math-eval/TAL-SCQ5K huggingface dataset:
Problem: There are three whole number $A$, $B$, $C$. $A\times B=35$, $B\times C=84$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$23$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$29$$ 
Answer: C
Analysis: $35=5\times 7$  $84=2\times 2\times 3\times 7$  Because $B$ is the factor both number contains, $B=7$  Thus, $A=5$, $C=12$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: If a whole number is divisible by $$111$$, then it must be divisible by ． 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$37$$ 
Answer: D
Analysis: Of the following choices, only $$37$$ is a factor of $$111$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: Which of the following is an odd number？ 
Options:
aoVal: A
content: $$490$$ 
aoVal: B
content: $$558$$ 
aoVal: C
content: $$36$$ 
aoVal: D
content: $$627$$ 
aoVal: E
content: $$452$$ 
Answer: D
Analysis: Odd numbers end with 1, 3, 5, 7, 9 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: Find the biggest prime number that is smaller than $$50$$. 
Options:
aoVal: A
content: $$43$$ 
aoVal: B
content: $$45$$ 
aoVal: C
content: $$47$$ 
aoVal: D
content: $$49$$ 
Answer: C
Analysis: We have $$3$$ prime numbers between $$40$$ and $$50$$: $$41$$, $$43$$, $$47$$, so we choose $$\text{C}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: Divide an odd number by $$4$$. The remainder is always． 
Options:
aoVal: A
content: odd　 
aoVal: B
content: even　 
aoVal: C
content: $$1$$ 
aoVal: D
content: prime　 
Answer: A
Analysis: Eliminate choices. Try several examples. A good example is $$7\div4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers
Problem: $$\left(\sqrt{64}+\sqrt{64}\right)^{2}=$$． 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$64$$ 
aoVal: C
content: $$128$$ 
aoVal: D
content: $$256$$ 
Answer: D
Analysis: $$\left(\sqrt{64}+\sqrt{64}\right)^{2}=\left(8+8\right)^{2}=\left(16\right)^{2}=256$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: The product of two whole numbers is $$5$$. What is the sum of these two numbers? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: If the product of two whole numbers is $$5$$, then one of the numbers is $$5$$ and the other is $$1$$. Their sum is $$6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: The thousands digit of the sum of 5+55+555+5555+55555 is ． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
Answer: A
Analysis: $$61725$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases
Problem: The sum of an odd number and an even number is always． 
Options:
aoVal: A
content: an odd number 
aoVal: B
content: an even number 
aoVal: C
content: a prime number 
aoVal: D
content: a multiple of $3$ 
Answer: A
Analysis: The sum is always odd. For example, $15 + 10 = 25$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers
Problem: Which of the following is correct?  $72=2^{x}\times3^{y}$  $539=z^{2}\times11$ 
Options:
aoVal: A
content: $x\textgreater y\textgreater z$ 
aoVal: B
content: $y\textgreater z\textgreater x$ 
aoVal: C
content: $x\textgreater y=z$ 
aoVal: D
content: $z\textgreater y\textgreater x$ 
aoVal: E
content: $z\textgreater x\textgreater y$ 
Answer: E
Analysis: $x=3$, $y=2$, $z=7.$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: The multiplication $$abc\times de=7632$$ uses each of the digits $$1$$ to $$9$$ exactly once. What is the value of $$b$$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$9$$ 
Answer: C
Analysis: Note first that $$7632 =2\times2\times2\times2\times3\times3\times53$$. Therefore either the two-digit number $$de = 53$$ or the three-digit number $$abc$$ is a multiple of $$53$$. Since the multiplication uses each of the digits $$1$$ to $$9$$ once and $$7632$$ contains a $$3$$, the option $$de= 53$$ is not allowable. Hence we need to find a three-digit multiple of $$53$$ that does not share any digits with $$7632$$ and divides into $$7632$$ leaving an answer that also does not share any digits with $$7632$$. We can reject $$2 \times 53 = 106$$ since it contains a $$6$$ but $$3 \times 53 = 159$$ is a possibility. The value of $$7632\div159$$ is $$2\times2\times2\times2\times3 = 48$$ which does not have any digits in common with $$7632$$ nor with $$159$$. We can also check that no other multiple of $$53$$ will work. Therefore the required multiplication is $$159 \times 48 = 7632$$ and hence the value of $$b$$ is $$5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization
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 $2022$? 
Options:
aoVal: A
content: $$2007$$ 
aoVal: B
content: $$2009$$ 
aoVal: C
content: $$2012$$ 
aoVal: D
content: $$2015$$ 
aoVal: E
content: $$2019$$ 
Answer: E
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 $3$ but leaves a remainder of $1$ when divided by $2$.  $2022$ can be divisible by both $2$ and $3$, so it is $2022-3=2019$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: ($$1999$$ Math League, Grade $$5$$, Question \#$$16$$)  The average of two odd numbers is always ． 
Options:
aoVal: A
content: odd$$ $$ 
aoVal: B
content: even$$ $$ 
aoVal: C
content: prime$$ $$ 
aoVal: D
content: whole$$ $$ 
Answer: D
Analysis: Avg of $$1$$ \& $$5$$ is $$3$$. Avg of $$1$$ and $$7$$ is $$4$$. Both are whole numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Odd and Even Applications
Problem: What is the missing number in the box?~\uline{~~~~~~~~~~}~  $$\square \div7=83 \rm R 4$$ 
Options:
aoVal: A
content: $$332$$ 
aoVal: B
content: $$339$$ 
aoVal: C
content: $$581$$ 
aoVal: D
content: $$585$$ 
Answer: D
Analysis: $$83\times7=581$$  $$581+4=585$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders->Relationship between Dividend, Divisor, Quotient and Remainder in Division
Problem: If the last $$2$$ digits of an integer are $$84$$, the integer must be divisible by． 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$16$$ 
Answer: A
Analysis: The sum of any multiple of $$100$$ and $$84$$ is divisible by $$4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: A $14$-digit. number $666666 XY 444444$ is a multiple of $26$. If $X$ and $Y$ are both positive, what is the smallest vaue of $X+ Y$? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: Since $1001$ is a multiple of $13$, $111111 = 111 \times 1001$ is also a multiple of $13$.  It follows that both $666666$ and $444444$ are both multiples of $26$.  $666666XY 444444 = 66666600000000 + XY 000000 + 444444$  $\Rightarrow XY$ must be divisible by $13$.  Smallest $X+Y=1+3=4$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: Which of the following numbers are prime numbers?  $$\textasciitilde$$  $103$~ ~ ~ ~ ~ ~ ~ ~ ~$115$~ ~ ~ ~ ~ ~ ~ ~ ~$127$~ ~ ~ ~ ~ ~ ~ ~ ~$139$  $$\textasciitilde$$ 
Options:
aoVal: A
content: Only $103$ 
aoVal: B
content: $103$ and $$127$$ 
aoVal: C
content: $103$ and $139$ 
aoVal: D
content: $103$, $127$, and $139$ 
Answer: D
Analysis: $115=5\times23$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Problem: Which of the following is a prime number? 
Options:
aoVal: A
content: $$39$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$79$$ 
aoVal: D
content: $$87$$ 
Answer: C
Analysis: We note that the other options are not prime numbers because:  $39 = 3 \times 13$  $50 = 5 \times 10$  $87 = 3 \times 29$  Hence by the process of elimination, $79$ is a prime number. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: It is known that the positive whole number $$n$$ is divisible by $$21$$ and by $$9$$. Which of the answers below can be the number of divisors of the number $$n$$? ($$2002$$ Math kangaroo Problems, Level $$7-8$$, Question \#$$21$$) 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: D
Analysis: $n$ is divisible by $63$. $63=3^{2}\times7$. $63$ has $(2+1)\times(1+1)=6$ divisors. Thus, $n$ has at least $6$ divisors, and $n$ can\textquotesingle t have an odd number of divisors. Option $D$ is the only possible answer. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number->Applying Theorem of the Number of Factors of a Number Directly->The Total Number of Factors
Problem: A prime number is a number greater than $$1$$ whose only whole number factors are itself and $$1$$. What is the smallest prime number greater than $$50$$? 
Options:
aoVal: A
content: $$51$$ 
aoVal: B
content: $$52$$ 
aoVal: C
content: $$53$$ 
aoVal: D
content: $$59$$ 
Answer: C
Analysis: A prime number is a number greater than $$1$$ whose only whole number factors are itself and $$1$$. Of the numbers greater than $$50$$, $$51 = 3\times17$$ and $$52 = 2\times26$$. The first prime is $$53$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: The $7$-digit numbers $\underline{74 A 52 B 1}$ and $\underline{326 A B 4 C}$ are each multiples of $3$ . Which of the following could be the value of $C$ ? (2014 amc 8 Problem, Question \#21) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$8$$ 
Answer: A
Analysis: Since both numbers are divisible by $3$ , the sum of their digits has to be divisible by three. $7+4+5+2+1=19$. To be a multiple of $3$ , $A+B$ has to be either $2$ or $5$ or $8 \ldots$ and so on. We add up the numerical digits in the second number; $3+2+6+4=15$. We then add two of the selected values, $5$ to $15$ , to get $20$ . We then see that $C=1,4$ or $7,10 \ldots$ and so on, otherwise the number will not be divisible by three. We then add 8 to 15 , to get 23 , which shows us that $C=1$ or $4$ or $7 \cdots$ and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1,4$, and $7$ . However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$ , but there is a $1$ , so $(\mathbf{A}) 1$ is our answer. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Find the greatest prime number that is smaller than $$50$$. 
Options:
aoVal: A
content: $$43$$ 
aoVal: B
content: $$45$$ 
aoVal: C
content: $$47$$ 
aoVal: D
content: $$49$$ 
Answer: C
Analysis: We have $$3$$ prime numbers between $$40$$ and $$50$$: $$41$$, $$43$$, $$47$$, so we choose $$\text{C}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: Which of the following is the correct expression of quinary (base $5$ numeral system)? 
Options:
aoVal: A
content: $\left (8231\right )\_5$ 
aoVal: B
content: $\left (2001\right )\_5$ 
aoVal: C
content: $\left (4341\right )\_7$ 
aoVal: D
content: $\left (2345\right )\_5$ 
Answer: B
Analysis: In quinary, the base number must be $5$, and all the digits in the brackets must be less than $5$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Properties and Applications of Number Bases
Problem: How many zeros does the number $$14^{3}\times15^{4}$$ 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: How many prime numbers are there between $90$ and $110$?  $$\textasciitilde$$  $$\textasciitilde$$ 
Options:
aoVal: A
content: $3$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: C
Analysis: $97$, $101$, $103$, $107$, $109$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: How many positive factors of $$36$$ are also multiples of $$4$$ ? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: B
Analysis: $$36={{2}^{2}}\times {{3}^{2}}$$, so the number of its factors would be $$\left( 2+1 \right)\times \left( 2+1 \right)=9$$, Among them, there are $$\left( 2+1 \right)\times 1=3$$ factors which has $${{2}^{2}}$$ as its factors. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: The product of two different primes hasdivisors. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: For any $$2$$ such primes, the factors are $$1$$, the primes, and their product. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: $$N$$ is a two$$-$$digit number. When $$N$$ is divided by $$9$$, the remainder is $$1$$. When $$N$$ is divided by $$10$$, the remainder is $$3$$. What is the remainder when $$N$$ is divided by $$11$$? ． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$7$$ 
Answer: E
Analysis: The smallest possible $$N$$ is $$73$$, and $$73 \div 11\rm R7$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: What are the last $2$ digits on the right in the expansion of the expression $2^{2018201}- 8$? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$44$$ 
aoVal: D
content: $$88$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: $2^{10}=1024\equiv24($mod$100)$  $\left (2^{10}\right )^{2}=1024^{2}\equiv76($mod$100)$  $\left (2^{10}\right )^{3}\equiv 76\times24\equiv24($mod$100)$  $\left (2^{10}\right )^{4}\equiv76($mod$100)$  $\cdots\cdots$  $$2^{2018201}-8=2\left (2^{2018200}\right )-8 \equiv 2\left (76\right )-8 \equiv 44($$mod$$100)$$. 
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: A student thinks of a natural number. She divides the number by $$9$$ and the remainder is $$7$$. What is the remainder when double 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: How many factors of $$36$$ are also multiples of $$4$$ ? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: B
Analysis: $$36={{2}^{2}}\times {{3}^{2}}$$, so the number of its factors would be $$\left( 2+1 \right)\times \left( 2+1 \right)=9$$, Among them, there are $$\left( 2+1 \right)\times 1=3$$ factors which has $${{2}^{2}}$$ as its factors. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: ♥ $$\times$$ ☺ $$=$$ ♦  ☺ is an even number.  which of the following gives an odd answer? 
Options:
aoVal: A
content: ♦ $$-\textasciitilde3$$ 
aoVal: B
content: ☺ $$+$$ ♦ 
aoVal: C
content: ☺ $$\times$$ ☺ 
aoVal: D
content: ♦ $$\times$$~♦ 
Answer: A
Analysis: ♥ $$\times$$ ☺ $$=$$ ♦  Since ☺ is an even number,~♦ must also be an even number.  ♦ $$-\textasciitilde3$$ is the only option to given an odd answer because even $$-$$ odd $$=$$ odd. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers
Problem: What is the remainder when we divide $19^{2021}$ by $4$? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: D
Analysis: We note that $19^{2} \equiv 1$ $( \text{mod} \textbackslash; 4)$ and hence:  $19^{2021} \equiv 1^{1010} \times 3 \equiv 3$ $( \text{mod} \textbackslash; 3)$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Congruence
Source: 其它
Problem: Two whole numbers differ by $$1$$. If one number has $$3$$ digits and the other has $$4$$ digits, what is their sum? 
Options:
aoVal: A
content: $$1001$$ 
aoVal: B
content: $$1100$$ 
aoVal: C
content: $$1999$$ 
aoVal: D
content: $$2001$$ 
Answer: C
Analysis: If $$1$$ more than a $$3$$-digit number is a $$4$$-digit number, then the numbers are $$999$$ and $$1000$$ and their sum is $$1999$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: What is the last digit of the smallest positive integer whose digits add to $$2022$$? ． 
Options:
aoVal: A
content: $$5 $$ 
aoVal: B
content: $$6 $$ 
aoVal: C
content: $$ 7 $$ 
aoVal: D
content: $$8 $$ 
aoVal: E
content: $$9$$ 
Answer: E
Analysis: For the number to be as small as possible, we need the number of digits to be as small as possible.  For instance, $$111\cdots 1 (2022\textasciitilde1\text{s})$$ has a digit sum of $$2022$$, but it is a much larger number than $$333\cdots 3 (674\textasciitilde3\text{s})$$, which also has a digit sum of $$2022$$. Clearly, to reduce the number of digits in the number, we need to make as many as possible of the digits in the number equal to $$9$$.  Now $$2022 \div9 = 224$$ remainder $$6$$, so the smallest positive integer with digit sum of $$2022$$ is $$699\cdots 9 (224\textasciitilde9\text{s})$$. Its last digit is $$9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders
Problem: The greatest odd factor of $$30$$ is ． 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$21$$ 
Answer: C
Analysis: The factors of $$30$$ are $$1$$, $$30$$, $$2$$, $$15$$, $$3$$, $$10$$, $$5$$, $$6$$, so the greatest odd factor of $$30$$ is $$15$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
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 $$ \begin{aligned} 16383 \& =2^{14}-1 \textbackslash\textbackslash{} \& =\left(2^{7}+1\right)\left(2^{7}-1\right) \textbackslash\textbackslash{} \& =129 \cdot 127 \end{aligned} $$ 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: Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$ ? (2018 AMC 8 Problem, Question \#14) 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$20$$ 
Answer: D
Analysis: If we start off with the first digit, we know that it can\textquotesingle t be $9$ since $9$ is not a factor of $120$ . We go down to the digit $8$ , which does work since it is a factor of $120$ . Now, we have to know what digits will take up the remaining four spots. To find this result, just divide $\frac{120}{8}=15$. The next place can be $5$ , as it is the largest factor, aside from $15$ . Consequently, our next three values will be $3,1$ and $1$ if we use the same logic. Therefore, our five-digit number is $85311$ , so the sum is $8+5+3+1+1=18 \Longrightarrow($ D) 18 . 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: Dividing a certain number by $$13$$ leaves a quotient of $$8$$ with a remainder of $$7$$. Find this number. 
Options:
aoVal: A
content: $$111$$ 
aoVal: B
content: $$121$$ 
aoVal: C
content: $$132$$ 
aoVal: D
content: $$115$$ 
Answer: A
Analysis: $$8\times 13+7=111$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders
Problem: A number can be xxxx and it will have a remainder of $2$ when divided by $4$. What is the maximum value of the number no larger than $2300$? 
Options:
aoVal: A
content: $$2288$$ 
aoVal: B
content: $$2290$$ 
aoVal: C
content: $$2294$$ 
aoVal: D
content: $$2098$$ 
aoVal: E
content: $$2300$$ 
Answer: D
Analysis: The number can be written as $$4n+2$$ and $$3x$$.  It must be a multiple of $3$ but leaves a remainder of $2$ when divided by $4$.  $2300$ can be divisible by both $2$ and $3$, so it is $2098$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: A $6$-digit number starting with $18$, $18ABCD$, is a multiple of $6$, $7$, $9$ and $10$. Find $\left (A +B + C+ D\right )$ for the smallest such number? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$28$$ 
aoVal: E
content: None of the above 
Answer: E
Analysis: The LCM of $6$, $7$, $9$ and $10$ is $2 \times 3^{2}\times 5 \times7= 630$.  $180 000=285 \times 630 + 450$,  ∴$$$$the$$$$ smallest number is $286 \times 630=180 180$,  Sum of the last 4 digits, $A + B + C+ D=9$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples