math-eval/TAL-SCQ5K huggingface dataset:
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  10 + a - b (10) a + b = a = 9 b - 9, 72-8 a = b, b = 9, a = 1, the original number is 19, last digit is 9. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: How many whole numbers greater than $$0$$ whose square is equal to its square root? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$4$$ 
Answer: B
Analysis: The only positive whole number whose square is equal to its square root is $$1$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: How many numbers of the following are divisible by $3$?  $$\textasciitilde$$  $314\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} 528\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~899\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1024\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1356\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~3336$  $\textasciitilde$  $\textasciitilde$ 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: A
Analysis: The sums of the digits of $528$, $1356$, and $3336$ are multiples of $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: Of the following, which has the largest odd factor? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$36$$ 
aoVal: D
content: $$40$$ 
Answer: A
Analysis: Keep dividing each by $$2$$ until you get an odd number.  $$\text{A}$$.$$2\times15$$.  $$\text{B}$$.$$2\times2\times 2 \times 2\times2\times1$$.  $$\text{C}$$.$$2\times2\times9$$.  $$\text{D}$$.$$2\times2 \times 2\times5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: Is it possible to find two numbers such that if you add up the sum and the difference of these two numbers, you get $$999$$? If yes, please write it out. If not, please give your reason. 
Options:
aoVal: A
content: Possible 
aoVal: B
content: Not possible 
Answer: B
Analysis: For example you have number A and B.  A+B = C  A-B = D  (A+B)+(A-B)=2A  Thus, C+D must always be even number. Cannot be odd number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Problem: Weili wanted to pack $60$ apples and $75$ pears into as many bags as possible, with no remainder. She packed the same number of fruit in each bag. The number of apples in each bag was the same. How many apples were there in each bag? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$4$$ 
Answer: D
Analysis: Factors of $60$: $1$, $2$, $3$, $4$, $5$, $6$, $10$, $12$, $15$, $20$, $30$ and $60$,  Factors of $75$: $1$, $3$, $5$, $15$, $25$ and $75$,  Since Weili wanted to pack the apples and pears into as many bags as possible, we need to choose the largest common factor of both $60$ and $75$.  Largest common factor $\rightarrow 15$,  Hence, Weili should use $15$ bags. Number of apples in each bag$\rightarrow60\div15$  $=4$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: Which of the following integers is not a multiple of $$45$$? 
Options:
aoVal: A
content: $$765$$ 
aoVal: B
content: $$675$$ 
aoVal: C
content: $$585$$ 
aoVal: D
content: $$495$$ 
aoVal: E
content: $$305$$ 
Answer: E
Analysis: Note that $$45=5\times 9$$. As $$5$$ and $$9$$ are coprime, a positive integer is a multiple of $$45$$ if and only if it is a multiple of both $$5$$ and $$9$$. The units digit of all five options is $$5$$, so they are all multiples of $$5$$. An integer is a multiple of $$9$$ if and only if the sum of its digits is also a multiple of $$9$$. The sums of the digits of the five options is $$18$$, $$18$$, $$18$$, $$18$$ and $$8$$. So $$305$$ is the only one of the options which is not a multiple of $$9$$ and hence is not a multiple of $$45$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: A whole number is a perfect square if it can be expressed as the product of two equal whole numbers. What is the sum of the first $10$ perfect squares? 
Options:
aoVal: A
content: $$384$$ 
aoVal: B
content: $$385$$ 
aoVal: C
content: $$386$$ 
aoVal: D
content: $$387$$ 
Answer: B
Analysis: We can list out the perfect squares:  $1\times 1 = 1$  $2\times 2 = 4$  $3\times 3 = 9$  $4\times 4 = 16$  $5\times 5 = 25$  $6\times 6 = 36$  $7\times 7 = 49$  $8\times 8 = 64$  $9\times 9 = 81$  $10\times 10 = 100$  Hence the sum of the first $10$ perfect squares is $385$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: How many zeros does the number $$12^{2}\times15^{3}$$ end with? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$1$$ 
Answer: D
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->The Number of Zeros at the end of a Product
Problem: Arthur writes down three two-digit integers. One is square, one is prime and one is triangular.  She uses the digits $$1$$, $$2$$, $$3$$, $$4$$, $$5$$ and $$6$$ exactly once each.  Which largest prime does he write? ． 
Options:
aoVal: A
content: $$13 $$ 
aoVal: B
content: $$23 $$ 
aoVal: C
content: $$31 $$ 
aoVal: D
content: $$41 $$ 
aoVal: E
content: $$43$$ 
Answer: D
Analysis: First note that Arthur can write down three squares, namely $$16$$, $$25$$ and $$36$$. Also, he can write down four triangular numbers, namely $$15$$, $$21$$, $$36$$ and $$45$$. If he chooses $$16$$ and $$45$$ for the square and triangular number respectively, then the remaining digits are $$2$$ and $$3$$, the prime is $$23$$. If he chooses $$25$$ and $$36$$ then the remaining digits are $$1$$ and $$4$$, the prime is $$41$$. If he chooses $$36$$ for the square number, the remaining difits can be a prime. So the largest prime he write is $$41$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers
Problem: How many three-digit positive integers have an odd number of even digits? 
Options:
aoVal: A
content: $$150$$ 
aoVal: B
content: $$250$$ 
aoVal: C
content: $$350$$ 
aoVal: D
content: $$450$$ 
aoVal: E
content: $$550$$ 
Answer: D
Analysis: There are only 2 ways for an odd number of even digits: 1 even digit or all even digits. Case 1: 1 even digit There are $5 \cdot 5=25$ ways to choose the odd digits, 5 ways for the even digit, and 3 ways to order the even digit. So, $25 \cdot 5 \cdot 3=375$. However, there are $5 \cdot 5=25$ ways that the hundred\textquotesingle s digit is 0 and we must subtract this from 375 , leaving us with 350 ways. Case 2: all even digits There are $5 \cdot 5 \cdot 5=125$ ways to choose the even digits, and $5 \cdot 5=25$ ways where the hundred\textquotesingle s digit is 0 . So, $125-25=100$. Adding up the cases, the answer is $100+350=$ (D) 450 . 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: How many different primes are in the prime factorization 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: A whole number is a perfect square if it can be expressed as the product of two equal whole numbers. For example, $$9$$ is a perfect square since $$9= 3\times 3$$. How many perfect squares are greater than $$0$$ and less than $$1000$$? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$33$$ 
Answer: B
Analysis: A whole number is a perfect square if it is the product of two equal whole numbers. Thus, $$1\times1$$ and $$2\times 2$$ and $$3\times 3$$ and $$4\times 4$$ are perfect squares. Continue until the product is bigger than $$1000: 30\times 30 = 900$$; $$31\times 31 = 961$$; $$32\times 32 = 1024$$ (too big). 
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 $C$. $A\times B=45$, $B\times C=50$. $A+B+C=$~\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: Eddie has some baseball cards from the $$1920$$s. If he divides the number of cards he has by $$4$$, then he will have $$3$$ remaining cards; if he divides the number of cards he has by $$5$$, he will have $$4$$ remaining cards; if he divides the number of cards he has by $$7$$, he will have $$6$$ remaining cards. How many cards does Eddie have at least? 
Options:
aoVal: A
content: $$139$$ 
aoVal: B
content: $$140$$ 
aoVal: C
content: $$141$$ 
aoVal: D
content: $$142$$ 
Answer: A
Analysis: The number of cards after adding $$1$$ is divisible by $$4$$, $$5$$, and $$7$$. Since the least common multiple of $$4$$, $$5$$, and $$7$$ is $$4\times5\times7=140$$, Eddie has $$140-1=139$$ cards at least. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: The product of three $2-$digit consecutive even numbers is $$\overline{\textbackslash\#\textbackslash\#\textbackslash\#2}$$. What is their sum? 
Options:
aoVal: A
content: $$42$$ 
aoVal: B
content: $$64$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$48$$ 
aoVal: E
content: $$52$$ 
Answer: D
Analysis: Considering the ones digit and the value of the product, which is less than $10000$, only $14, 16,$ and $18$ match the condition. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: The square root of $$16$$ is ． 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$64$$ 
Answer: B
Analysis: The square root of $$16$$ is $$4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Find the smallest prime number that is greater than $$90$$. 
Options:
aoVal: A
content: $$91$$ 
aoVal: B
content: $$96$$ 
aoVal: C
content: $$97$$ 
aoVal: D
content: $$99$$ 
Answer: C
Analysis: We only have one prime number between $$90$$ and $$100$$, which is $$97$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: Find the smallest whole number between $$14$$ and $$40$$ that is divisible by $$3$$ and by $$4$$. 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$36$$ 
Answer: D
Analysis: Multiple of $$3$$ and $$4$$ also a multiple of $$12$$.  $$12, 24, 36, 48, \cdots $$  Smallest multiple between $$14$$ and $$40$$ is $$24$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Source: 其它
Problem: The sum of the prime factors of $$231$$ is． 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$152$$ 
aoVal: D
content: $$383$$ 
Answer: A
Analysis: Since $$231=3\times7\times11$$, the sum of its prime factors is $$3 +7+11=21$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)
Problem: I add up all even numbers between $$1$$ and $$101$$. Then from my total I subtract all odd numbers between $$0$$ and $$100$$.  What is the result? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$100$$ 
aoVal: D
content: $$255$$ 
aoVal: E
content: $$2525$$ 
Answer: B
Analysis: Let the required result be $$S$$. Then  $$S= \left( {2+4+6+\cdots +100} \right)- \left( {1+3+5+\cdots +99} \right)~ $$  $$\textasciitilde\textasciitilde= \left( {2-1} \right) + \left( {4-3} \right) + \left( {6-5} \right) +\cdots +\left( {100-99} \right) $$  $$\textasciitilde\textasciitilde=50\times 1=50$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers
Problem: The product of two whole numbers is $$42$$. Their sum \emph{cannot} be． 
Options:
aoVal: A
content: $$43$$ 
aoVal: B
content: $$33$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$13$$ 
Answer: B
Analysis: All but $$33$$ can be represented as required, as shown below.  A. $$43=1+42$$  B. $$33$$  C. $$23=2+21$$  D. $$13=6+7$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: The sum of two prime numbers is $$99$$. What is the difference between the two prime 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: Andy writes down the largest two-digit prime such that each of its digits is prime.  Baker writes down the smallest three-digit prime such that each of its digits is prime.  Carl adds Andy\textquotesingle s number and Baker\textquotesingle s number.  What answer does Carl obtain? ． 
Options:
aoVal: A
content: $$174 $$ 
aoVal: B
content: $$185 $$ 
aoVal: C
content: $$198 $$ 
aoVal: D
content: $$209 $$ 
aoVal: E
content: $$296$$ 
Answer: E
Analysis: The prime digits are $$2$$, $$3$$, $$5$$ and $$7$$. So the largest two-digit integer whose digits are both prime is $$77$$. However, $$77$$ is not prime, nor is $$75$$, but $$73$$ is prime. So Andy writes down $$73$$. The smallest three-digit integer whose digits are both prime is $$222$$. However, $$222$$ is not prime, but $$223$$ is prime.  So Baker writes down $$223$$. Therefore the answer which Carl obtains is $$73 +223 = 296$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers
Problem: Which of the following numbers cannot be written as the sum of $$4$$ consecutive whole numbers? 
Options:
aoVal: A
content: $$1994$$ 
aoVal: B
content: $$2042$$ 
aoVal: C
content: $$2050$$ 
aoVal: D
content: $$2060$$ 
Answer: D
Analysis: $$497+498+499+500 =1994$$; $$509+510+511+512 = 2042$$; $$511+512+513+514=2050$$. (Divide by $$4$$, and "start"~near the quotient.) 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: A multi-digit number$$\underbrace{20092009\cdot \cdot \cdot 2009}\_{n 2009s}736$$, can be divisible by ~$$11$$ . The smallest value of $$n$$ is． 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: B
Analysis: $$3$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: There are some flowers along the corridor arranged in the following order:  $5$ purple, $3$ red, $2$ yellow, $2$ pink, $3$ red, $2$ yellow, $2$ pink $\cdots$  If there are $100$ flowers altogether, how many red flowers are there altogether? 
Options:
aoVal: A
content: $$13$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$39$$ 
aoVal: D
content: $$42$$ 
Answer: D
Analysis: $$3+2+2=7$$;  $100-5=95$  $95\div7=13R4$;  $$13$$$\times$$$3$$+$$3$$=$$42$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Problem: In the $${{9}^{\text{th}}}$$ Century BC, an Indian mathematician named Al Khwarizmi wrote a book about math calculations. Since his calculations were always written on a clay tablet, he was afraid the calculation process might be lost. So, he created a system to determine whether his calculations were correct.  Example: $$1234+18983+18922=39039$$,  $$1234$$ divided by $$9$$ yields a remainder of $$1$$,  $$18983$$ divided by $$9$$ yields a remainder of $$2$$,  $$18922$$ divided by $$9$$ yields a remainder of $$4$$,  When the remainders are added up and divided by $$9$$, the remainder is $$7$$. Dividing the number on the right-hand side of the equals sign by $$9$$ yields a remainder of $$6$$. Therefore, $$7$$ does not equal $$6$$, making the above equation incorrect.  Use the method above to determine whether the following equations are correct or not:．  ①$$2638457+3521983+6745785=12907225$$,  ②$$7832145-2167953=5664192$$. 
Options:
aoVal: A
content: √√ 
aoVal: B
content: √$$\times $$ 
aoVal: C
content: $$\times $$√ 
aoVal: D
content: $$\times $$$$\times $$ 
Answer: C
Analysis: $$(1)$$ Dividing the numbers on the left-hand side of the equals sign by $$9$$ yields remainders of $$8$$, $$4$$, and $$6$$. The sum of the remainders divided by $$9$$ is $$0$$. Dividing the right-hand side of the equals sign by $$9$$ yields a remainder of $$1$$. Therefore, $$0$$ does not equal $$1$$, making the equation incorrect.  $$(2) $$The remainders on both sides of the equation are $$6$$. We can check by performing the calculation normally. It is correct. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Casting Out Nines
Problem: If a four-digit number $$5A2A$$ can be divisible by both $$4 $$ and $$5$$, the digit that $$A$$ represents is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$8$$ 
Answer: A
Analysis: We consider the last two-digit to see if a number is divisible by $4$. We consider the last one digit to see if a number is divisible by $5$. If $$\overline{\textasciitilde5A2A\textasciitilde}$$ is divisible by $5$, the ones digit can only be $5$ or $0$. When the ones digit is $0$, the number that is formed by the last two digits, $20$, is divisible by $4$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: If my school has four times as many girls as boys, then the number of girls minus the number of boys \emph{could} be． 
Options:
aoVal: A
content: $$2013$$ 
aoVal: B
content: $$2011$$ 
aoVal: C
content: $$2009$$ 
aoVal: D
content: $$2008$$ 
Answer: A
Analysis: For every $$4$$ girls and $$1$$ boy, the difference is $$3$$. The difference is always divisible by $$3$$. Of the choices, only $$2013$$ is divisible by $$3$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: The values of length and width of a rectangle are both prime numbers. If the area of that rectangle is $10$, what is its perimeter? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$14$$ 
Answer: D
Analysis: $10=2\times 5$,  the length is $5$ and the width is $2$.  $(2+5)\times 2=14$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
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: $$56\times67$$ 
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: John's age is a multiple of $$7$$ this year. His age next year will be a multiple of $$6$$. What is John's age this year? 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$28$$ 
aoVal: C
content: $$35$$ 
aoVal: D
content: $$42$$ 
Answer: C
Analysis: $\textasciitilde$  $\textasciitilde$  $\textasciitilde$  $\textasciitilde$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: What is the least possible remainder when an even number is divided by $$7$$? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: A
Analysis: When $$14$$ (or any other even multiple of $$7$$) is divided by $$7$$, the remainder is $$0$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders
Problem: Which of the following is a multiple of $8$? 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$28$$ 
aoVal: D
content: $$30$$ 
Answer: B
Analysis: $3\times 8=24$, so $24$ is a multiple of $8$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它