content: $$47$$ 
aoVal: B
content: $$48$$ 
aoVal: C
content: $$49$$ 
aoVal: D
content: $$50$$ 
Answer: C
Analysis: The whole-number factors of $$49$$ are $$1$$, $$7$$, and $$49$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers
Problem: How many positive integer factors of 2020 have more than 3 factors? (As an example, 12 has 6 factors, namely $1,2,3,4,6$, and 12 .) (2020 AMC 8 Problem, Question \#17) 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$10$$ 
Answer: B
Analysis: Since $2020=2^{2} \cdot 5 \cdot 101$, we can simply list its factors: $$ 1,2,4,5,10,20,101,202,404,505,1010,2020 . $$ There are 12 of these; only $1,2,4,5,101$ (i.e. 5 of them) don\textquotesingle t have over 3 factors, so the remaining $12-5=(\text{B}) 7$ factors have more than 3 factors. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: From 202 to 2020 (including these two numbers), how many multiples of 9 are there? 
Options:
aoVal: A
content: $$201$$ 
aoVal: B
content: $$202$$ 
aoVal: C
content: $$101$$ 
aoVal: D
content: $$102$$ 
Answer: B
Analysis: $[\frac{2020}{9}]-[\frac{201}{9}]=224-22=202$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: Amy runs a lap around the track in $4$ minutes and Pawel in $5$ minutes. Amy and Pawel start to run around the track at the same time. After how many minutes will the boys meet at the starting point again? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: It depends on the distance around the track 
Answer: D
Analysis: The answer is the least common multiple of $4$ and $5$, which is $20$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: How many positive factors of $$144$$ are also multiples of $$4$$? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$12$$ 
Answer: D
Analysis: $$144={{2}^{4}}\times {{3}^{2}}=4\times2^{2}\times {{3}^{2}}$$. Among them, there are $$\left( 2+1 \right)\times (2+1)=9$$ factors which has $$4$$ as its factor. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: John, Emily and Nick want to buy some apples from Walmart where the apples are sold in pack of four. If they buy six packs and share the apples evenly, how many apples will each one of them get? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$10$$ 
Answer: C
Analysis: $4\times 6=24$;  $24=3\times 8$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: Each of the following results in an even number except． 
Options:
aoVal: A
content: $$952+136$$ 
aoVal: B
content: $$952-136$$ 
aoVal: C
content: $$952\div 136$$ 
aoVal: D
content: $$952\times 136$$ 
Answer: C
Analysis: The sum, difference, and product of even numbers are even, but $$952\div 136=7$$ is odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Odd and Even Applications
Problem: What is the smallest prime number greater than $$47$$? 
Options:
aoVal: A
content: $$59$$ 
aoVal: B
content: $$57$$ 
aoVal: C
content: $$53$$ 
aoVal: D
content: $$51$$ 
Answer: C
Analysis: $$49 =7\times7$$; $$51 = 3\times17$$. The first prime is $$53$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: For a positive integer $n$, the factorial notation $n!$ repressents the product of the integer from $n$ to $1$, (For example, $6!=6\times5\times4\times3\times2\times1$.) What value of $N$ satisfies the following equation?  $$6!\times8!=8\times N!$$ 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$13$$ 
Answer: B
Analysis: $6!=10\times9\times8$, $$6!\times8!=8\times 10!$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: Find the smallest multiple of $$7$$ that is greater than or equal to $$20 $$. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$21$$ 
aoVal: C
content: $$22$$ 
aoVal: D
content: $$28$$ 
Answer: B
Analysis: $$2\times7=14$$, $$3\times7=21$$, since $$14$$ is smaller than $$20$$ and $$21$$ is the closest multiple of $$7$$ and greater than $$20$$, we choose $$\text B$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: There are three whole number $A$, $B$, $C$. $A\times B=77$, $B\times C=132$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$33$$ 
Answer: A
Analysis: $77=7\times 11$  $132=2\times 2\times 3\times 11$  Because $B$ is the factor both number contains, $B=11$  Thus, $A=7$, $C=12$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: Which of the following number is a multiple of $9$? 
Options:
aoVal: A
content: $$234$$ 
aoVal: B
content: $$136$$ 
aoVal: C
content: $256$ 
aoVal: D
content: $$418$$ 
Answer: A
Analysis: $2+3+4=9$, so $234$ is divisible by 9. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders
Source: 其它
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: First, list number that leaves a remainder of $2$ when divided by $6$: $8,14,20,26,32,38,44,50,56,62\cdots $. Meanwhile, divide each number with $5$ to see it the number meets the condition. If the number divided by $5$ leaves a remainder of $2$, divide it by $4$ see if it still works. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: The product of $$2$$ odd numbers is always.　 
Options:
aoVal: A
content: divisible by $$3$$ 
aoVal: B
content: odd　 
aoVal: C
content: prime　 
aoVal: D
content: even　 
Answer: B
Analysis: The product of $$2$$ odd numbers, such as $$5\times7=35$$, is always odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Problem: What does the "3"~in 1234 mean? 
Options:
aoVal: A
content: 3 hundreds 
aoVal: B
content: 3 tens 
aoVal: C
content: 3 thousands 
aoVal: D
content: 3 ones 
aoVal: E
content: 3 millions 
Answer: B
Analysis: 3 tens 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers->Understanding Numbers and Digits
Source: 其它
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: $$9$$ 
aoVal: E
content: $$10$$ 
Answer: D
Analysis: $$11$$，$$13$$，$$17$$，$$31$$，$$37$$，$$71$$，$$73$$，$$79$$，$$97$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: Fill in the blanks below with the largest possible numbers:  ~\uline{~~~~~~~~~~}~$$\div16 = 5\rm R$$~\uline{~~~~~~~~~~}~.  ~\uline{~~~~~~~~~~}~$$\div16 = 5\rm$$ $\cdots\cdots$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: 105, 15 
aoVal: B
content: 95, 15 
aoVal: C
content: 88, 13 
aoVal: D
content: 64, 16 
Answer: B
Analysis: The largest possible remainder should be at least $$1$$ smaller than the divisor. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders
Problem: What is the missing number in the box?  $$7063000 =7000000+\boxed{?}+3000$$ 
Options:
aoVal: A
content: $$600$$ ones 
aoVal: B
content: $$600$$ tens 
aoVal: C
content: $$600$$ hundreds 
aoVal: D
content: $$600$$ thousands 
Answer: C
Analysis: $600$ hundreds has the same value as $600\times100=60000$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: Which of the following cases can\textquotesingle t make an even number? 
Options:
aoVal: A
content: An odd number$$+$$An even number 
aoVal: B
content: An even number$$+$$An even number 
aoVal: C
content: An odd number$$+$$An odd number 
Answer: A
Analysis: The sum of an odd number and an even number is always an odd~ number. The sum of two numbers which are both even or both odd is always an even number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Problem: Divide 8 numbers $$15$$、$$18$$、$$21$$、$$22$$、$$42$$、$$44$$、$$50$$ and $$60$$ into two groups with 4 numbers in each group to make the product of numbers in each group the same, so the two groups are ． 
Options:
aoVal: A
content: （$$15$$, $$22$$, $$21$$, $$60$$），（$$18$$, $$44$$, $$42$$, $$50$$） 
aoVal: B
content: （$$15$$, $$42$$, $$44$$, $$60$$），（$$18$$, $$22$$, $$21$$, $$50$$） 
aoVal: C
content: （$$15$$, $$44$$, $$21$$, $$60$$），（$$18$$, $$22$$, $$42$$, $$50$$） 
aoVal: D
content: （$$15$$, $$44$$, $$21$$, $$50$$），（$$18$$, $$22$$, $$42$$, $$60$$） 
Answer: C
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->Finding Factors Given the Product
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, two plants will grow on the second day, and on the twenty-sixth days, they will just fill the pond. If eight water plants are put into the pond on the first day, how many days it will 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 develop to eight plants. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: What is the remainder when $$12+34+56+89+90$$ is divided by $$10$$? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$9$$ 
Answer: B
Analysis: Add ones digits, then divide by $$10$$: $$2+4+6+9+0=21$$;~$$21\div 10=2\text{R}1$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders
Problem: Only one of the following four numbers is a perfect square. Which one is it? 
Options:
aoVal: A
content: $$76186$$ 
aoVal: B
content: $$750235$$ 
aoVal: C
content: $$921438$$ 
aoVal: D
content: $$2660161$$ 
Answer: D
Analysis: $$\text{A}$$: $$76186\div2=28093$$. But $$28093$$ is not divisible by $$2$$. So, $$76186$$ is divisible by $$2$$ but not $$4$$, $$76186$$ is not a perfect square.  $$\text{B}$$: $$750235\div5=150047$$. But $$150047$$ is not divisible by $$5$$. So, $$750235$$ is divisible by $$5$$ but not $$25$$, $$750235$$ is not a perfect square.  $$\text{C}$$: $$921438\div2=460719$$. But $$460719$$ is not divisible by $$2$$. So, $$921438$$ is divisible by $$2$$ but not $$4$$, $$921438$$ is not a perfect square.  Thus, answer must be $$\text{D}$$. In fact, $$2660161=7^{2}\times233^{2}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: What is the sum of the prime factors of $2010$? (2010 AMC 8 Problems, Question \#14) 
Options:
aoVal: A
content: $$67$$ 
aoVal: B
content: $$75$$ 
aoVal: C
content: $$77$$ 
aoVal: D
content: $$201$$ 
aoVal: E
content: $$210$$ 
Answer: C
Analysis: First, we must find the prime factorization of $2010.2010=2 \cdot 3 \cdot 5 \cdot 67$. We add the factors up to get $(\mathbf{C}) 77$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: The whole numbers from $$1$$ to $$2016$$ inclusive are written on a blackboard. Moritz underlines all the multiples of two in red, all the multiples of three in blue and all the multiples of four in green. How many numbers does Moritz underline exactly twice? 
Options:
aoVal: A
content: $$1008$$ 
aoVal: B
content: $$1004$$ 
aoVal: C
content: $$504$$ 
aoVal: D
content: $$336$$ 
aoVal: E
content: $$168$$ 
Answer: C
Analysis: There is no number that is both a multiple of three and a multiple of four without also being a multiple of two. Hence, the numbers underlined exactly twice are those that are a multiple of two and of three but not of four and those that are a multiple of two and four but not of three. The first set of numbers consists of the set of odd multiples of six. Since $$2016 \div 6 = 336$$, there are $$336$$ multiples of $$6$$ in the list of numbers and hence $$336 \div 2 = 168$$ odd multiples of six that would be underlined in red and blue but not green. The second set of numbers consists of two out of every three multiples of four and, since $$2016 \div 4 = 504$$, there are $$3\times 504 = 336 $$ numbers that would be underlined in red and green but not blue. Hence there are $$168+ 336 = 504$$ numbers that Moritz would underline exactly twice. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: Which of the following is a prime number? 
Options:
aoVal: A
content: $$38$$ 
aoVal: B
content: $$45$$ 
aoVal: C
content: $$53$$ 
aoVal: D
content: $$57$$ 
Answer: C
Analysis: We note that the other options are not prime numbers because:  $38 = 2 \times 19$  $45 = 4 \times 9$  $57 = 3 \times 19$  Hence by the process of elimination, $53$ is a prime number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: A student wrote down a natural number. When she divided the number by $$9$$, the remainder was $$7$$. What is the~~remainder when twice that number is divided by $$9$$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
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: is a factor of $$100110011001$$. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$12$$ 
Answer: A
Analysis: $$100110011001$$ is divisible by $$3$$, since the sum of its digits is $6$ which is divisible by $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: The product of any $$3$$-digit whole number and any $$2$$-digit whole number can contain at most how many digits? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: C
Analysis: The largest product of a $$3$$-digit number and a $$2$$-digit number is $$999\times99 = 98901$$. That product has $$5$$ digits. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: What is the greatest natural number $$n$$ such that $$n+27$$ and $$n-62$$ are squares of natural number? 
Options:
aoVal: A
content: $$598$$ 
aoVal: B
content: $$1598$$ 
aoVal: C
content: $$3998$$ 
aoVal: D
content: $$1998$$ 
aoVal: E
content: Such a number does not exist 
Answer: D
Analysis: Let $a^{2} = n+27$ and $b^{2} = n-62$. Taking the difference:  $a^{2}-b^{2} = (a+b)(a-b) = n+27-(n-62) = 89$  Since $89$ is prime, that means $a+b=89$ and $a-b=1$, which means $a=45$ and $b=44$.  That means $n=a^{2}-27 = 45^{2}-27 = 2025-27=1998$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Questions involving Square Numbers
Problem: If I write all the whole numbers from $$1$$ to $$100$$ in words, how many times will I write the letter V? 
Options:
aoVal: A
content: $$19$$ 
aoVal: B
content: $$29$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$31$$ 
aoVal: E
content: $$32$$ 
Answer: D
Analysis: There are $$10$$ sevens in the unit digit, $$10$$ seventys, $$9$$ fives in the units (as fifteen is not fiveteen) plus eleven and twelve. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: How many multiples of $5$ are there from $1$ to $50$? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$15$$ 
Answer: B
Analysis: $5$, $10$, $15$, $20$, $25$, $30$, $35$, $40$, $45$, $50$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: Which of the following prime factorizations below is correct? 
Options:
aoVal: A
content: $$97=1\times97$$ 
aoVal: B
content: $$85=5\times 17$$ 
aoVal: C
content: $$64=8\times 8$$ 
aoVal: D
content: $$52=1\times 2\times 2\times 13$$ 
Answer: B
Analysis: ①A prime number is a natural number greater than $$1$$ that has no other factors than $$1$$ and itself.  ②Decomposing a composite number into the product of several prime factors is called decomposing prime factors.  ③$$1$$ is neither a prime number nor a composite number. So $$\text{D}$$ is not correct.  The $$\text{A}$$ option is also incorrect. $$8$$ is not a prime number, so $$\text{C}$$ is incorrect. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Problem: Which of the followings is the least common multiple for $$25$$ and $$125$$? 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$125$$ 
aoVal: C
content: $$130$$ 
aoVal: D
content: $$600$$ 
Answer: B
Analysis: We do prime factorization for $$25$$ and $$125$$ first. $$25={{5}^{2}}$$ and $$125={{5}^{3}}$$. We have three $$5$$'s, thus the least common multiple for $$25$$ and $$125$$ is $$125$$. We choose $$\text{B}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: Of the multiples of $$7$$ that exceed $$7$$, how many are factors of $$700$$? 
Options:
aoVal: A
content: $$99$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$5$$ 
Answer: B
Analysis: Multiply $$7$$ by any of these: $$2$$, $$2^{2}$$, $$5$$, $$5^{2}$$, $$2\times 5$$, $$2^{2}\times 5$$, $$2\times 5^{2}$$, or $$2^{2}\times 5^{2}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number
Problem: There are three whole numbers $A$, $B$ and $C$. $A\times B=55$, $B\times C=100$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$34$$ 
aoVal: D
content: $$36$$ 
Answer: D
Analysis: $55=5\times 11$  $100=2\times 2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=11$, $C=20$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: What is the difference between the largest single-digit prime number and the smallest three-digit prime number? 
Options:
aoVal: A
content: $$94$$ 
aoVal: B
content: $$95$$ 
aoVal: C
content: $$96$$ 
aoVal: D
content: $$97$$ 
aoVal: E
content: $$98$$ 
Answer: A
Analysis: The largest single-digit prime number is $$7$$ and the smallest three-digit prime number is $$101$$; their difference is $$101 -7=94$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)
Problem: The product of \emph{all} the prime numbers less than $$10$$ is divisible by ． 
Options:
aoVal: A
content: $$70$$ 
aoVal: B
content: $$60$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$4$$ 
Answer: A
Analysis: The primes less than $$10$$ are $$2$$, $$3$$, $$5$$, and $$7$$. Their product is $$210$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers