math-eval/TAL-SCQ5K huggingface dataset:
Problem: A bag contains $$50$$ straws of two different colours. $$10$$ straws are yellow and the rest are red. What percentage of the straws are red? 
Options:
aoVal: A
content: $$10\textbackslash\%$$ 
aoVal: B
content: $$20\textbackslash\%$$ 
aoVal: C
content: $$40\textbackslash\%$$ 
aoVal: D
content: $$80\textbackslash\%$$ 
Answer: D
Analysis: 50-10=40；$$40\div$$ 50=0.8；0.8=80\% 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Properties and Applications of Number Bases->Mixed Operations of Number Bases
Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Word Problems Involving Fractions and Percentages->Finding the Percentage Given a Part and a Whole
Problem: How many solutions that can be expressed with positive integers does the equation below have? ($$1999$$ Math kangaroo Problems, Level $$7-8$$, Question \#$$26$$)  $$a^{2}b-1=1999$$ 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: D
Analysis: $a^{2}b=2000=2^{4}\times5^{3}$. Thus, the value of $a$ can be $1$, $2$, $4$, $5$, $10$, and $20$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->Finding Factors Given the Product
Problem: Find the value for lcm $$[5,7,11]$$. 
Options:
aoVal: A
content: $$385$$ 
aoVal: B
content: $$1155$$ 
aoVal: C
content: $$77$$ 
aoVal: D
content: $$35$$ 
Answer: A
Analysis: Since $$5=1\times5$$, $$7=1\times7$$, and $$11=1\times11$$, we have one $$5$$, one $$7$$, and one $$11$$, thus the least common multiple for $$5$$, $$7$$, and $$11$$ is $$5\times7\times11=385$$. We choose $$\rm A$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: A computer can do $$7\times {{10}^{9}}$$ operations per second. Then, it can do~\uline{~~~~~~~~~~}~operations in $$5\times {{10}^{2}}$$ seconds. 
Options:
aoVal: A
content: $$35\times {{10}^{10}}$$ 
aoVal: B
content: $$3.5\times {{10}^{11}}$$ 
aoVal: C
content: $$3.5\times {{10}^{12}}$$ 
aoVal: D
content: $$3.5\times {{10}^{19}}$$ 
aoVal: E
content: $$3.5\times {{10}^{18}}$$ 
Answer: C
Analysis: $$7\times {{10}^{9}}\times 5\times {{10}^{2}}=3.5\times {{10}^{12}}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->The Relationship between Exponents and the Number of Factors
Problem: If you were to work out the answer to the sum  $$2^{2016}+0^{2016}+1^{2016}+6^{2016}$$  you would get a number with $$1569$$ digits, starting with $$566$$ $$136$$ $$001$$ $$\cdots$$  What is the last digit of this number? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$9$$ 
Answer: B
Analysis: We can find the last (units) digit by looking at the units digits of the four parts. We can quickly note that the units digits of $$0^{2016}$$ and $$1^{2016}$$ and $$6^{2016}$$are $$0$$, $$1$$ and$$6$$ respectively. Now then we have to look at powers of $$2:{{2}^{1}}=2$$, $${{2}^{2}}=4$$, $${{2}^{3}}=8$$, $${{2}^{4}}=16$$, and after this the units digits repeat ($${{2}^{5}}=3\underline{2}$$, $${{2}^{6}}=6\underline{4}$$, $${{2}^{7}}=12\underline{8}$$, $${{2}^{8}}=25\underline{6}$$, $$\cdots $$). We notice that when the indices are multiples of 4 (eg. $${{2}^{4}}=16$$, $${{2}^{8}}=256$$, $$\cdots $$) the units digit of the power of $$2$$ is $$6$$. Hence the units digit of $$2^{2016}+0^{2016}+1^{2016}+6^{2016}$$ is the same as the units digit of $$6 + 0 + 1 + 6$$, that is $$3$$. 
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: The $5$-digit number $\overline{2018U}$ is divisible by $9$. What is the remainder when this number is divided by $8$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
Answer: B
Analysis: We use the property that the digits of a number must sum to a multiple of $9$ if it is divisible by $9$. This means $2+0+1+8+U$ must be divisible by $9$. The only possible value for U then must be $7$. Since we are looking for the remainder when divided by $8$, we can ignore the thousands. The remainder when $187$ is divided by $8$ is $(\rm B)$ $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: Which of the options below is equivalent to $${{({{2}^{2}}\times {{3}^{3}}\times {{5}^{5}})}^{4}}$$? 
Options:
aoVal: A
content: $${{2}^{6}}\times {{3}^{7}}\times {{5}^{9}}$$ 
aoVal: B
content: $${{2}^{2}}\times {{3}^{3}}\times {{5}^{20}}$$ 
aoVal: C
content: $${{2}^{2}}\times {{3}^{3}}\times {{5}^{9}}$$ 
aoVal: D
content: $${{2}^{8}}\times {{3}^{12}}\times {{5}^{20}}$$ 
Answer: D
Analysis: Answer $$={{2}^{2\times 4}}\times {{3}^{3\times 4}}\times {{5}^{5\times 4}}={{2}^{8}}\times {{3}^{12}}\times {{5}^{20}}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->The Relationship between Exponents and the Number of Factors
Problem: What is the sum of all factors of $$24$$? 
Options:
aoVal: A
content: $$52$$ 
aoVal: B
content: $$60$$ 
aoVal: C
content: $$62$$ 
aoVal: D
content: $$72$$ 
aoVal: E
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: Add any two odd numbers. The ones\textquotesingle{} digit of the sum is always． 
Options:
aoVal: A
content: $$2 $$ 
aoVal: B
content: $$$$prime 
aoVal: C
content: $$$$odd 
aoVal: D
content: $$$$even 
Answer: D
Analysis: Add two numbers from: $$1$$, $$3$$, $$5$$, $$7$$, $$9$$. The result is always even. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers
Problem: What is the smallest integer $$n$$ for which the number $$\left(2^{2}-1\right)\cdot \left(3^{2}-1\right)\cdot \left(4^{2}-1\right)\cdots \left(n^{2}-1\right)$$ is the square of an integer? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$27$$ 
aoVal: E
content: None of these. 
Answer: B
Analysis: Using the difference of two squares, we can get:  $$\left(2^{2}-1\right)\cdot \left(3^{2}-1\right)\cdot \left(4^{2}-1\right)\cdots \left(n^{2}-1\right) = 1\times 3\times 2\times 4\times 3\times 5\times \cdots \times (n-1) \times (n+1)$$  you will notice that this is equal to:  $$1\times 2\times 3^{2} \times 4^{2} \times \cdots \times (n-1)^{2} \times n \times (n+1)$$  So this will only be a square if $2n(n+1)$ is a square, which happens when $n=8$ as $2n(n+1) = 144=12^{2}$.  Answer: $B$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Questions involving Square Numbers
Problem: What is the greatest number of consecutive integers such that the sum of the digits of none of them is divisible by $$5$$? ． 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: C
Analysis: Five consecutive numbers can be: $$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$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: The number $$95$$~\uline{~~~~~~~~~~}~$$94775998$$ is divisible by $$198$$. What is the missing digit? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$9$$ 
Answer: E
Analysis: Already divisible by $$2$$, so we must check it divides by $$9$$ by adding the digits together:  $$9+5+x+9+4+7+7+5+9+9+8=54+x$$, this must be a multiple of $$9$$ so $$x$$ is $$0$$ or $$9$$.  It should also be divisible by $$11$$, so the alternating sum of digits should too.  $$9-5+x-9+4-7+7-5+9-9+8=2+x$$ is divisible by $$11$$.  Hence $$x$$ is $$9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: Lucas buys $24$ machines. But two of the digits of the total price on the invoice are not clear, which only show $\overline{\square 8\square 2}$ dollars. The purchasing agent says the two unclear digits on the invoice are the same. Given that the unit price of this kind of machine is an integer, how much is it? 
Options:
aoVal: A
content: $$76$$ 
aoVal: B
content: $$160$$ 
aoVal: C
content: $$202$$ 
aoVal: D
content: $$328$$ 
aoVal: E
content: $$412$$ 
Answer: D
Analysis: The number of total price can be divided by $3$ and $8$. It can be divided by $8$ which means the tens digit must be $3$ or $7$. And when the tens digit is $7$, the number can be divided by $3$.  Thus, the total price is $7872$ dollars, and each of the machine is $328$ dollars. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: How many two-digit numbers can be divided by both 2 and 3 at the same time. 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$75$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$60$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: *and* = 2 x 3 = 6  Multiple of 6 with 2 digits = 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: How many of the whole numbers less than $$100$$ are $$10$$ greater than an odd whole number?　 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$46$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$91$$ 
Answer: A
Analysis: Add $$10$$ to $$1$$, $$3$$, $$5$$, $$7$$, $$\cdots $$, $$87$$, and $$89$$. None of these sums is more than $$99$$. There are $$45$$ such sums. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers
Problem: Find the remainder when $$\frac{{{2}^{2019}}}{{{4}^{982}}-{{2}^{1963}}}$$ is divided by $$5$$． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
Answer: A
Analysis: $$\frac{{{2}^{2019}}}{{{2}^{1964}}-{{2}^{1963}}}$$  $$={{2}^{2019-1963}}$$  $$={{2}^{56}}$$，  ∴$${{2}^{56}}:6$$，  $$6\div 5:1$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->The Relationship between Exponents and the Number of Factors
Problem: How many of these four expressions are perfect squares?  $$1^{3}+2^{3}$$~ ~ ~ ~ ~ ~$$1^{3}+2^{3}+3^{3}$$~ ~ ~ ~ ~$$1^{3}+2^{3}+3^{3}+4^{3}$$~ ~ ~ ~ ~$$1^{3}+2^{3}+3^{3}+4^{3}+5^{3}$$ 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: E
Analysis: All four expressions are perfect squares:  $$1^{3} + 2^{3} = 1 + 8 = 9 = 3^{2}$$;  $$1^{3}+2^{3}+3^{3}=1+8+27=36=6^{2}$$;  $$1^{3}+2^{3}+3^{3}+4^{3}=1+8+27+64=100=10^{2}$$;  $$1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} = 1 + 8 + 27 + 64 + 125 = 225 = 15^{2}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: $$135 798 642$$ is not divisible by~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$9$$ 
Answer: B
Analysis: $$42$$ is not divisible by $$4$$, so neither is $$135 798 642$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: $$3^{2}+3^{2}+3^{2}+3^{2}=$$． 
Options:
aoVal: A
content: $$4^{2}$$ 
aoVal: B
content: $$6^{2}$$ 
aoVal: C
content: $$12^{2}$$ 
aoVal: D
content: $$33^{2}$$ 
Answer: B
Analysis: $$3^{2}+3^{2}+3^{2}+3^{2}=9+9+9+9=36=6^{2}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Which of the following is \emph{not} a factor of $$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$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)
Problem: What is the sum of the two smallest prime factors of $250$? (2007 AMC 8 Problems, Question \#3) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$12$$ 
Answer: C
Analysis: The smallest prime factors of $250$ are $2$ and $5$. Thus, the sum is $2 + 5 = 7$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: What is the remainder when $$16+16+16+16$$ is divided by $$4$$ ． 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$0$$ 
Answer: D
Analysis: Since $$16\div 4$$ has a remainder of $$0$$, the remainder is $$0+0+0+0=0$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: A $$2-$$digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number? ． 
Options:
aoVal: A
content: $$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$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: Lee counted by $$7$$\textquotesingle s beginning with one of the whole numbers from $$1$$ through $$7$$, until Lee passed $$1000$$. If Lee counted three of the following numbers, which number did Lee not count? 
Options:
aoVal: A
content: $$107$$ 
aoVal: B
content: $$184$$ 
aoVal: C
content: $$534$$ 
aoVal: D
content: $$641$$ 
Answer: D
Analysis: Each number on Lee\textquotesingle s list must have the same remainder when divided by $$7$$. Divide each choice by $$7$$. The respective remainders are $$2$$, $$2$$, $$2$$, and $$4$$. Thus, $$107$$, $$184$$, and $$534$$ work. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: Leo prepares more than $400$ cupcakes for a party. Now he can divided all of them equally into $5$ piles. After his pet cat eats one of the cupcakes, he finds that now he can divide the remaining cupcakes equally into $6$ piles. Then, the naughty pet cat eat another one, and Leo divides the remaining cupcakes into $7$ piles. How many cupcakes did Leo prepare at least at the beginning? Find the sum of the three digits. 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$13$$ 
aoVal: E
content: $$14$$ 
Answer: A
Analysis: The number is a multiple of $5$. When it is divided by $6$, it will have a remainder of $1$; when it is divided by $7$, it will have a remainder of $2$.  Thus, if we add another $5$ to the number, it can be multiple of all the three numbers, which is $5\times6\times7=210$ at least. But the number is more than $400$, so we can get $210\times2-5=415.$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Source: 其它
Problem: Ms Lee has many students. They can be split into $$5$$ groups, $$6$$ groups or $$9$$ groups equally. How many students might she possibly have? 
Options:
aoVal: A
content: $$120$$ 
aoVal: B
content: $$150$$ 
aoVal: C
content: $$180$$ 
aoVal: D
content: $$240$$ 
aoVal: E
content: $$300$$ 
Answer: C
Analysis: $6=2\times3, 9=3\times3, 2\times3\times3\times5=90$. The number of students is a multiple of 90. $180\div90=2$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Word Problems Involving Factors and Multiples->Multiples Word Problems
Problem: (US $$1998$$ Math kangaroo Problems, Level $$5-6$$, Question \#$$15$$)When from any three-digit number we subtract that number written backwards, the difference will always be a number that is divisible by:． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$13$$ 
Answer: D
Analysis: No matter how we write the three-digit number, the sum of its three digits is always the same. Therefore, the remainders of the two numbers divided by $$9$$ are also the same. The difference will always be a number that is divisible by $$9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Casting Out Nines
Problem: There are three whole number $A$, $B$, $C$. $A\times B=45$, $B\times C=50$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$23$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$29$$ 
Answer: C
Analysis: $45=3\times 3\times 5$  $50=2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=9$, $C=10$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: Miss Angel want to pack some candy for her students. She has a total of $$104$$ pieces of candy. Each of the goody bag will contain $$7$$ pieces of candy. In order to ensure all her students get a full bag of candy, how many more pieces of candy she needed? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: B
Analysis: $$104\div7=14R6$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders
Source: 其它
Problem: If the sum of two whole numbers equals twice their difference, this sum cannot be． 
Options:
aoVal: A
content: $$222$$ 
aoVal: B
content: $$444$$ 
aoVal: C
content: $$888$$ 
aoVal: D
content: $$1000$$ 
Answer: A
Analysis: If the sum is twice the difference, one \# is triple the other. The sum is divisible by $$4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: The product of $$2860$$ and $$m$$ is a square number. Find the smallest possible value of $$m$$. 
Options:
aoVal: A
content: $$65$$ 
aoVal: B
content: $$572$$ 
aoVal: C
content: $$715$$ 
aoVal: D
content: $$2860$$ 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers
Source: 其它
Problem: Think Lab buys $24$ experimental apparatus. But two of the digits of the total price on the invoice are not clear, which only show $\overline{\square 8\square 2}$ dollars. The purchasing agent says the two unclear digits on the invoice are the same. Given that the unit price of this kind of apparatus is an integer, how much is it? 
Options:
aoVal: A
content: $$76$$ 
aoVal: B
content: $$160$$ 
aoVal: C
content: $$202$$ 
aoVal: D
content: $$328$$ 
aoVal: E
content: $$412$$ 
Answer: D
Analysis: The number of total price can be divided by $3$ and $8$. It can be divided by $8$ which means the tens digit must be $3$ or $7$. And when the tens digit is $7$, the number can be divided by $3$.  Thus, the total price is $7872$ dollars, and each of them is $328$ dollars. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: There are over $$1000$$ students at my school. When~ the number of students is divided by $$10$$， the remainder is $$3$$． When~ the number is divided by $$13$$， the remainder is $$3$$. What is the remainder when the number is divided by $$130$$？ 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$9$$ 
Answer: B
Analysis: When $$10$$ students or$$13$$ students are in a group, there are always $$3$$ students remained. It means that if we subtract $$3$$ from the total number, the result is a common multiple of $$10$$ and $$13$$. Therefore when there are $$130$$ in one group, there are still$$\textasciitilde3$$ students remained. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: Olivia is thinking of a two-digit number. She says, "If I divide the number by $$6$$, the remainder is $$3$$. If I divide the number by $$8$$, the remainder is also $$3$$."  What is the smallest possible number that Olivia could be thinking of? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$27$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$51$$ 
Answer: B
Analysis: The Lowest Common Multiple of $$6$$ and $$8$$ is $$24$$.  $$24+3=27$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Source: 其它
Problem: Let $N=34 \cdot 34 \cdot 63 \cdot 270$. What is the ratio of the sum of the odd divisors of $N$ to the sum of the even divisors of $N$ ? 
Options:
aoVal: A
content: $1:16$ 
aoVal: B
content: $1:15$ 
aoVal: C
content: $1:14$ 
aoVal: D
content: $1:8$ 
aoVal: E
content: $1:3$ 
Answer: C
Analysis: Prime factorizing $N$, we see $N=2^{3} \cdot 3^{5} \cdot 5 \cdot 7 \cdot 17^{2}$. The sum of $N$ \textquotesingle s odd divisors are the sum of the factors of $N$ without 2 , and the sum of the even divisors is the sum of the odds subtracted by the total sum of divisors. The sum of odd divisors is given by $$ a=\left(1+3+3^{2}+3^{3}+3^{4}+3^{5}\right)(1+5)(1+7)\left(1+17+17^{2}\right) $$ and the total sum of divisors is $$ (1+2+4+8)\left(1+3+3^{2}+3^{3}+3^{4}+3^{5}\right)(1+5)(1+7)\left(1+17+17^{2}\right)=15 a . $$ Thus, our ratio is $$ \frac{a}{15 a-a}=\frac{a}{14 a}=\text { (C) } 1: 14 \text {. } $$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: A student wrote down a natural number. When she divided the number by $$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->Factors and Multiples
Source: 其它