If a four-digit number $$5A2A$$ can be divisible by $20$, the digit that $$A$$ represents is  .

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules

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 $4$.


If $$1994$$ is added to any odd number, the sum will always be．

odd

even

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers

even number $$+$$ odd number $$=$$ odd number.


How many whole numbers are greater than $$9$$ and less than $$60$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers

There are $$60$$ whole numbers from $$0$$ to $$59$$. That\textquotesingle s $$50$$ without $$0$$ to $$9$$.


If $$a$$, $$b$$ are prime numbers, and $$3a+7b=41$$, then $$a+b=$$．

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers

Based on the laws relating to the parity in addition and multiplication, either $$a$$ or $$b$$ must be $$2$$. If $$a = 2$$, then $$b = 5$$, and $$a + b = 7$$; if $$b = 2$$, then $$a = 9$$; $$9$$ is not a prime number, which doesn\textquotesingle t match the conditions in the question. Therefore, we choose B.


The hundreds digit of a three-digit number is $$2$$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value

$$\rm Method$$ $$1$$:  Let the hundreds, tens, and units digits of the original three-digit number be $$a$$, $$b$$, and $$c$$, respectively. We are given that $$a=c+2$$. The original three-digit number is equal to $$100a+10b+c=100(c+2)+10b+c=101c+10b+200$$. The hundreds, tens, and units digits of the reversed three-digit number are $$c$$, $$b$$, and $$a$$, respectively. This number is equal to $$100c+10b+a=100c+10b+(c+2)=101c+10b+2$$. Subtracting this expression from the expression for the original number, we get $$(101c+10b+200)-(101c+10b+2)=198$$ . Thus, the units digit in the final result is $$8$$.  $$\rm Method$$ $$2$$:  The result must hold for any three-digit number with its hundreds digit being $$2$$ more than the units digit. $$301$$ is such a number. Evaluating, we get $$301-103=198$$. Thus, the units digit in the final result is $$8$$.


Cindy prepares burgers with two slices of beef each. A box of beef has $20$ slices. How many burgers can she prepare with all the three and a half box of beef? (Adapted from 2013 Math Kangaroo Problem, Level 3-4, Question \#7)

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders

There are $3\times20+20\div2=70$ slices of beef. $70\div2=35$


$$\overline{**45}$$, $$\overline{19*8}$$, $$\overline{23*1}$$, and $$\overline{3*49}$$ are four $4-$digit numbers with some unknown digits. Which number is possible to be a perfect square?

Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers

According to the ones digit of perfect squares, we can eliminate choice $B$.  A perfect square ending with $5$ must be a multiple of $25$, so we can eliminate choice $A$.  $${{48}^{2}}=2304$$ and $${{49}^{2}}=2401$$ so we can eliminate choice $C$.  ${57}^{2}=3249$. We choose option $D$.


Today is $$1^{st}$$ January $$2023$$, which is a Sunday. Teacher Angel has $$5$$ candies in her bag. Every time she completed her Tuesday and Thursday class, she will award herself with $$2$$ candies. On which date would she have $$15$$ candies in her bag?

$$15^{th}$$ January $$2023$$

$$16^{th}$$ January $$2023$$

$$17^{th}$$ January $$2023$$

$$18^{th}$$ January $$2023$$

$$19^{th}$$ January $$2023$$

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems


$$N$$ is a two$$-$$digit number. When $$N$$ is divided by $$5$$, the remainder is $$1$$. When $$N$$ is divided by $$11$$, the remainder is $$1$$. The smallest possible value of $N$ is~\uline{~~~~~~~~~~}~.(Adapted from $$2016$$ AMC $$8$$ Problem, Question \#$$5$$)

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder

Among $1, 56, 111$, \ldots~the smallest possible $$N$$ that satisfies the two conditions is $$56$$.


Let $$N$$ be the greatest five$$-$$digit number whose digits have a product of $$120$$. What is the sum of the digits of $$N$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization

To make the largest possible five$$-$$digit $$N$$, you should make the number in the biggest digit as large as possible. $$120=8\times5\times3\times1\times1$$. So, the sum of digits is $$8+5+3+1+1=18$$.


The first $2018$ integers ($1$, $2$, $3$, $\cdots$, $2017$, $2018$) are written on the blackboard. What is the minimum number of integers that should be erased from the blackboard, so that the last digit of the product of the remaining integers is $2$?

None of the above

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders

First, we need to remove all the integers which are multiples of $5$, otherwise the last digit of the products is $0$ or $5$.  Hence, $403$ integers need to be removed.  Next, note that the last digit of each of the products below is $6$.  $1\times2\times3\times4\times6\times7\times8\times9$,  $11\times12\times13\times14\times16\times17\times18\times19$,  $\cdots\cdots$  $2001\times2002\times2003\times2004\times2006\times2007\times2008\times2009$,  and the last digit of the product $2011\times2012\times2013\times2014\times2016\times2017\times2018$ is $4$.  Hence, we need to remove one more "$2$"and the last digit of the product will be $2$.  So the answer is $404$.


Summer has some nuts and wants to divide them equally to $5$ kids. Everyone can get $7$ nuts at most. How many nuts does Summer have at most?

There are $5-1=4$ nuts left after dividing at most. Thus, Summer has $5\times7+4=39$ nuts at most.


The ones digit of $$9\times 8\times 7\times 6\times 5\times 4\times 3\times 3\times 4\times 5\times 6\times 7\times 8\times 9$$ is．

Since $$5\times 4 = 20$$, the ones digit of the given product must be $$0$$.


Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $$\overline{1A2}$$. What is the missing digit $A$ of this $3$-digit number? (2014 AMC 8 Problem, Question \#8)


How many 0s are there at the end of the product $$2\times3\times5\times7\times8\times12\times25$$. ．

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization


What is the sum of the ten-thousands\textquotesingle{} digit and the millions\textquotesingle{} digit of $$1234567890$$?

The ten-thousands\textquotesingle digit plus the millions\textquotesingle~digit of $$1234567890$$ is $$6 + 4 = 10$$.


Which of the following numbers is the odd one out?

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers

$$1138 →1\times1\times3\times8=24$$  $$1226 →1\times2\times2\times6=24$$  $$1324 →1\times3\times2\times4=24$$  $$1416~→1\times4\times1\times6=24$$  $$1854 →1\times8\times5\times4=160$$  $$1854$$ is the odd one out as the product of its digits is not $$24$$.


The greatest odd factor of the product $$1\times 2\times 3\times 4\times 5\times 6$$ is．

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples

$$1\times 2\times 3\times 4\times 5\times (2\times 3)=2\times 4\times 2\times (1\times 3\times 5\times 3)=16\times 45$$.


The product of any two multiples of $$3$$ must be a multiple of．

Since $$3\times3 =9$$, the product of two multiples of 3 is divisible by $$9$$.


There are three whole numbers $A$, $B$ and $C$. $A\times B=35$, $B\times C=84$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$35=5\times 7$  $84=2\times 2\times 3\times 7$  Because $B$ is the factor both number contains, $B=7$  Thus, $A=5$, $C=12$.


Determine whether the following calculations give odd or even numbers.  (a) $$14327+21462-3583$$  (b) $$9377\times1525$$

Both Odd

Both Even

Odd, Even

Even, Odd

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers


When $$106$$ is divided by $$3$$, the remainder is．

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders


Find the value for lcm $$\left[ 4,6,8\right]$$.

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples

List the prime factorization for $$4$$, $$6$$ and $$8$$ first. $$4=2^{2}$$, $$6=2\times3$$, $$8=2^{3}$$. The largest exponent for $$2$$ is $$3$$,~~and the largest exponent for $$3$$ is $$1$$, thus the least common multiple for $$4$$, $$6$$, and $$8$$ is $$2^{3}\times3=24$$. We choose $$\rm C$$．


John, Mike, Emily and Nick want to buy some apples from Walmart where the apples are sold in pack of six. How many packs should they buy to get the same amount of apples for each one of them?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples

For each one of them to get the same amount of apples, the total amount should be a multiple of $4$, because there are $4$ of them.  The apples are sold in pack of six. $3$ packs contain $18$ apples; $5$ packs contain $30$; $6$ packs contain $36$; $7$ packs contain $42$.  Among the number of $18, 30, 36, 42$, only $36$ can be divided by $4$. $36$ is a multiple of $4$. Therefore, they should buy $6$ packs.


There is a book with 650 pages. Henry tears 31 paper from the book, each paper contains two pages. Is it possible that the sum of their page number equals to 2000?

$$Yes.$$

$$No.$$

If there are odd number of odd page number, the sum of page numbers is odd.


The product of all $$4$$ sides of a square is $$1296$$. The sum of all $$4$$ sides of the square is．

Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers

Let $x$ represent the length of each side of the square, then:  $x^{4} = 1296 \implies x^{2} = \sqrt{1296} = 36$.  Hence $x = \sqrt{36} = 6$ and $4x = 4\times 6 = 24$.


Change a digit of the number $45879$ to make the new five-digit number be divisible by $125$. What is the new five-digit number?

$45875$ and $45870$

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules

The last three digits must be divisible by $125$, so it can only be $875$.


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?

None of the above.


Divide 5 numbers $$2$$、$$3$$、$$12$$、$$15$$ and $$30$$ into two groups to make the product of numbers in each group the same, so the two groups are ．

（$$2$$, $$3$$, $$15$$），（$$12$$, $$30$$）

（$$2$$, $$12$$, $$15$$），（$$3$$, $$30$$）

（$$2$$, $$3$$, $$30$$），（$$12$$, $$15$$）

（$$12$$, $$3$$, $$15$$），（$$2$$, $$30$$）

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->Finding Factors Given the Product


How many prime numbers are there between $120$ and $140$?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers


Which of the following is not a prime number?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers

Since $$51=3\times17$$, $$51$$ is not a prime number.


Mom was cooking at home with the lights on when suddenly the power went out. Dad came home and pressed the switch four times, and then Mason came home and pressed the switch three times. When the power came back on, were the lights off or on?

On

Off


Mike has $$17$$ apples and $$19$$ pears. He puts every $5$ fruits in each box. At least how many more fruits does he need to get $8$ boxes of fruits? (Adapted from 2001 Math Kangaroo Problem, Level 3-4, Question \#7)


A square has an area of $144\text{cm}^{2}$, what is the length of each side of the square?

The length of each side of the square is $\sqrt{144} = 12\text{cm}$.


Vicky bought $26$ apples to divide to her $5$ cousins. If Vicky wants to give them all of the apples and also wants to make every cousin gets the same amount, at least how many more apples should Vicky buy?


If $$1\times2\times3\times4\times5\times6\times7\times8\times$$$$9\times10\times11\times12\times13\times$$$$14\times15 = 1307 674 368000$$, how many times does the digit "$$0$$" appear in the product $$10\times20\times30\times40\times50\times60\times$$$$70\times80\times90\times100\times110\times120\times130\times140\times150$$?

If $$1\times2\times3\times4\times5\times6\times7\times$$$$8\times9\times10\times11\times12\times13\times14\times15 = $$$$1307 674 368000$$, and we multiply each of these $$15$$ numbers by $$10$$, the new product will have an additional $$15$$ zeroes, and $$15+4=19$$.


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 $5789$?

Overseas Competition->Knowledge Point->Number Theory Modules

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 $2$ but leaves a remainder of $1$.  $5790$ can be divisible by both $2$ and $3$, so it is $5790-3=5787$.


The sum of three $2-$digit consecutive numbers is the maximum 2-digit number. What is their product?


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 $$9$$?

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 $$5$$.


Salah has collected more than $$20$$ football cards.  When he puts his cards in piles of four, he has three cards left over.  When he puts the cards in piles of five, he has four cards left over.  Which of the following could be the number of cards he has in total?

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem

The number of Salah\textquotesingle s cards is $$3$$ more than a multiple of $$4$$, and $$4$$ more than a multiple of $$5$$. Of the two-digit options given, the only one that satisfies both criteria is $$39(=9\times4+3 =7\times5+4)$$.


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?

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.


How many whole numbers greater than $$0$$ whose square is equal to its square root?

The only positive whole number whose square is equal to its square root is $$1$$.


How many numbers of the following are divisible by $3$?  $$\textasciitilde$$  $314\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} 528\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~899\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1024\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1356\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~3336$  $\textasciitilde$  $\textasciitilde$

The sums of the digits of $528$, $1356$, and $3336$ are multiples of $3$.


Of the following, which has the largest odd factor?

Keep dividing each by $$2$$ until you get an odd number.  $$\text{A}$$.$$2\times15$$.  $$\text{B}$$.$$2\times2\times 2 \times 2\times2\times1$$.  $$\text{C}$$.$$2\times2\times9$$.  $$\text{D}$$.$$2\times2 \times 2\times5$$.


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.

Possible

Not possible

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.


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?

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$\rightarrow60\div15$  $=4$.


Which of the following integers is not a multiple of $$45$$?

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 $$45$$.


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?

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 $385$.


How many zeros does the number $$12^{2}\times15^{3}$$ end with?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->The Number of Zeros at the end of a Product


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? ．

$$13 $$

$$23 $$

$$31 $$

$$41 $$

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers

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 $$41$$.


How many three-digit positive integers have an odd number of even digits?

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 .


How many different primes are in the prime factorization of $$2016$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)

$$2016=2\times2\times2\times2\times2\times3\times3\times7$$; there are $$3$$ different primes.


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 $$1000$$?

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).


There are three whole numbers $A$, $B$ and $C$. $A\times B=45$, $B\times C=50$. $A+B+C=$~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization

$45=3\times 3\times 5$  $50=2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=9$, $C=10$.


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?

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.


The product of three $2-$digit consecutive even numbers is $$\overline{\textbackslash\#\textbackslash\#\textbackslash\#2}$$. What is their sum?

Considering the ones digit and the value of the product, which is less than $10000$, only $14, 16,$ and $18$ match the condition.


Find the smallest prime number that is greater than $$90$$.

We only have one prime number between $$90$$ and $$100$$, which is $$97$$.


Find the smallest whole number between $$14$$ and $$40$$ that is divisible by $$3$$ and by $$4$$.

Multiple of $$3$$ and $$4$$ also a multiple of $$12$$.  $$12, 24, 36, 48, \cdots $$  Smallest multiple between $$14$$ and $$40$$ is $$24$$.


The sum of the prime factors of $$231$$ is．

Since $$231=3\times7\times11$$, the sum of its prime factors is $$3 +7+11=21$$.


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?

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers

Let the required result be $$S$$. Then  $$S= \left( {2+4+6+\cdots +100} \right)- \left( {1+3+5+\cdots +99} \right)~ $$  $$\textasciitilde\textasciitilde= \left( {2-1} \right) + \left( {4-3} \right) + \left( {6-5} \right) +\cdots +\left( {100-99} \right) $$  $$\textasciitilde\textasciitilde=50\times 1=50$$.


The product of two whole numbers is $$42$$. Their sum \emph{cannot} be．

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples


The sum of two prime numbers is $$99$$. What is the difference between the two prime numbers?

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 $$\rm C$$ .


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? ．

$$174 $$

$$185 $$

$$198 $$

$$209 $$

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 $$73 +223 = 296$$.


Which of the following numbers cannot be written as the sum of $$4$$ consecutive whole numbers?

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder


A multi-digit number$$\underbrace{20092009\cdot \cdot \cdot 2009}\_{n 2009s}736$$, can be divisible by ~$$11$$ . The smallest value of $$n$$ is．


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?

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems


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$$7832145-2167953=5664192$$.

√$$\times $$

$$\times $$√

$$\times $$$$\times $$

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Casting Out Nines

$$(1)$$ 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.


If a four-digit number $$5A2A$$ can be divisible by both $$4 $$ and $$5$$, the digit that $$A$$ represents is~\uline{~~~~~~~~~~}~.

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 $4$.


If my school has four times as many girls as boys, then the number of girls minus the number of boys \emph{could} be．

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders

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 $$3$$.


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?

$10=2\times 5$,  the length is $5$ and the width is $2$.  $(2+5)\times 2=14$.


Which of the following is divisible by all of the integers from $$1$$ to $$10$$ inclusive?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization

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.


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?

$\textasciitilde$  $\textasciitilde$  $\textasciitilde$  $\textasciitilde$


What is the least possible remainder when an even number is divided by $$7$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders

When $$14$$ (or any other even multiple of $$7$$) is divided by $$7$$, the remainder is $$0$$.


Which of the following is a multiple of $8$?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples

$3\times 8=24$, so $24$ is a multiple of $8$.


There are three whole number $A$, $B$, $C$. $A\times B=35$, $B\times C=84$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$35=5\times 7$  $84=2\times 2\times 3\times 7$  Because $B$ is the factor both number contains, $B=7$  Thus, $A=5$, $C=12$.


If a whole number is divisible by $$111$$, then it must be divisible by ．

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules

Of the following choices, only $$37$$ is a factor of $$111$$.


Which of the following is an odd number？

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers


Find the biggest prime number that is smaller than $$50$$.

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers


Divide an odd number by $$4$$. The remainder is always．

odd

even

prime

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers

Eliminate choices. Try several examples. A good example is $$7\div4$$.


Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers


The product of two whole numbers is $$5$$. What is the sum of these two numbers?

None of the above

If the product of two whole numbers is $$5$$, then one of the numbers is $$5$$ and the other is $$1$$. Their sum is $$6$$.


The thousands digit of the sum of 5+55+555+5555+55555 is ．

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases


The sum of an odd number and an even number is always．

an odd number

an even number

a prime number

a multiple of $3$

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers

The sum is always odd. For example, $15 + 10 = 25$.


Which of the following is correct?  $72=2^{x}\times3^{y}$  $539=z^{2}\times11$

$x\textgreater y\textgreater z$

$y\textgreater z\textgreater x$

$x\textgreater y=z$

$z\textgreater y\textgreater x$

$z\textgreater x\textgreater y$

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization


The multiplication $$abc\times de=7632$$ uses each of the digits $$1$$ to $$9$$ exactly once. What is the value of $$b$$?

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 $$5$$.


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 $2022$?

Overseas Competition->Knowledge Point->Number Theory Modules

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 $2022-3=2019$.


($$1999$$ Math League, Grade $$5$$, Question \#$$16$$)  The average of two odd numbers is always ．

odd$$ $$

even$$ $$

prime$$ $$

whole$$ $$

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Odd and Even Applications

Avg of $$1$$ \& $$5$$ is $$3$$. Avg of $$1$$ and $$7$$ is $$4$$. Both are whole numbers.


What is the missing number in the box?~\uline{~~~~~~~~~~}~  $$\square \div7=83 \rm R 4$$

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders->Relationship between Dividend, Divisor, Quotient and Remainder in Division


If the last $$2$$ digits of an integer are $$84$$, the integer must be divisible by．

The sum of any multiple of $$100$$ and $$84$$ is divisible by $$4$$.


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 $X+ Y$?

None of the above

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 $X+Y=1+3=4$.


Which of the following numbers are prime numbers?  $$\textasciitilde$$  $103$~ ~ ~ ~ ~ ~ ~ ~ ~$115$~ ~ ~ ~ ~ ~ ~ ~ ~$127$~ ~ ~ ~ ~ ~ ~ ~ ~$139$  $$\textasciitilde$$

Only $103$

$103$ and $$127$$

$103$ and $139$

$103$, $127$, and $139$

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers


Which of the following is a prime number?

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.


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 \#$$21$$)

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

$n$ 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.


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 $$50$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples

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 $$53$$.


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)

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.


Find the greatest prime number that is smaller than $$50$$.


Which of the following is the correct expression of quinary (base $5$ numeral system)?

$\left (8231\right )\_5$

$\left (2001\right )\_5$

$\left (4341\right )\_7$

$\left (2345\right )\_5$

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Properties and Applications of Number Bases

In quinary, the base number must be $5$, and all the digits in the brackets must be less than $5$.


How many zeros does the number $$14^{3}\times15^{4}$$ end with?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->The Number of Zeros at the end of a Product


How many prime numbers are there between $90$ and $110$?  $$\textasciitilde$$  $$\textasciitilde$$


How many positive factors of $$36$$ are also multiples of $$4$$ ?

$$36={{2}^{2}}\times {{3}^{2}}$$, 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.


The product of two different primes hasdivisors.

For any $$2$$ such primes, the factors are $$1$$, the primes, and their product.


$$N$$ 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 $$11$$? ．

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder

The smallest possible $$N$$ is $$73$$, and $$73 \div 11\rm R7$$.


What are the last $2$ digits on the right in the expansion of the expression $2^{2018201}- 8$?

None of the above

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders


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 $$9$$?

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 $$5$$.


How many factors of $$36$$ are also multiples of $$4$$ ?


♥ $$\times$$ ☺ $$=$$ ♦  ☺ is an even number.  which of the following gives an odd answer?

♦ $$-\textasciitilde3$$

☺ $$\times$$ ☺

♦ $$\times$$~♦

♥ $$\times$$ ☺ $$=$$ ♦  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.


What is the remainder when we divide $19^{2021}$ by $4$?

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Congruence

We note that $19^{2} \equiv 1$ $( \text{mod} \textbackslash; 4)$ and hence:  $19^{2021} \equiv 1^{1010} \times 3 \equiv 3$ $( \text{mod} \textbackslash; 3)$


Two whole numbers differ by $$1$$. If one number has $$3$$ digits and the other has $$4$$ digits, what is their sum?

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers

If $$1$$ more than a $$3$$-digit number is a $$4$$-digit number, then the numbers are $$999$$ and $$1000$$ and their sum is $$1999$$.


What is the last digit of the smallest positive integer whose digits add to $$2022$$? ．

$$5 $$

$$6 $$

$$ 7 $$

$$8 $$

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders

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 $$9$$.


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 ?

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 $$ 1+2+7=\text { (C) } 10 $$


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)

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 .


Dividing a certain number by $$13$$ leaves a quotient of $$8$$ with a remainder of $$7$$. Find this number.


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 $2300$?

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 $2098$.


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?

None of the above

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples


Which of the followings is not a multiple of $3$?


Which of the following products is an odd number?

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers

The product is odd if and only if every factor is odd.


How many positive factors of $$36$$ are also multiples of $$4$$ ?


A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? (2006 AMC 8 Problem, Question \#23)

The counting numbers that leave a remainder of 4 when divided by 6 are $4,10,16,22,28,34, \cdots$ The counting numbers that leave a remainder of 3 when divided by 5 are $3,8,13,18,23,28,33, \cdots$ So 28 is the smallest possible number of coins that meets both conditions. Because $4 \cdot 7=28$, there are (A) 0 coins left when they are divided among seven people.


Among numbers like $5$, $55$, $555$, $5555$, $$\cdots$$, how many of them are perfect squares?

Countless

Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers


Dividing a certain two$$-$$digit number by $$7$$ leaves a remainder of $$5$$; dividing it by $$11$$ leaves a remainder of $$9$$. What is the smallest possible value of this number?

The number when added by $$2$$ is divisible by $$7$$ and by $$11$$. Hence, the smallest value is $$7\times 11-2=75$$.


If a four-digit number $$\overline{5ab4}$$ is a perfect square number, then $$a+b=$$．


The natural numbers from $$1$$ to $$99$$ inclusive are divided into n groups such that the following conditions hold:  Each number belongs to exactly one group.  Each group contains at least two numbers.  If two numbers belong to the same group, then their sum is not divisible by $$3$$.  What is the smallest number $$n$$ which satisfies the above conditions? ．

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules

We can divide the $$99$$ numbers into $$3$$ groups according to their remainder when divided by $$3$$. $$1$$, $$4$$, $$7\_\cdots$$ they belong to the group $$A$$, since they leave a remainder of $$1$$ when divided by $$3$$. $$2$$, $$5$$, $$8\_\cdots$$ they belong to the group $$B$$, since they leave a remainder of $$2$$ when divided by $$3$$. $$3$$, $$6$$, $$9\_\cdots$$ they belong to the group $$C$$, since they leave a remainder of $$0$$ when divided by $$3$$. As we know, the members in group $$C$$ cannot get together. Also, when the member of group $$A$$ and the member of group $$B$$ get together, they will create the sum which is a multiple of $$3$$. So the $$33$$ members of group $$C$$ will be in $$33$$ groups. Group $$A$$ and group $$B$$ can join in the $$33$$ groups however they want.


If the four-digit number $$28X8$$ is divisible by $$3$$, how many possible values are there for $$X$$?

$$2+8+8=18$$, $$18$$ is multiple of $$3$$  Thus, $$X$$ itself must be a multiple of $$3$$, or $$0$$.  We have $$3, 6, 9$$ and $$0$$. Four possible value.


How many three-digit numbers have an odd number of factors?

Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers

It can be shown that a positive integer has an odd number of factors if and only if it is square. ~The smallest three-digit square number is $${{10}^{2}}=100$$ and the largest is $${{31}^{2}}=961$$. Hence there are $$31-9=22$$ three-digit numbers which have an odd number of factors.


♥ $$\times$$ ☺ $$=$$ ♦  ☺ is an even number.  Which of the following gives an odd answer?

♦ $$-\textasciitilde7$$

☺ $$\times$$ ☺

♦ $$\times$$~♦

♥ $$\times$$ ☺ $$=$$ ♦  Since ☺ is an even number,~♦ must also be an even number.  ♦ $$-\textasciitilde7$$ is the only option to given an odd answer because even $$-$$ odd $$=$$ odd.


Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textbackslash$ 1 A 2$. What is the missing digit $A$ of this $3$ -digit number? (2014 AMC 8 Problem, Question \#8)

Overseas Competition->Knowledge Point->Number Theory Modules

We know that a number is divisible by $11$ if the odd numbers added together minus the even numbers added together(or vice versa) is a multiple of $11$ . So, we have $1+2-A=$ a multiple of $11$ . The only multiple that works here is $0$ , as $11 \cdot 0=0$. Thus, $A=(D)$


Which of the following is not a factor of $$2016$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)


Jam has some pieces of candy. He wants to share with some kids. If he shares the candy among $8$ kids on average, there will be $2$ pieces left. If he shares the candy among $9$ kids on average, there will be $3$ pieces left. If he shares the candy among $10$ kids on average, there will be $4$ pieces left. How many pieces of candy are there?

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem


Chloe is working on this equation: $$475+17\times 58+990-19\times 32+33\times 111$$.  Her answer is $$5681$$.  Without calculating, do you think Chloe\textquotesingle s answer is correct or wrong? Explain why.

Correct

Wrong

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers

Her calculation was wrong. The answer is supposed to be an even number.  $$117\times32$$，$$19\times12$$ are even numbers and $$133\times11$$ is an odd number.  So in that case, we can think of the equation as:  odd + even + even - even + odd = even  Hence, the answer should be an even number! However, Chloe\textquotesingle s answer is an odd number.


If the four-digit number $3P78$ is divisible by $3$, how many possible values are there for $P$?

None of the above

use divisibility rule.  3+ 7+8+R = divisible by 3.  18 + R = divisible by 3.  smallest possible number for R is 0, max possible amount for R is 9  3x6=18 - 18 =0 (can be divisible by 3)  3x7=21 -18 = 3  3x 8= 24 -18 =6  3x 9= 27 - 18 = 9  Total = 4 ways.


John loves collecting stamps! If he divides the number of stamps he has by $$32$$, then he will have $$30$$ remaining stamps; if he divides the number of stamps he has by $$9$$, he will have $$7$$ remaining stamps; if he divides the number of stamps he has by $$7$$, he will have $$5$$ remaining stamps. How many stamps, at least, does John have?

The total number of stamps when divided by $$32$$, $$9$$, and $$7$$ leaves remainders of $$30$$, $$7$$, and $$5$$, respectively. In other words, the number when added by $$2$$ is divisible by $$32$$, $$9$$, and $$7$$. Hence, the smallest such number is $$32\times9\times7-2=2014$$. The answer is $$\rm A$$.


$${{2}^{3}}\times {{3}^{2}}\times 5\times 7$$

$${{2}^{4}}\times {{3}^{2}}\times 5$$

$${{2}^{4}}\times {{3}^{2}}\times 5\times 7$$

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers


Which of the following is the correct expression of quinary (base-$5$ numeral system)?

$\left (8231\right )\_5$

$\left (2001\right )\_5$

$\left (4341\right )\_7$

$\left (2345\right )\_5$

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Properties and Applications of Number Bases

In quinary, the base number must be $5$, and all the digits in the parentheses must be less than $5$.


$$2010$$ is divided by $$N$$ and gets a remainder of $$15$$. There are~\uline{~~~~~~~~~~}~possible values of $$N$$.

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number

$$2010-15=1995=3\times 5\times 7\times 19$$  $$\left( 1+1 \right)\times \left( 1+1 \right)\times \left( 1+1 \right)\times \left( 1+1 \right)=16$$ factors.  Remove $$1$$, $$3$$, $$5$$, $$7$$, and $$15$$ from them, we can get $$16-5=11$$ possible values.


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?

Let the original number be ab, and reverse it to be ba, calculated by the place value principle, ba-ab=72  10b+a-(10a+b)=72, 9b-9a=72, b-a=8, b=9 and a=1. So the original number is 19, last digit is 9.


How many three-digit numbers are divisible by $13$ ? (2004 AMC 8 Problem, Question \#18)

Let $k$ be any positive integer so that $13 k$ is a multiple of 13 . For the smallest three-digit number, $13 k\textgreater100$ and $k\textgreater\frac{100}{13} \approx 7.7$. For the greatest three-digit number, $13 k\textless999$ and $k\textless\frac{999}{13} \approx 76.8$. The number $k$ can range from 8 to 76 so there are $(\mathbf{C}) 69$ threedigit numbers.


Which of the following numbers are prime numbers?  $$\textasciitilde$$  $179$~ ~ ~ ~ ~ ~ ~ ~ ~$129$~ ~ ~ ~ ~ ~ ~ ~ ~$187$~ ~ ~ ~ ~ ~ ~ ~ ~$157$  $$\textasciitilde$$

Only $179$

$179$ and $$187$$

$179$ and $157$

$179$, $157$, and $187$


The square root of the square root of $$16$$ is ．

The square root of $$16$$ is $$4$$ and the square root of $$4$$ is $$2$$.


Candy buys $$15$$ bottles of drinks, and they are placed in order on the counter in the following way: $$2$$ bottles of milk, a bottle of juice, a bottle of coke, $$2$$ bottles of milk, a bottle of juice, a bottle of coke$$\ldots $$ According to the pattern, the $$15$$\textsuperscript{th} bottle is．

$$\text{milk}$$

$$\text{juice}$$

$$\text{coke}$$

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems


How many factors $2000$ that have more than $2$ factors are there? (As an example, $12$ has $6$ factors, namely $1$, $2$, $3$, $4$, $6$ and $12$. But $2$ and $3$ only two factors.)

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization

$2000=2^{3}\times5^{3}$, the number of its factors would be $(4+1)(3+1)=20$, but $1$, $2$, and $5$ do not meet the condition.


Which of the following prime factorization is incorrect?


In Walmart, apples are sold in pack of eight. How many apples will you get, if you buy five packs?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples


The sum of the first four square numbers is $$30$$.  What is the sum of the first five square numbers?


In the division expression $$28\div$$~\uline{~~~~~~~~~~}~$$=$$~\uline{~~~~~~~~~~}~$$\text{r}4$$, how many different pairs of number are there to fill the gaps?

We can use the equation: divisor $$\times$$ quotient $$=$$ dividend $$-$$ remainder, so here we can get divisor $$\times$$ quotient $$=28-4=24$$. Therefore, the only possibilities are $$1$$ and $$24$$, $$2$$ and $$12$$, $$3$$ and $$8$$, and $$4$$ and $$6$$ for a total of four possible combinations.


There are three whole numbers $A$, $B$ and $C$ ($B\neq 1$). $A\times B=45$, $B\times C=50$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$45=3\times 3\times 5$  $50=2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=9$, $C=10$.


Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder


Which of the followings is divisible by $$8$$ and $$7$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders


$a$ can only be $1$

$a$ can only be $37$

$a$ can be 1 or $37$

$a$ can be any number

"$$a$$ is a factor of $37$" means $37$ is divisible by $a$.


How many $$2$$-digit whole numbers have no odd factor except $$1$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers


Of the integers from $$1$$ to $$1000$$, how many are multiples of $$3$$, $$4$$, \emph{and} $$5$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples

A multiple of $$3$$, $$4$$, and $$5$$ is a multiple of $$60$$, and $$16\times60\textless{}1000 \textless{}17\times60$$.


There are three whole numbers $A$, $B$ and $C$ $(B\ne 1)$. $A\times B=15$, $B\times C=35$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$15=3\times 5$  $35=5\times 7$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=3$, $C=7$,  $A+B+C=15$.


In the division expression $$28\div$$~\uline{~~~~~~~~~~}~$$=$$~\uline{~~~~~~~~~~}~$$\text{R}4$$, how many different combinations are there for the quotient and the divisor ?


The sum of two prime numbers is $$99$$. What is the difference between the two numbers?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers

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 $$\rm C$$ .


The first $2018$ integers ($1$, $2$, $3$, $\cdots$, $2017$, $2018$) are written on the blackboard. What is the minimum number of integers that should be erased from the blackboard, so that the last digit of the product of the remaining integers is $2$?

None of the above

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders

First, we need to remove all the integers which are multiples of $5$, otherwise the last digit of the products is $0$ or $5$.  Hence, $403$ integers need to be removed.  Next, note that the last digit of each of the products below is $6$.  $1\times2\times3\times4\times6\times7\times8\times9$,  $11\times12\times13\times14\times16\times17\times18\times19$,  $\cdots\cdots$  $2001\times2002\times2003\times2004\times2006\times2007\times2008\times2009$,  and the last digit of the product $2011\times2012\times2013\times2014\times2016\times2017\times2018$ is $4$.  Hence, we need to remove one more "$2$"and the last digit of the product will be $2$.  So the answer is $404$.


odd

even

$88-29+1=60$  $60\div2=30$  Odd numbers all paired.


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?

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


How many solutions that can be expressed with positive integers does the equation below have? ($$1999$$ Math kangaroo Problems, Level $$7-8$$, Question \#$$26$$)  $$a^{2}b-1=1999$$

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->Finding Factors Given the Product


Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples

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 $$\rm A$$．


A computer can do $$7\times {{10}^{9}}$$ operations per second. Then, it can do~\uline{~~~~~~~~~~}~operations in $$5\times {{10}^{2}}$$ seconds.

$$35\times {{10}^{10}}$$

$$3.5\times {{10}^{11}}$$

$$3.5\times {{10}^{12}}$$

$$3.5\times {{10}^{19}}$$

$$3.5\times {{10}^{18}}$$

Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->The Relationship between Exponents and the Number of Factors


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?

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 $$3$$.


The $5$-digit number $\overline{2018U}$ is divisible by $9$. What is the remainder when this number is divided by $8$?

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules

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 $(\rm B)$ $3$.


Which of the options below is equivalent to $${{({{2}^{2}}\times {{3}^{3}}\times {{5}^{5}})}^{4}}$$?

$${{2}^{6}}\times {{3}^{7}}\times {{5}^{9}}$$

$${{2}^{2}}\times {{3}^{3}}\times {{5}^{20}}$$

$${{2}^{2}}\times {{3}^{3}}\times {{5}^{9}}$$

$${{2}^{8}}\times {{3}^{12}}\times {{5}^{20}}$$


What is the sum of all factors of $$24$$?


Add any two odd numbers. The ones\textquotesingle{} digit of the sum is always．

$$2 $$

$$$$prime

$$$$odd

$$$$even

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers

Add two numbers from: $$1$$, $$3$$, $$5$$, $$7$$, $$9$$. The result is always even.


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?

None of these.

Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Questions involving Square Numbers

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: $B$


What is the greatest number of consecutive integers such that the sum of the digits of none of them is divisible by $$5$$? ．

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: $$56$$, $$57$$, $$58$$, $$59$$, $$60$$, $$61$$, $$62$$, $$63$$.


The number $$95$$~\uline{~~~~~~~~~~}~$$94775998$$ is divisible by $$198$$. What is the missing digit?

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 $$9$$.


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?

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.


How many two-digit numbers can be divided by both 2 and 3 at the same time.

None of the above


How many of the whole numbers less than $$100$$ are $$10$$ greater than an odd whole number?


Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers


$$42$$ is not divisible by $$4$$, so neither is $$135 798 642$$.


Which of the following is \emph{not} a factor of $$2016$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)


What is the sum of the two smallest prime factors of $250$? (2007 AMC 8 Problems, Question \#3)

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers

The smallest prime factors of $250$ are $2$ and $5$. Thus, the sum is $2 + 5 = 7$.


What is the remainder when $$16+16+16+16$$ is divided by $$4$$ ．

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder

Since $$16\div 4$$ has a remainder of $$0$$, the remainder is $$0+0+0+0=0$$.


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? ．

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value

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 $$9$$.


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?

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders

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.


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 $X+ Y$?

None of the above

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 $X+Y=1+3=4$.


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.

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem

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 $210\times2-5=415.$


If $$1994$$ is added to any odd number, the sum will always be．

odd

even

even number $$+$$ odd number $$=$$ odd number.


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?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Word Problems Involving Factors and Multiples->Multiples Word Problems

$6=2\times3, 9=3\times3, 2\times3\times3\times5=90$. The number of students is a multiple of 90. $180\div90=2$.


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?

None of the above


(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:．

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Casting Out Nines

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 $$9$$.


There are three whole number $A$, $B$, $C$. $A\times B=45$, $B\times C=50$. $A+B+C=$~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization

$45=3\times 3\times 5$  $50=2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=9$, $C=10$.


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?

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders


If the sum of two whole numbers equals twice their difference, this sum cannot be．

If the sum is twice the difference, one \# is triple the other. The sum is divisible by $$4$$.


The product of $$2860$$ and $$m$$ is a square number. Find the smallest possible value of $$m$$.

Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers


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?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples

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.


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 $$130$$？

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.


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?

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems

The Lowest Common Multiple of $$6$$ and $$8$$ is $$24$$.  $$24+3=27$$.


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 $N$ ?

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 $$ \frac{a}{15 a-a}=\frac{a}{14 a}=\text { (C) } 1: 14 \text {. } $$


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 $$9$$?

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 $$5$$.


Which of the following numbers has the smallest prime factor? (2003 AMC 8 Problem, Question \#2)

The smallest prime factor is $2$, and $58$ is the only multiple of $2$ among these five numbers.


Given that $$6$$ and $$9$$ are multiples of $$3$$, which of the followings might not be a multiple of $$3$$?

$$9 + 6$$

$$9 - 6$$

$$2 \times 9$$

$$2 \times 9 + 1$$

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules

$$\text{A}$$: corresponding to $$\text{a}+\text{b}$$; $$\text{B}$$: corresponding to $$\text{a}-\text{b}$$; $$\text{C}$$: $$\text{na}$$, where $$\text{n}$$\emph{\emph{~}}is an integer. We choose $$\text{D}$$.


The ones digit of $$9\times 8\times 7\times 6\times 5\times 4\times 3\times 3\times 4\times 5\times 6\times 7\times 8\times 9$$ is．

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers

Since $$5\times 4 = 20$$, the ones digit of the given product must be $$0$$.


Which of the following is a composite number?


How many numbers of the following are divisible by $9$?  $$\textasciitilde$$  $452 \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} 387\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1057\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~108\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~496\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1233$  $\textasciitilde$  $\textasciitilde$


What is the product of the least common multiple of $$6$$ and $$18$$ and the greatest common factor of $$6$$ and $$18$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples

The least common multiple of $$6$$ and $$18$$ is $$18$$. The greatest common factor of $$6$$ and $$18$$ is $$6$$. Finally, $$6\times18 = 108$$.


$137$ and $237$

$137$ and $301$

$237$ and $301$

$137$ and $151$

$3$ is the factor of $237$ because $3 \times79 = 237$; $7$ is the factor of $301$ because $7 \times 43 = 301$.


The expression $$1\times 2\times 3\times \cdots \times n$$ has exactly $$128$$ consecutive zeros at the end of its result, then the maximum value of $$n$$ is~\uline{~~~~~~~~~~}~．

Overseas Competition->Knowledge Point->Number Theory Modules

There are $$128$$ zeros at the end of the result, and the estimated answer is close to $$500$$.  $$\left[ \frac{500}{5} \right]+\left[ \frac{500}{25} \right]+\left[ \frac{500 }{125} \right]=100+20+4=124$$,  Just add $$4$$ numbers including $$5$$: $$505$$, $$510$$, $$515$$, $$520$$, so the maximum value of $$n$$ that satisfies the condition is $$524$$ .


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 numbers on the invoice are the same. Given that the single price of this kind of apparatus is an integer, how much is it?


The hundreds digit of a three-digit number is $$2$$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?

$$\rm Method$$ $$1$$:  Let the hundreds, tens, and units digits of the original three-digit number be $$a$$, $$b$$, and $$c$$, respectively. We are given that $$a=c+2$$. The original three-digit number is equal to $$100a+10b+c=100(c+2)+10b+c=101c+10b+200$$. The hundreds, tens, and units digits of the reversed three-digit number are $$c$$, $$b$$, and $$a$$, respectively. This number is equal to $$100c+10b+a=100c+10b+(c+2)=101c+10b+2$$. Subtracting this expression from the expression for the original number, we get $$(101c+10b+200)-(101c+10b+2)=198$$ . Thus, the units digit in the final result is $$8$$.  $$\rm Method$$ $$2$$:  The result must hold for any three-digit number with its hundreds digit being $$2$$ more than the units digit. $$301$$ is such a number. Evaluating, we get $$301-103=198$$. Thus, the units digit in the final result is $$8$$.  ($$2010$$ AMC $$8$$ Problem, Question \#$$22$$)


I think of a number.  When I divide it by $$2$$, the remainder is $$1$$.  When I divide it by $$3$$, the remainder is $$2$$.  What is the smallest possible value of the number?~\uline{~~~~~~~~~~}~


Jam has some pieces of candy. He wants to share with some kids. If he shares the candy among $8$ kids equally, there will be $2$ pieces left. If he shares the candy among $9$ kids equally, there will be $3$ pieces left. If he shares the candy among $10$ kids equally, there will be $4$ pieces left. How many pieces of candy are there?


Which of the following number pairs do not contain any divisible relationship?

$(12+78), 2$

$(39+61), 3$

$(44+82), 4$

$(25+5100), 5$

If both numbers in the parentheses are multiple of the smaller number, the number in the parentheses should be divisible by the smaller number.


Becky is separating 53 cherries into bag A, bag B and bag C. It is known that there are even number of cherries in both bag A and bag B. Could you tell whether the number of cherries in bag C is even or odd number?

Odd

Even

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers

Number of cherries in bag C: 53 (odd number) - even - even  Number of odd character = 1  Therefore, the number of cherries in bag C is odd.


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 single price of this kind of apparatus is an integer, how much is it?


The product of $$4$$ odd numbers is always．

even

odd

less than $$20$$

greater than $$20$$

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers

The product of odd numbers is always odd.


There are three whole numbers $A$, $B$ and $C$ ($B\neq1$). $A\times B=21$, $B\times C=57$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$21=3\times 7$  $57=3\times 19$  Because $B$ is the factor both number contains, $B=3$  Thus, $A=7$, $C=19$.


Micky and Marcy want to make dumplings. They start with different speeds. There are two kinds of boxes. The smaller one of them could contain $9$ dumplings, and the bigger one could contain $17$ dumplings. Micky works with the smaller one, and Marcy works with the bigger one. What is the smallest number of dumplings that Micky needs to make until they start with a new box respectively at the same time?

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

The least common multiple of $9$ and $17$: $9 \times 17 = 153$


The $5$-digit number $\overline{2018U}$ is divisible by $9$. What is the remainder when this number is divided by $8$?

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 $(\rm B)$ $3$.


Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers


How many prime numbers are less than $$10$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)


A two-digit prime number is still prime when the digits of its first and tenth digits are exchanged. There are  such prime numbers.


In a Fibonacci-like sequence $$1,3,4,7,11,18\cdots $$(where each term is the sum of the two previous terms, starting from the third term), what is the remainder when the $$5555^{}\text{th}$$ term is divided by $$5$$？


If the product of an even number and an odd number is $$840$$, what is the largest possible value of this odd number?

If the product of an even and an odd number is $$840={2^{3}}\times105$$, then the largest possible value of the odd number is $$105$$.


Modify a digit in $$675479$$ so that this six-digit number is divisible by $$25$$. What is the modified six-digit number?

change the number on ones digit from $$9$$to$$5$$，$$675475$$．


Which of the following is a factor of $$380$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples


$$N$$ 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 $$11$$? ．

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder

The smallest possible $$N$$ is $$73$$, and $$73 \div 11\rm R7$$.


What is the least possible remainder when an even number is divided by $$7$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders

When $$14$$ (or any other even multiple of $$7$$) is divided by $$7$$, the remainder is $$0$$.


$$18$$ apples are equally given to $$9$$ kids. How many apples can every kid get?


Which of the following numbers is not prime?

$$13 $$

$$19 $$

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers


If $n$ and $m$ are integers and $n^{2}+m^{2}$ is even, which of the following is impossible? (2014 AMC 8 Problem, Question \#13)

$n$ and $m$ are even

$n$ and $m$ are odd

$n+m$ is even

$n+m$ is odd

none of these are impossible

The question asks which one is impossible, all we need to do is find one possible way that the others are possible. After trying, when $n$ and $m$ are both even or odd, the calculation works, so $D$ is not correct.


Which of the following is not a factor of $48$?


Find the value for lcm $$\left[ 12,18\right]$$.

List the prime factorization for $$12$$ and $$18$$ first. $$12=2^{2}\times3$$ and $$18=3^{2}\times2$$. The largest exponent for $$2$$ is $$2$$, and the larger exponent for $$3$$ is $$2$$, thus the least common multiple for $$12$$ and $$18$$ is $$2^{2}\times3^{2}=36$$. We choose $$\text{A}$$.


If $$a=1$$, $$b=2$$, and $$c=3$$, then $$\overline{abc}=123$$. Find the value of $$m$$ according to this rule.  $$\overline{m21}=8\times \overline{m9}+3m$$.

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value


Answer the question below:  If $$a$$, $$b$$ are prime numbers, and $$3a+7b=41$$, then $$a+b=$$．

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers

Based on the laws relating to the parity in addition and multiplication, either $$a$$ or $$b$$ must be $$2$$. If $$a = 2$$, then $$b = 5$$, and $$a + b = 7$$; if $$b = 2$$, then $$a = 9$$; $$9$$ is not a prime number, which doesn\textquotesingle t match the conditions in the question. Therefore, we choose B.


A kind of water plant grows so fast that it doubles every day. If one plant is put into the pond on the first day, it will turn to two plants on the second day, and on the $26$\textsuperscript{th} days, they can fill the pond. If $8$ water plants are put into the pond on the first day, how many days will it take to fill the pond?

It takes three days for one plant to turn to $8$ plants.


A number has the same digit in its tens place and its hundredth place. How many times greater is the value of the digit in the tens place than the value of the digit in the hundredth place?

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases

Let the digit be $d$  $10d\div 0.01d=1000$


If an equilateral triangle has in-teger sides, its perimeter \emph{cannot} be．

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders

The perimeter of an equilateral $$\triangle $$ with integer sides is divisible by $$3$$.


How many different primes are in the prime factorisation of $$2016$$?

$$2016=2\times2\times2\times2\times2\times3\times3\times7$$; there are $$3$$ different primes.


If $$1\times2\times3\times4\times5\times6\times7\times8\times$$$$9\times10\times11\times12\times13\times$$$$14\times15 = 1307 674 368000$$, how many times does the digit "$$0$$" appear in the product $$10\times20\times30\times40\times50\times60\times$$$$70\times80\times90\times100\times110\times120\times130\times140\times150$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization

If $$1\times2\times3\times4\times5\times6\times7\times$$$$8\times9\times10\times11\times12\times13\times14\times15 = $$$$1307 674 368000$$, and we multiply each of these $$15$$ numbers by $$10$$, the new product will have an additional $$15$$ zeroes, and $$15+4=19$$.


The $5$-digit number $\overline{2018U}$ is divisible by $9$. What is the remainder when this number is divided by $8$?

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules


If add $$1994$$ to any odd number, the sum will always be．

odd

even

prime

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers

even number $$+$$ odd number $$=$$ odd number.


Kate and Kerry are two dogs. Every $3$-hour, Kate goes to the dog park. Every $2$-hour, Kerry gos to the dog park. Today, Kate and Kerry first meet at $10:00\text{am}$. When will they meet each other again?


The characteristic of numbers that are divisible by $$4$$ is．

Last digit is divisible by $$4$$.

Last two digits are divisible by $$4$$.

Last three digits are divisible by $$4$$.

The sum of digits is divisible by $$4$$.

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders


$$\sqrt {2\times 4\times 8}\times \sqrt {8\times 8}=$$．

$$\sqrt {2\times 4\times 8}\times \sqrt {8\times 8}=8\times 8=64$$.


Which of the following numbers is not the square of a whole number?


Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples


When from any three-digit number we subtract that number written backwards, the difference will always be a number that is divisible by:．

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Casting Out Nines

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 $$9$$.


Which of the following has an odd number of whole-number factors?

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers

The whole-number factors of $$49$$ are $$1$$, $$7$$, and $$49$$.


How many positive integer factors of 2020 have more than 3 factors? (As an example, 12 has 6 factors, namely $1,2,3,4,6$, and 12 .) (2020 AMC 8 Problem, Question \#17)

Overseas Competition->Knowledge Point->Number Theory Modules

Since $2020=2^{2} \cdot 5 \cdot 101$, we can simply list its factors: $$ 1,2,4,5,10,20,101,202,404,505,1010,2020 . $$ There are 12 of these; only $1,2,4,5,101$ (i.e. 5 of them) don\textquotesingle t have over 3 factors, so the remaining $12-5=(\text{B}) 7$ factors have more than 3 factors.


From 202 to 2020 (including these two numbers), how many multiples of 9 are there?


Amy runs a lap around the track in $4$ minutes and Pawel in $5$ minutes. Amy and Pawel start to run around the track at the same time. After how many minutes will the boys meet at the starting point again?

It depends on the distance around the track

The answer is the least common multiple of $4$ and $5$, which is $20$.


How many positive factors of $$144$$ are also multiples of $$4$$?

$$144={{2}^{4}}\times {{3}^{2}}=4\times2^{2}\times {{3}^{2}}$$. Among them, there are $$\left( 2+1 \right)\times (2+1)=9$$ factors which has $$4$$ as its factor.


John, Emily and Nick want to buy some apples from Walmart where the apples are sold in pack of four. If they buy six packs and share the apples evenly, how many apples will each one of them get?


Each of the following results in an even number except．

$$952\div 136$$

$$952\times 136$$

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Odd and Even Applications

The sum, difference, and product of even numbers are even, but $$952\div 136=7$$ is odd.


What is the smallest prime number greater than $$47$$?


For a positive integer $n$, the factorial notation $n!$ repressents the product of the integer from $n$ to $1$, (For example, $6!=6\times5\times4\times3\times2\times1$.) What value of $N$ satisfies the following equation?  $$6!\times8!=8\times N!$$


Find the smallest multiple of $$7$$ that is greater than or equal to $$20 $$.

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples

$$2\times7=14$$, $$3\times7=21$$, since $$14$$ is smaller than $$20$$ and $$21$$ is the closest multiple of $$7$$ and greater than $$20$$, we choose $$\text B$$.


There are three whole number $A$, $B$, $C$. $A\times B=77$, $B\times C=132$. $A+B+C=$~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization

$77=7\times 11$  $132=2\times 2\times 3\times 11$  Because $B$ is the factor both number contains, $B=11$  Thus, $A=7$, $C=12$.


Which of the following number is a multiple of $9$?


The smallest number greater than $2$ that leaves a remainder of $2$ when divided by $3,4,5$, or $6$ lies between what numbers? (2012 AMC 8 Problem, Question \# 15)

$40$ and $50$

$51$ and $55$

$56$ and $60$

$61$ and $65$

$66$ and $99$

First, list number that leaves a remainder of $2$ when divided by $6$: $8,14,20,26,32,38,44,50,56,62\cdots $. Meanwhile, divide each number with $5$ to see it the number meets the condition. If the number divided by $5$ leaves a remainder of $2$, divide it by $4$ see if it still works.


The product of $$2$$ odd numbers is always.

divisible by $$3$$

odd

prime

even

Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers

The product of $$2$$ odd numbers, such as $$5\times7=35$$, is always odd.


3 hundreds

3 tens

3 thousands

3 ones

3 millions

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers->Understanding Numbers and Digits


A two-digit prime number is still prime when the digits of its first and tenth digits are exchanged. There are  such prime numbers.


Fill in the blanks below with the largest possible numbers:  ~\uline{~~~~~~~~~~}~$$\div16 = 5\rm R$$~\uline{~~~~~~~~~~}~.  ~\uline{~~~~~~~~~~}~$$\div16 = 5\rm$$ $\cdots\cdots$~\uline{~~~~~~~~~~}~.

105, 15

95, 15

88, 13

64, 16

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders

The largest possible remainder should be at least $$1$$ smaller than the divisor.


What is the missing number in the box?  $$7063000 =7000000+\boxed{?}+3000$$

$$600$$ ones

$$600$$ tens

$$600$$ hundreds

$$600$$ thousands

Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers

$600$ hundreds has the same value as $600\times100=60000$


Which of the following cases can\textquotesingle t make an even number?

An odd number$$+$$An even number

An even number$$+$$An even number

An odd number$$+$$An odd number

The sum of an odd number and an even number is always an odd~ number. The sum of two numbers which are both even or both odd is always an even number.


Divide 8 numbers $$15$$、$$18$$、$$21$$、$$22$$、$$42$$、$$44$$、$$50$$ and $$60$$ into two groups with 4 numbers in each group to make the product of numbers in each group the same, so the two groups are ．

（$$15$$, $$22$$, $$21$$, $$60$$），（$$18$$, $$44$$, $$42$$, $$50$$）

（$$15$$, $$42$$, $$44$$, $$60$$），（$$18$$, $$22$$, $$21$$, $$50$$）

（$$15$$, $$44$$, $$21$$, $$60$$），（$$18$$, $$22$$, $$42$$, $$50$$）

（$$15$$, $$44$$, $$21$$, $$50$$），（$$18$$, $$22$$, $$42$$, $$60$$）

Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->Finding Factors Given the Product


A kind of water plant grows so fast that it doubles every day. If one plant is put into the pond on the first day, two plants will grow on the second day, and on the twenty-sixth days, they will just fill the pond. If eight water plants are put into the pond on the first day, how many days it will take to fill the pond?

It takes three days for one plant to develop to eight plants.


What is the remainder when $$12+34+56+89+90$$ is divided by $$10$$?


Only one of the following four numbers is a perfect square. Which one is it?

Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers


What is the sum of the prime factors of $2010$? (2010 AMC 8 Problems, Question \#14)

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers

First, we must find the prime factorization of $2010.2010=2 \cdot 3 \cdot 5 \cdot 67$. We add the factors up to get $(\mathbf{C}) 77$.


The whole numbers from $$1$$ to $$2016$$ inclusive are written on a blackboard. Moritz underlines all the multiples of two in red, all the multiples of three in blue and all the multiples of four in green. How many numbers does Moritz underline exactly twice?

There is no number that is both a multiple of three and a multiple of four without also being a multiple of two. Hence, the numbers underlined exactly twice are those that are a multiple of two and of three but not of four and those that are a multiple of two and four but not of three. The first set of numbers consists of the set of odd multiples of six. Since $$2016 \div 6 = 336$$, there are $$336$$ multiples of $$6$$ in the list of numbers and hence $$336 \div 2 = 168$$ odd multiples of six that would be underlined in red and blue but not green. The second set of numbers consists of two out of every three multiples of four and, since $$2016 \div 4 = 504$$, there are $$3\times 504 = 336 $$ numbers that would be underlined in red and green but not blue. Hence there are $$168+ 336 = 504$$ numbers that Moritz would underline exactly twice.


Which of the following is a prime number?

Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers

We note that the other options are not prime numbers because:  $38 = 2 \times 19$  $45 = 4 \times 9$  $57 = 3 \times 19$  Hence by the process of elimination, $53$ is a prime number.


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 $$9$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder

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 $$5$$.


$$100110011001$$ is divisible by $$3$$, since the sum of its digits is $6$ which is divisible by $3$.


The product of any $$3$$-digit whole number and any $$2$$-digit whole number can contain at most how many digits?

The largest product of a $$3$$-digit number and a $$2$$-digit number is $$999\times99 = 98901$$. That product has $$5$$ digits.