math-eval/TAL-SCQ5K huggingface dataset:
Problem: The students of a class can be divided into groups of $5$ or groups of $7$ when there is a groupwork to do. How many students at least are in that class? 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$21$$ 
aoVal: D
content: $$35$$ 
Answer: D
Analysis: The least common multiple of $5$ and $7$ is $35$. 
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: A two-digit prime number is still a prime number when its first and tenth digits are exchanged. There aresuch 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: Avril likes collecting baseball cards. The number of cards she has is divisible by $$2$$, $$3$$, and $$5$$. How many baseball cards does Avril have at least? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$60$$ 
Answer: C
Analysis: The total number of baseball cards is the least common multiple of $$2$$, $$3$$, and $$5$$, i.e. $$\left[ 2,3,5\right]=2\times3\times5=30$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: $$\sqrt {3\times 12}\times \sqrt {4\times 9}=$$． 
Options:
aoVal: A
content: $$2\times 18$$ 
aoVal: B
content: $$18\times 18$$ 
aoVal: C
content: $$3\times 2$$ 
aoVal: D
content: $$36\times 36$$ 
Answer: A
Analysis: $$\sqrt {36}\times \sqrt {36}=6\times 6=36 = 2\times 18$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Vivian has many barbie cards. If she puts $8$ cards in each group, there will be $5$ cards left. If she puts $9$ cards in each group, there will be $3$ cards left. If she puts $10$ cards in each group, there will be $1$ card left. How many cards at least should Vivian take away, so that the remaining number of cards can be divisible by $8,$ $9,$ and $10$? 
Options:
aoVal: A
content: $$36$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$21$$ 
aoVal: E
content: $$9$$ 
Answer: D
Analysis: $8+8+5=9+9+3=10+10+1=21$, so after we removing $21$ cards, the number of the remaining cards can be divisible by $8, 9,$ and $10.$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Source: 其它
Problem: The sum of the first $$5$$ \textbf{odd} positive numbers is $$25$$. What is the sum of the first $$5$$ \textbf{even} positive numbers? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$45$$ 
aoVal: E
content: $$50$$ 
Answer: A
Analysis: Each of five positive even numbers is $$1$$ larger than the corresponding odd number in $$1$$,$$3$$, $$5$$, $$7$$ and $$9$$, which we are told have a sum of $$25$$. So the sum of the first five positive even numbers is $$25 +5 =30$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers
Problem: Cayden is drawing some words on a mural in his school. Everyday, he is only able to draw out one letter. He wants to draw:  $$I$$ $$love$$ $$Mathematics$$ $$and$$ $$English$$  If he starts on a Tuesday, and during the weekend, he is not able to draw. When will he finish the drawing? 
Options:
aoVal: A
content: Monday 
aoVal: B
content: Tuesday 
aoVal: C
content: Wednesday 
aoVal: D
content: Thursday 
aoVal: E
content: Friday 
Answer: B
Analysis: $$26\div 5 = 5R1$$  Start from Tuesday, thus, it is Tuesday 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Source: 其它
Problem: $$123+234+345$$ divided by $$4$$ has a remainder of． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: C
Analysis: The sum of each numbers remainder can be divided by $$4$$ to get $$2$$: $$123\div 4$$ R $$3$$；$$234\div 4$$ R $$2$$；$$345\div 4$$ R $$1$$; Therefore, $$\left( 3+2+1 \right)\div 4$$ R $$2$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: The sum of the squares of the first $$20$$ positive integers is $$2870$$. What is the sum of the squares of the first $$19$$ positive integers? 
Options:
aoVal: A
content: $$2350$$ 
aoVal: B
content: $$2361$$ 
aoVal: C
content: $$2470$$ 
aoVal: D
content: $$2850$$ 
Answer: C
Analysis: The sum of the squares of the first $$20$$ positive integers is $$2870$$.  The sum of the squares of the first $$19$$ is $$2870-20^{2}=2870-400=2470$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: What is the remainder of $$(223\times311+198\times273)\div 5$$? ． 
Options:
aoVal: A
content: $1$ 
aoVal: B
content: $2$ 
aoVal: C
content: $3$ 
aoVal: D
content: $4$ 
aoVal: E
content: $5$ 
Answer: B
Analysis: According to the additive and multiplicative properties of remainders, the remainder of this expression equals the remainder of $$(3\times1+3\times3)\div 5=12\div5=2 \text{R} 2$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder 
Source: 其它
Problem: How many numbers of the following are divisible by $3$?  $$\textasciitilde$$  $213 \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} 422\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~741\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~971\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1197\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~2937$  $\textasciitilde$  $\textasciitilde$ 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: The sums of the digits of $422$ and $971$ are not multiple of $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: A positive number is called a perfect square whenever it is the square of a whole number. The first three perfect squares are $$1$$, $$4$$, and $$9$$. The $$100$$th perfect square is． 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$1000$$ 
aoVal: C
content: $$10000$$ 
aoVal: D
content: $$100000$$ 
Answer: C
Analysis: The first perfect square is $$1^{2}$$, the $$2$$nd is $$2^{2}$$, and the $$3$$rd is $$3^{2}$$.  With this pattern, the $$100$$th perfect square is $$100^{2} = 100\times100= 10000$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: There are three whole number $A$, $B$, $C$. $A\times B=55$, $B\times C=100$. $A+B+C=$~\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: If I multiply four different prime numbers, the product must have  positive divisors. 
Options:
aoVal: A
content: $$ 8 $$ 
aoVal: B
content: $$12 $$ 
aoVal: C
content: $$16 $$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$10 $$ 
Answer: C
Analysis: $$2\times2 \times2\times2=16$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number
Problem: In this fictional "Old Island", all the numbers contain only odd digits. The order of the counting numbers is as follows:  $1, 3, 5, 7, \cdots , 19, 31, 33, \cdots $  What is the 31st counting number in the island? 
Options:
aoVal: A
content: $$101$$ 
aoVal: B
content: $$111$$ 
aoVal: C
content: $$99$$ 
aoVal: D
content: $$113$$ 
aoVal: E
content: None of the above. 
Answer: B
Analysis: 1,3,5,7,9 -\/-\/-\/-\/-\/-\/-\/-\/-\/-⑤  11,13,15,17,19-\/-\/-\/-\/-\/-\/-⑤  31,33,35,37,39-\/-\/-\/-\/-⑤  51,53,55,57,59-\/-\/-\/-\/-⑤  71,73,75,77,79-\/-\/-\/-\/-⑤  91,93,95,97,99-\/-\/-\/-⑤  $$111$$  the 31st number is 111. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: What is the remainder of $$(223\times311+198\times273)\div 5$$? 
Options:
aoVal: A
content: $1$ 
aoVal: B
content: $2$ 
aoVal: C
content: $3$ 
aoVal: D
content: $4$ 
aoVal: E
content: $5$ 
Answer: B
Analysis: According to the additive and multiplicative properties of remainders, the remainder of this expression equals the remainder of $$(3\times1+3\times3)\div 5=12\div5=2 \text{R} 2$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder 
Source: 其它
Problem: Bill writes down all the numbers from $1$ to $60$ inclusive. How many times does he use the digit $5$? 
Options:
aoVal: A
content: $16$ 
aoVal: B
content: $$17$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$19$$ 
aoVal: E
content: $$20$$ 
Answer: A
Analysis: In tens place: there are ten $$5$$s: $$50\sim59$$  In ones place: there are six $$5$$s: $$5, 15, 25, 35, 45, 55$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: The greatest prime number that is a divisor of $16,384$ is $2$ because $16,384=2^{14}$. What is the sum of the digits of the greatest prime number that is a divisor of $16,383$ ? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$16$$ 
aoVal: E
content: $$22$$ 
Answer: C
Analysis: We have $$16383=2^{14}-1=\left(2^{7}+1\right)\left(2^{7}-1\right) =129 \cdot 127 $$. Since $129$ is composite, $127$ is the largest prime divisible by $16383$. The sum of $127$\textquotesingle s digits is $$ 1+2+7=\text { (C) } 10 $$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: If a natural number can be written as the sum of both two and three consecutive natural numbers, then we can call it a Think Number. What is the largest Think Number no larger than $5789$? 
Options:
aoVal: A
content: $$5786$$ 
aoVal: B
content: $$5787$$ 
aoVal: C
content: $$5788$$ 
aoVal: D
content: $$5789$$ 
aoVal: E
content: $$5784$$ 
Answer: B
Analysis: The number can be written as $$n+(n+1)=2n+1(n\geqslant 1)$$ and $$x+(x+1)+(x+2)=3x+3$$.  It must be a multiple of $3$ .  $5790$ can be divisible by both $2$ and $3$, so it is $5790-3=5787$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: Change a digit of the number $98760$ to make the new five-digit number be divisible by $250$. What is the new five-digit number? 
Options:
aoVal: A
content: $$98765$$ 
aoVal: B
content: $$98750$$ 
aoVal: C
content: $$98755$$ 
Answer: B
Analysis: The last three digits must be divisible by $125$, and the ones place must be divisible by $2$, so it can only be $98750$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: $$(14\times 9\times 8)\div \left( 9\times 7\times 8 \right)=$$． 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$0$$ 
aoVal: D
content: $$3$$ 
Answer: A
Analysis: $$(14\times 9\times 8)\div \left( 9\times 7\times 8 \right)=14\times 9\times 8\div 9\div 7\div 8=14\div 7\times (9\div 9)\times (8\div 8)=14\div 7=2$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders
Problem: A two-digit number can be divided by both $$1$$ and $$5$$. It is also an even number. How many such numbers are there? 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$19$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$9$$ 
Answer: E
Analysis: $$10, 20, 30, 40, 50, 60, 70, 80, 90$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Source: 其它
Problem: How many $$0$$\textquotesingle s does the product of $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$ end with? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$12$$ 
Answer: B
Analysis: There are $$6$$ factors of $$5$$ when you prime factorise~ $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Problem: The largest possible sum of two different two$$-$$digit numbers is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$99$$ 
aoVal: C
content: $$197$$ 
aoVal: D
content: $$198$$ 
Answer: C
Analysis: $$99 + 98 = 197$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: Julie shares a bag of 70 carrots among some rabbits. Each rabbit has exactly the same number of carrots. If Julie doesn\textquotesingle t have any carrots left, how many rabbits might she have fed? ． 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$8$$ 
Answer: C
Analysis: Among $$3,4,5,6,8$$, only $$5$$ is a factor of $$70$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: Jason has some baseball cards from the $$1920$$s. If he divides the number of cards he has by $$3$$, then he will have $$1$$ remaining cards; if he divides the number of cards he has by $$5$$, he will have $$3$$ remaining cards; if he divides the number of cards he has by $$7$$, he will have $$5$$ remaining cards. How many cards does Jason have at least? 
Options:
aoVal: A
content: $$101$$ 
aoVal: B
content: $$102$$ 
aoVal: C
content: $$103$$ 
aoVal: D
content: $$104$$ 
aoVal: E
content: $$105$$ 
Answer: C
Analysis: The number of cards after adding $$2$$ is divisible by $$3$$, $$5$$, and $$7$$. Since the least common multiple of $$3$$, $$5$$, and $$7$$ is $$3\times5\times7=105$$, Jason has $$105-2=103$$ cards at least. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: Two(different) numbers are selected from $$0$$，$$1$$，$$3$$，$$5$$，$$8$$ and $$9$$. How many two-digit even mumbers can be formed？ 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$19$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$30$$ 
Answer: A
Analysis: $$10$$、$$30$$、$$50$$、$$80$$、$$90$$、$$18$$、$$38$$、$$58$$、$$98$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Problem: The sum of $$2016$$ integers is even. At mostof them can be odd. 
Options:
aoVal: A
content: $$2016$$ 
aoVal: B
content: $$2015$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$0$$ 
Answer: A
Analysis: The sum of any even number of odd integers is always even. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers
Problem: The greatest common factor of $$23$$ and $$24$$ is． 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$1$$ 
Answer: D
Analysis: The only common factor of any two consecutive whole numbers is $$1$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: Which of the following statements is incorrect? 
Options:
aoVal: A
content: $$2350$$ is a multiple of $$2$$, $$5$$ and $$10$$ 
aoVal: B
content: $$1284$$ is a multiple of $$4$$ 
aoVal: C
content: $$9972$$ is not a multiple of $$9$$ 
aoVal: D
content: $$5742$$ is a multiple of $$6$$ and $$9$$ 
Answer: C
Analysis: A: $$2350$$ is even, thus it is a multiple of $$2. 2350$$ ends with $$0$$, thus it is a multiple of $$5$$ and $$10$$;  B: $$1284$$ ends with $$84$$, and $$84$$ is a multiple of $$4$$, thus $$1284$$ is a multiple of $$4$$;  C: $$9 + 9 + 7 + 2 = 27$$ is a multiple of $$9$$, thus $$9972$$ is a multiple of $$9$$;  D: $$5 + 7 + 4 + 2 = 18$$, it is a multiple of $$9$$ and $$3$$; since it ends with $$2$$, it is also a multiple of $$2$$, and thus a multiple of $$6$$;  Hence the incorrect statement is C. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: There are three whole number $A$, $B$, $C$. $A\times B=77$, $B\times C=364$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$70$$ 
aoVal: D
content: $$90$$ 
Answer: C
Analysis: $77=7\times 11$  $364=2\times 2\times 7\times 13$  Because $B$ is the factor both number contains, $B=7$  Thus, $A=11$, $C=52$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: Which number between $$60$$ and $$80$$ is both a multiple of $$3$$ and $$8$$?~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$63$$ 
aoVal: C
content: $$72$$ 
aoVal: D
content: $$96$$ 
Answer: C
Analysis: $$72$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Common Multiples and Least Common Multiples->Least Common Multiple of Two Numbers
Problem: Divide $$2000$$ by an odd number. The remainder must be． 
Options:
aoVal: A
content: $$$$even 
aoVal: B
content: $$$$odd 
aoVal: C
content: $$$$prime 
aoVal: D
content: $$$$whole 
Answer: D
Analysis: $$2000\div1001$$ and $$2000\div999$$ leave whole \# remainders $$999$$ and $$2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders
Problem: What is the remainder when 7,999,999,999 is divided by 8 ? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$9$$ 
Answer: C
Analysis: omitted  PMC 2021 \#7 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: Given that $$M=\overline{3abcd}$$, $$N=\overline{abcd3}$$ , and $$M-N=3177$$, what is $$\overline{abcd}$$? 
Options:
aoVal: A
content: $$2940$$ 
aoVal: B
content: $$2960$$ 
aoVal: C
content: $$2980$$ 
aoVal: D
content: $$3000$$ 
aoVal: E
content: $$3020$$ 
Answer: C
Analysis: $$M-N=3177$$  $$(30000+\overline{abcd})-(10\times \overline{abcd}+3)=3177$$  $$30000-3177=9\times \overline{abcd}$$  $$\overline{abcd}=2980$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: I have equal numbers of quarters, dimes, and nickels. These coins could have a total value of any of the following $$EXCEPT$$． 
Options:
aoVal: A
content: ＄$$2.40$$ 
aoVal: B
content: ＄$$3.80$$ 
aoVal: C
content: ＄$$4.40$$ 
aoVal: D
content: ＄$$5.20$$ 
Answer: B
Analysis: The value of $$1$$ quarter, $$1$$ dime, and $$1$$ nickel is $$40$$￠. My coins must have a total value divisible by $$40$$, but ＄$$3.80$$ is not divisible by $$40$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: The thousands digit of the sum of 5+55+555+5555 is ． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$6$$ 
Answer: E
Analysis: $$6170$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases
Problem: The thousands digit of the sum of 8+88+888+8888+88888 is ． 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
Answer: D
Analysis: $$8+88+888+8888+88888=98760$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases
Problem: Which of the following is not a composite number? 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$47$$ 
aoVal: C
content: $$49$$ 
aoVal: D
content: $$51$$ 
Answer: B
Analysis: $47$ is a prime number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: Students guess that Norb's age is $28$, $30$, $34$, $36$, $38$, and $41$. Norb says, "At least half of you guessed too low, two of your guesses are off by one, and my age is a prime number."~How old is Norb? (adapted from 2011 AMC 8 Problem, Question \#21) 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$29$$ 
aoVal: C
content: $$31$$ 
aoVal: D
content: $$37$$ 
aoVal: E
content: $$14$$ 
Answer: D
Analysis: If at least half the guesses are too low, the Norb\textquotesingle s age must be greater than $34$. If two of the guesses are off by one, then his age is between two guesses whose difference is $2$. It could be $29$, $35$ or $37$, but because his age is greater than $34$, it can only be $35$ or $37$. Lastly, Norb\textquotesingle s age is a prime number so the answer must be $37$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: Among numbers like $5$, $55$, $555$, $5555$, $$\cdots$$, how many of them are perfect squares? 
Options:
aoVal: A
content: $0$ 
aoVal: B
content: $1$ 
aoVal: C
content: $2$ 
aoVal: D
content: Countless 
Answer: A
Analysis: None of them 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers
Source: 其它
Problem: What is the remainder when $$222 222 222$$ is divided by $$4$$? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$0$$ 
Answer: B
Analysis: In division by $$4$$, the last $$2$$ digits determine the remainder, use $$22\div4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders