What is the greatest natural number $$n$$ such that $$n+27$$ and $$n-62$$ are squares of natural number?

Such a number does not exist

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


If I write all the whole numbers from $$1$$ to $$100$$ in words, how many times will I write the letter V?

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

There are $$10$$ sevens in the unit digit, $$10$$ seventys, $$9$$ fives in the units (as fifteen is not fiveteen) plus eleven and twelve.


How many multiples of $5$ are there from $1$ to $50$?

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


Which of the following prime factorizations below is correct?

$$85=5\times 17$$

$$64=8\times 8$$

$$52=1\times 2\times 2\times 13$$

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

①A prime number is a natural number greater than $$1$$ that has no other factors than $$1$$ and itself.  ②Decomposing a composite number into the product of several prime factors is called decomposing prime factors.  ③$$1$$ is neither a prime number nor a composite number. So $$\text{D}$$ is not correct.  The $$\text{A}$$ option is also incorrect. $$8$$ is not a prime number, so $$\text{C}$$ is incorrect.


Which of the followings is the least common multiple for $$25$$ and $$125$$?

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

We do prime factorization for $$25$$ and $$125$$ first. $$25={{5}^{2}}$$ and $$125={{5}^{3}}$$. We have three $$5$$'s, thus the least common multiple for $$25$$ and $$125$$ is $$125$$. We choose $$\text{B}$$．


Of the multiples of $$7$$ that exceed $$7$$, how many are factors of $$700$$?

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


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 $$49$$.


There are three whole numbers $A$, $B$ and $C$. $A\times B=55$, $B\times C=100$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$55=5\times 11$  $100=2\times 2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=11$, $C=20$.


What is the difference between the largest single-digit prime number and the smallest three-digit prime number?

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

The largest single-digit prime number is $$7$$ and the smallest three-digit prime number is $$101$$; their difference is $$101 -7=94$$.


The product of \emph{all} the prime numbers less than $$10$$ is divisible by ．

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


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


Danni wants to use the numbers $1$, $3$, $7$, and $9$ to make prime numbers less than $100$. How many different prime numbers can she make? (She can use the same number more than once.)  $$\textasciitilde$$

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

Using these $4$ numbers, we can make $20$ different numbers which are less than $100$. However, only these $12$ numbers are prime numbers. $3; 7; 11; 13; 17; 19; 31; 37; 71; 73; 79; 97$.


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

Add the digits to test: for $$603$$, $$6+0+3 =9$$.


Ivan is puting 13 cakes into box A, box B and box C. It is known that there are odd number of cakes in both box A and box B. Could you tell whether the number of cakes in box C is even or odd number?

Odd

Even

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

Number of cakes in box C: 13 (odd number) - odd - odd  Number of odd character = 3  Therefore, the number of cakes in box C is odd.


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 $$0$$.


The largest whole-number multiple of $$7$$ less than $$200$$ is ．

$$200\div7\textgreater28$$, so the largest such multiple is $$28\times7 =196$$.


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 $99$

Overseas Competition->Knowledge Point->Number Theory Modules

To find the answer to this problem, we need to find the least common multiple of $3,4,5,6$ and add $2$ to the result. The least common multiple of the four numbers is $60$ , and by adding $2$ , we find that that such number is $62$ . Now we need to find the only given range that contains $62$ . The only such range is answer (D), and so our final answer is $(\text{D}) 61$ and $65$


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


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 $$1$$ million is less than the number．

$$1$$ thousand

$$10$$ thousand

$$100$$ thousand

$$1$$ billion

The number $$1$$ billion is $$1$$ thousand times $$1$$ million.


$\overline{392AB}$ is a multiple of $45$, and $\overline{B34}$ is a three-digit even number. What is the sum of $A$ and $B$?

$4$ or $13$

$45=5\times9$  $\overline{392AB}$ should be ended with $0$ or $5.$ But $0$ cannot be the first digit. Thus, $B$ can only be $5$ and $A$ should be $8.$


There are three whole number $A$, $B$, $C$. $A\times B=21$, $B\times C=60$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$21=3\times 7$  $60=2\times 2\times 3\times 5$  Because $B$ is the factor both number contains, $B=3$  Thus, $A=7$, $C=20$.


Two whole numbers have a least common multiple of $60$.  -Each number is less than or equal to $12$.  -The greatest common factor of the two number is $2$.  What are the two numbers?

$6$ and $10$

$5$ and $12$

$10$ and $12$

$12$ and $16$

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

A: The least common multiple of $6$ and $10$ is $30$.  B: Two numbers are relatively prime.  C: The least common multiple is $60$, and the greatest common factor is $2$.  D: The least common multiple is $48$, and the greatest common factor is $4$.


There are some books in the central library. If Adam divides the number of the books in the library by $32$, there will be $30$ books remained; if he divides the number of books in the library by $9$, there will be $7$ books remained; if he divides the number of books in the library by $7$, there will be $5$ books remained. How many books, at least, are there in the central library?

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

The total number of books 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 $$\rm A$$.


Order all prime numbers from least to greatest. What is the sum of the next two prime numbers after $43$?


Pip swam $$3$$ back and forth in the lane for a total of $$156$$ meters, how long is the lane in this pool?

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

one back and forth $$156\div3=52$$meters  length of the lane$$52\div2=26$$meters


How many positive factors of $$360$$ are also multiples of $$12$$?

$$360={{2}^{3}}\times {{3}^{2}}\times5$$, so the number of its factors would be $$\left( 3+1 \right)\times \left( 2+1 \right)\times\left( 1+1 \right)=24$$. Among them, there are $$2\times 2\times(1+1)=8$$ factors which have $${{2}^{2}}\times 3$$ as its factors.


What is the ones digit of the product $$80\times 70\times 60\times 50\times 40\times 30\times 20\times 10\times 5\times 2$$?

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

Since $$10$$ is a factor of this product, the ones digit is $$0$$.


A recurring decimal can also be written by putting a bar over the digits that repeat. Thus, $$0.\overline {123}$$ means $$0.123123\cdots $$. The sum of $$0.\overline {234}$$, $$0.\overline {342}$$, and $$0.\overline {432}$$ is~\uline{~~~~~~~~~~}~.

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


Fill in the blank:~\uline{~~~~~~~~~~}~is $$4$$ tens $$5$$ ones greater than $$2$$ tens $$7$$ ones.

None of the above


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

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


If I multiply three different prime numbers, the product must have  positive divisors.

$$ 3 $$

$$5 $$

$$6 $$


Students\textquotesingle~seat numbers are from $$1$$ to $$35$$. How many times does digit $$3$$ appear? (adapted from $$2002$$ Math Kangaroo Problem, Level $$3$$ - $4$, Question \#$9$)

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

The digit $3$ appears in the ones place: $3$, $13$, $23$,  The digit $3$ appears in the tens place: $30, 31, 32, 33, 34, 35$  In total, $3 + 6 = 9$.


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? ．

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 $$9$$.


An unknown number is an odd number greater than $$50$$ but less than $$100$$. The number is a multiple of $$3$$ and $$7$$. What is the value of the unknown number?

None of the above

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

A multiple of $$3$$ and $$7$$ also a multiple of $$21$$  $$21\times3=63$$  $$21\times4=even$$  $$21\times5=105$$, exceed  Thus, only $$63$$


Among the numbers below, how many numbers can be divisible by $4$?  $234$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $789$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $7756$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $8865$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $3728$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $\textasciitilde$ $8064$


Della has a box of ping-pong balls. No matter she counts the balls $8$ by $8$, $10$ by $10$, or $12$ by $12$, there are always $3$ balls left. How many ping-pong balls are there in the box at least?

If the ping pong balls in the box are reduced by $3$, then there will be no extra balls when you count them $8$ by $8$, $10$ by $10$, or $12$ by $12$. It means that the number of the ping pong balls is a common multiple of $8$, $10$, and $12$ after being reduced by $3$. If you want to know how many ping pong balls there are at least, you can first find the least common multiple of $8$, $10$ and $12$, and then add $3$ to get the answer.  $$\begin{array}{l} {2\left\textbar{} \underline{\textasciitilde8\textasciitilde\textasciitilde10\textasciitilde\textasciitilde12}\right. }\textbackslash\textbackslash{\textasciitilde2\left\textbar{} \underline{4\textasciitilde\textasciitilde\textasciitilde5\textasciitilde\textasciitilde\textasciitilde6}\right. }\textbackslash\textbackslash{\textasciitilde\textasciitilde\textasciitilde\textasciitilde\textasciitilde2\textasciitilde\textasciitilde\textasciitilde5\textasciitilde\textasciitilde\textasciitilde3} \end{array}$$  $2\times2\times2\times5\times3=120$  $120+3=123$


$$1^{st}$$ November $$2016$$ is a Tuesday, what day will it be $$30^{th}$$ November of the same year?

Monday

Wednesday

Thursday

Saturday

None of the above

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

$$30-1+1=30$$  $$30\div7=4R2$$  Tuesday -\/-\textgreater~Wednesday.


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

$$123123123$$ is divisible by $$3$$, since the sum of its digits is $18$ which is divisible by $3$.


How many different prime numbers, when multiplied by $$11$$, have an even number as their product?

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

The product of two odd numbers is odd, so there is only one even product, $$2\times 11$$.


Which of the following statement is correct?

All prime numbers are odd numbers.

In every $3$ consecutive numbers, there must be a composite number.

All even numbers are composite numbers.

The sum of two different odd number must be a composite number.

A: Any non-$3$ multiple of 3 are composite  B: $1$, $2$, $3$ does not have composite number  C: $2$ is prime number  D: $1+$any odd number is multiple of $2$.


Let $a$ and $b$ be positive integers such that $a+a b=1443$ and $a b+b=1444$. Find $10 a+b$.

$$ \left\textbackslash{\begin{array}{l} a(b+1)=1443=3 \times 13 \times 37=37 \times 39 \textbackslash\textbackslash{} b(a+1)=1444=38^{2} \end{array}\right. $$  Compare and $a=37, b=38 .$ So $10 a+b=370+38=408$.


When Ringo places his marbles into bags with $6$ marbles per bag, he has $4$ marbles left over. When Paul does the same with his marbles, he has $3$ marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with $6$ marbles per bag. How many marbles will be leftover? (2012 AMC 10B Problems, Question \#4)

In total, there were $3+4=7$ marbles left from both Ringo and Paul.We know that $7 \equiv 1(\bmod 6)$. This means that there would be $1$ marble leftover.


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.


When two numbers are divided and the quotient is $$11$$ and remainder is $$5$$, the smallest value of the dividend is ．


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 make at least at the beginning? Find the sum of the three digits.

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 $210\times2-5=415.$


There are exactly $$3$$ prime numbers between．

$$10$$ and $$20 $$

$$20$$ and $$30$$

$$30$$ and $$40$$

$$40$$ and $$50$$

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


The $4$-digit number $32B9$ is divisble by $3$. If $B$ is even, find the digit $B$.

$$3+2+B+9$$  = 14+B is divisble by 3.  Max single digit possible is 9. Then 14+9 = 23. Range is between 15 to 23.  Multiple of 3 - 3 x 7=21, 3 x 6 =18, 3 x 5 = 15.  Since, it should be even number, 18-14=4.


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? (2010 AMC 8 Problem, Question \#22)

The result must hold for any three-digit number with 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 $(\mathbf{E}) 8$


Which of the following is a factor of $30$ ?


A $$2-$$digit number is such that the product of the digits adding the sum of the digits is equal to the number itself. What is the ones digit of the number?


Of the first $$100$$ whole numbers,~\uline{~~~~~~~~~~}~use the digit $$2$$ at least once.


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 have $${{2}^{2}}$$ as its factors.


What are the last three digits of the answer to the calculation below?  $$123\times 124\times 125\times 126\times 127$$

The product $$123 \times124\times125\times126 \times127$$ is a multiple of $$125$$; moreover, it also has a factor of $$2$$ three times, from $$124\left( 2\times 2\times 31 \right)$$ and from $$126\left( =2\times 63 \right)$$. Therefore it is a multiple of $$125 \times2\times2\times2= 1000$$, and so it must end in $$000$$. Alternatively, working from the options, it is easily seen that the product is a certainly an even multiple of $$5$$-so its unit digit is $$0$$.


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 $$15+4=19$$.


Malcolm wants to visit Isabella after school today and he knows the street where she lives but does not know her house number. She tells him,~"My house number has two digits, and all of the following four statements about it are true."  $(1)$ It is a prime number.  $(2)$ It is less than $40$.  $(3)$ One of its digits is $3$.  $(4)$ Another digit is an even number.  This information allows Malcolm to determine Isabella\textquotesingle s house number. What is the house number? (adapted from 2017 AMC 8 problem, Question \#8)

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

According to information $(3)$ and $(4)$, we know $3$ must be in the ones place and the digit in the tens place is an even number. Thus, it should be one of the numbers from $23$, $43$, $63$ and $83$. According to information $(1)$, it can be $23$, $43$, and $83$. According to information $(2)$, it can only be $23$.


Malcolm wants to visit Isabella after school today and he knows the street where she lives but doesn't know her house number. She tells him, ``My house number has two digits, and all of the following four statements about it are true.''  $(1)$ It is a prime number.  $(2)$ It is less than $40$.  $(3)$ One of its digits is $3$.  $(4)$ Another digit is an even number.  This information allows Malcolm to determine Isabella's house number. What is the house number? (adapted from 2017 AMC 8 problem, Question \#8)

According to information $(3)$ and $(4)$, we know $3$ must be in the ones place and the digit on the tens place is an even number. Thus, it should be one of the numbers from $23$, $43$, $63$ and $83$. According to information $(1)$, it can be $23$, $43$, and $83$. According to information $(2)$, it can only be $23$.


Without calculating, can you quickly choose the correct answer?  $$1236 + 3217$$ = ?


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 953?

$$Yes.$$

$$No.$$

If there are odd number of odd page number, the sum of page numbers is odd. Therefore, it could be 953.


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

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


What is the smallest prime number greater than $$59$$?


The sum of the thousands\textquotesingle{} digit and the tens\textquotesingle{} digit of $$12345$$ is．

The thousands\textquotesingle{} digit is $$2$$ and the tens\textquotesingle{} digit is $$4$$. The sum is $$6$$.


What is the sum of the digits in the number one million?

one

one hundred

one thousand

one million


Vivian bakes a cuboid cake with a dimension of $6\times8\times10$ and cuts it into many small cubical cakes with a dimension of $2\times2\times2$. She wants to equally divide the cake without remainder among $x$ people so that each person can get at least one small cake. How many possible values of $x$ are there?

$6\times8\times10 \div(2\times2\times2)=60.$ $60=2^{2}\times3\times5$, which means it has $3\times2\times2=12$ factors.


Four numbers are chosen from the set $3$, $6$, $12$, $27$, $48$ so that the product is a perfect square. What is the number that was not chosen?

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

Obsever that $3\times27=9^{2}$ and $12\times27=24^{2}$ are both squares. Hence, their product is also a perfect square and the fifth number must be $6$.


Which of the following statements in incorrect?

$$108$$ is a multiple of $$3$$ and $$9$$

$$100$$ is a multiple of $$10$$, but it is not a multiple of $$5$$

$$164$$ is a multiple of $$4$$

$$132$$ is a multiple of $$6$$

$$\rm A$$: $$1+0+8=9$$, it is a multiple of $$3$$ and $$9$$;  $$\rm B$$: $$100$$ ends with $$0$$, thus it is a multiple of $$10$$ and $$5$$;  $$\rm C$$: $$164$$ ends with $$64$$, and $$64$$ is a multiple of $$4$$, thus $$164$$ is also a multiple of $$4$$;  $$\rm D$$: $$132$$ is a multiple of $$2$$ and $$3$$, thus it is a multiple of $$6$$.  So we choose $$\rm B$$.


The positive integers from $$1$$ to $$150$$ inclusive are placed in a $$10$$ by $$15$$ grid so that each cell contains exactly one integer. Then the multiples of $$3$$ are given a red mark, the multiples of $$5$$ are given a blue mark, and the multiples of $$7$$ are given a green mark.  How many cells have more than $$1$$ mark?

As $$3$$ and $$5$$ are coprime, the squares that have more than one mark are multiples of both $$3$$ and $$5$$, $$\left( {} \right.$$multiples of $$15$$$$\left. {} \right)$$; or multiples of both $$3$$ and $$7$$, $$\left( {} \right.$$multiples of $$21$$$$\left. {} \right)$$; or multiples of $$5$$ and $$7$$, $$\left( {} \right.$$multiples of $$35$$$$\left. {} \right)$$; or multiples of $$3$$, $$5$$ and $$7$$, $$\left( {} \right.$$multiples of $$105$$$$\left. {} \right)$$. However, the latter will be included in all of the first three categories. Between $$1$$ and $$150$$ inclusive, there are ten multiples of $$15$$, seven multiples of $$21$$ and four multiples of $$35$$, making a total of $$21$$ multiples. However, there is one multiple of $$3$$, $$5$$ and $$7$$ between $$1$$ and $$150$$, namely $$105$$. So $$105$$ has been counted three times in those $$21$$ multiples, but corresponds to exactly one marked square. Therefore the total number of marked squares is $$21−2=19$$.


Yes

No

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


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

The sum of each numbers remainder can be divided by $$4$$ to get $$2$$: $$123\div 4$$ R $$3$$；$$234\div 4$$ R $$2$$；$$345\div 4$$ R $$1$$; Therefore, $$\left( 3+2+1 \right)\div 4$$ R $$2$$．


When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of a two-digit score. By which of the following might her incorrect sum have differed from the correct one?

Let the two digits be $$a$$ and $$b$$.  The correct score was $$10a+b$$. Clara misinterpreted it as $$10b+a$$. The difference between the two is $$\left\textbar{} 9a-9b\right\textbar$$ which factors into $$\left\textbar{} 9(a-b)\right\textbar$$. Therefore, since the difference is a multiple of $$9$$, the only answer choice that is a multiple of $$9$$ is $$45$$ .  ( $$2013$$ AMC $$8$$ Problem, Question \#$$13$$)


Which of the following prime factorization below is correct?

$$85=5\times 17$$

$$64=8\times 8$$

$$52=1\times 2\times 2\times 13$$

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

①A prime number is a natural number greater than $$1$$ that has no other factors than $$1$$ and itself.  ②Decomposing a composite number into the product of several prime factors is called decomposing prime factors.  ③$$1$$ is neither a prime number nor a composite number. So $$\text{D}$$ is not correct.  The $$\text{A}$$ option is also incorrect. $$8$$ is not a prime number, so $$\text{C}$$ is incorrect.


A whole number divisible by $$6$$ and by $$14$$ need \emph{not} be divisible by．

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

As an example, $$42$$ is divisible by $$6$$ and by $$14$$ but not by $$12$$.


Allen has some apples, and the number of apples is more than $2200$ but less than $2300$. If he distributes them among $12$ children evenly, there will be $11$ apples left. If he distributes them among $13$ children evenly, there will be $7$ apples left.~ If he distributes them among $14$ children evenly, there will be $3$ apples left. Which is the correct range of the number of pens?

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

If he distributes them among $12$ children evenly, there will be $11+12\times4=59$ apples left. If he distributes them among $13$ children evenly, there will be $7+13\times4=59$ apples left.~ If he distributes them among $14$ children evenly, there will be $3+14\times 4=59$ apples left. So at least he takes $59+12\times 13\times 14=2243$.


Hasan writes down a two-digit number. He then writes the same two-digit number next to his original number to form a four-digit number. What is the ratio of his four-digit number to his two-digit number?

It depends on his number

Let Hasan\textquotesingle s two-digit number be \textquotesingle$$ab$$\textquotesingle, which is equal to $$10 a + b$$. The four-digit number he forms is therefore \textquotesingle$$abab$$\textquotesingle, which is equal to $$1000 a + 100 b + 10 a + b$$ and hence to $$100 (10 a + b)+ 10 a + b = 101 \times (10 a + b)$$. Therefore the ratio of his four-digit number to his two-digit number is $$101:1$$.


Which of the following statement is correct?

All prime numbers are odd numbers

In every $3$ consecutive numbers, there must be a composite number

All even numbers are composite numbers

The sum of two different odd number must be a composite number

A: Any non-$3$ multiple of 3 are composite  B: $1$, $2$, $3$ does not have composite number  C: $2$ is prime number  D: $1+$any odd number is multiple of $2$.


What is the square root of the number whose square is $$16$$?

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

The number whose square is $$16$$ is $$4$$. The square root of $$4$$ is $$2$$.


Teacher divides $$365$$ coins among $$3$$ kids. One of them gets an odd number of coins, and another gets also an odd number of coins. Which of the following would be the possible number of coins that the last kid gets?

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

Odd number $+$ Odd number $=$ Even number  Odd number $-$ Even number $=$ Odd number


One day Randy was bored and he wrotre all the whole numbers from $49$ to $97$ on his paper. How many times did he write the digit \textquotesingle\textquotesingle$8$\textquotesingle\textquotesingle{} on his paper?

None of the above

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


The sum of two whole numbers is $$12$$, and the product is $$32$$. Find the difference between these numbers.

If the sum of $$2$$ whole numbers is $$12$$, and the product is $$32$$, then the numbers are $$8$$ and $$4$$; $$8-4 = 4$$.


If the sum of $$7$$ whole numbers is even, at mostof the numbers could be odd.

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

If the sum of $$7$$ whole numbers is even, there must be an even number of odd numbers. The total number of odd numbers could be $$6$$.


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

None of the above

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

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.


How many prime numbers are there between $130$ and $150$?  $$\textasciitilde$$  $$\textasciitilde$$


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

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


When $$1001$$ is divided by a certain one-digit number, the remainder is $$5$$. What is the remainder when the same one-digit number divides $$2006$$? ．

Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder ->The additive property of the remainder

$$1001\div a=b~~\text {R} 5$$, $a$ is a one-digit number bigger than $$5$$, so $a$ =6;  $$1005\div a=c~ \text {R} 9$$;  $$2006\div a=d \text {R} 14$$;  $$14=2\times6 + 2$$, so the remainder is $$2$$.


How many perfect cubes lie between $2^{8}+1$ and $2^{18}+1$, inclusive? (2018 AMC 8 Problem, Question \#25)

Overseas Competition->Knowledge Point->Number Theory Modules

We compute $2^{8}+1=257$. We\textquotesingle re all familiar with what $6^{3}$ is, namely, which is too small. The smallest cube greater than it is $7^{3}=343$. $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^{6})^{3}=64^{3}$, which therefore clearly will be the largest cube less than $2^{18}+1$. Therefore, the required number of cubes is $64-7+1=58$.


When I rounded $$142.857$$ to the nearest hundred, ten, one, tenth and hundredth,was not one of my rounded numbers.

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


Which of the following is not a multiple of $12$?

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


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


Suppose it is now the month of December. What month will it be $$100$$ calendar months from now?

January

February

March

April

May

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

$$100\div12=8R4$$  $$4$$ months after December will be April.


There are three ropes. One of them is $10$ inches long, the other two are $28$ and $30$ inches respectively. If we cut those ropes into small pieces with nothing left and each piece has the same length, how long is each piece at most?

Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Common Factors and the Greatest Common Factors->The Greatest Common Factor of Multiple Numbers

The greatest common factor of $10$, $28$ and $30$ is $2$.


A number divisible by both $$6$$ and $$20$$ must also be divisible by.


Every positive integer is congruent modulo $9$ to the sum of its decimal digits.  Now, let $S(n)$ equal the sum of the digits of positive integer $n$. For example, $S(1507)=13$. For a particular positive integer $n, S(n)=691$. Which of the following could be the value of $S(n+2)$? (Adapted From 2017 AMC 12A Problems, Question \#18)

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

Note that $n \equiv S(n) \bmod 9$, so $S(n+2)-S(n) \equiv n+2-n\equiv2 \bmod 9$. So, since $S(n)=691 \equiv 7 \bmod 9 $, we have that $S(n+2) \equiv 9 \equiv 0\bmod 9$. Then, only one of the answer choices is congruent to $0 \bmod 9$, which is $(A)=9$.


What is the number between $$37$$ and $$49$$ is exactly divisible by both $$3$$ and $$4$$?


What is the length of the largest square that can be made from $$50$$ one-centimetre square tile?

None of the above

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


There are three whole numbers $A$, $B$ and $C$ ($B\neq1$). $A\times B=35$, $B\times C=84$. $A+B+C=$~\uline{~~~~~~~~~~}~.

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


Emily writes down the largest two-digit prime such that each of its digits is prime.  Krish writes down the smallest two-digit prime such that each of its digits is prime.  Kirsten subtracts Krish\textquotesingle s number from Emily\textquotesingle s number.  What answer does Kirsten obtain? ．

$$14 $$

$$20 $$

$$36 $$

$$45 $$

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

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 Emily writes down $$73$$. The smallest two-digit integer whose digits are both prime is $$22$$. However, $$22$$ is not prime, but $$23$$ is prime.  So Krish writes down $$23$$. Therefore the answer which Kirsten obtains is $$73 - 23 = 50$$.


The $5$-digit number "$2018U$" is divisible by $9$, where $$U$$ is the ones digit. 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 $(\rm B)$ $3$.


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 it will take to fill the pond?

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


What is the total value of all the odd numbers between $$34$$ and $$42$$?

None of the above

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


How many numbers can be divided by both $$3$$ and $$5$$ from $$50$$ to $$500$$?

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


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

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


A whole number is a perfect square if it can be expressed as the product of two equal whole numbers. How many perfect squares are greater than $$0$$ and less than $$1000$$?

We note that $$31\times 31 = 961$$ and $$32\times 32 = 1024$$, hence there are $31$ perfect squares greater than $$0$$ and less than $$1000$$.


How many positive integer factors of $2020$ have more than $3$ factors? (2020 AMC 8 Problems, Question \#17)

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

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 $12-5=(\mathbf{B}) 7$ factors have more than $3$ factors.


How many whole numbers between $$1$$ and $$100$$ are $$3$$ times a prime?

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


For how many positive integer values of $$N$$ is the expression $$\frac{36}{N+2}$$ an integer? ($$1994$$ AHSME Problem, Question \#$$10$$)

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

$$36={{2}^{2}}\times {{3}^{2}}$$, so the number of its factors would be $$\left( 2+1 \right)\times \left( 2+1 \right)=9$$. But $$\textasciitilde N+2$$ cannot be $$1$$ or $$2$$, so the number of possible values of $$N$$ is$$\textasciitilde9-2=7$$.


The $$5$$-digit number $$2018U$$ is divisible by $$9$$. What is the remainder when this number is divided by $$4$$?

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 $4$, we can ignore the hundreds. The remainder when $87$ is divided by $4$ is $3$.


The result of the calculation $$123456789\times8$$ is almost the same as $$987654321$$ except that two of the digits are in a different order. What is the sum of these two digits?

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

The units digit of $$123456789\times8$$ is $$2$$, since $$9\times 8=72$$ . So, if the statement in the question is correct then the two digits which are in a different order are $$1$$ and $$2$$, whose sum is $$3$$. As a check, $$123456789\times8$$ is indeed $$987654312$$ .


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

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

Since $$8=8\times 1$$.  We choose $$\rm A$$.


The product of $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$ ends withzeros．

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


How many numbers in the number set $16,17,18,19$ and $20$ have more than $3$ factors?


$$N$$ is a two$$-$$digit number. When $$N$$ is divided by $$9$$, the remainder is $$1$$. When $$N$$ is divided by $$10$$, the remainder is $$4$$. What is the remainder when $$N$$ is divided by $$11$$? (Adapted from $$2016$$ AMC $$8$$ Problem, Question \#$$5$$)

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

Among $64, 154$\ldots~the smallest possible $$N$$ that satisfies the two conditions is $$64$$, and $$64 \div 11\rm R9$$.


Find the digit in the ones place for $$2\times2\times5\times5\times5$$. ．

The product includes the factor $$5\times2=10$$, so the digit in the ones place is $$0$$.


I am thinking of a whole number greater than $$0$$ whose square equals its square root. How many such numbers are there?

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

The only positive whole number whose square is equal to its square root is $$1$$, so the answer is $$\text{B}$$.


Some pirates found lots of gold coins. They divided all the coins equally into $5$ groups. Then, the leader of each group would divide the coins he got with his teammates. The $5$ groups had $4,$ $8,$ $9,$ $10,$ and $11$ pirates, respectively. In addition, all the leaders found that after they divided the coins evenly, there were always $2$ coins left. Which of the following could be the possible range where the number of coins of each group was in?

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

$[4, 8, 9, 10, 11]=3960$, so the possible number of coins could be $3962$ or $7922$.


On one side of Long Street the houses are numbered with the consecutive odd numbers from $$1$$ to $$19$$. On the other side of that street, the houses are numbered with the consecutive even numbers from $$2$$ to $$14$$. How many houses are there on Long Street? (2006 Math Kangaroo Problem, Level 3-4, Question \#9)

Overseas Competition->Knowledge Point->Number Theory Modules


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


How many $$0$$\textquotesingle s does the product of $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$ end with?

There are $$6$$ factors of $$5$$ when you prime factorise~ $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$.


Among the eleven numbers from $$121$$ to $$131$$, how many prime numbers are there?

Only two prime numbers: $$127$$, $$131$$. Therefore, we choose $$\rm C$$.


Connie counts from $$1$$ to $$20$$. What is the sum of the prime numbers she counts?

Connie counts from $$1$$ to $$20$$. The sum of the prime numbers she counts is $$2+3+5+7+11+13+17+19=77$$. (Note: $$1$$ is not prime.)


Divide $$4$$、$$9$$、$$10$$、$$14$$、$$15$$ and $$21$$ into 2 groups with 3 numbers in each group to make the product of numbers in each group the same. How can we divide the numbers?

$$(14,9,10), (21,15,4)$$

$$(14,21,10), (9,15,4)$$

$$(14,4,10), (21,15,9)$$

$$(14,15,9), (21,10,4)$$


Which of the following numbers leaves a remainder of $$1$$ when divided by $$2$$ and $$3$$?

When we divide by $$2$$, $$3$$, the remainder is $$1$$. Hence, if we first subtract $$1$$, the result will be divisible by $$2$$, $$3$$.  $2\times3+1=7$


Teacher Angel has $$87$$ apples. She want to pack the apples into container so she can keep them nicely in the fridge. Each fo the container can only hold $$8$$ apples. What is the least number of containers needed so that teacher Angel can pack all the apples in the fridge?

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


How many 0s are there at the end of the product $$2\times3\times5\times2\times5\times3\times5\times5$$. ．


A square of a positive number is $$500\textbackslash\%$$ greater than that number. What number is it?

Let the number be $x$. Since $x^{2} = 6x$, therefore $x=6$.


What are the last three digits of the answer to the calculation below?  $$123\times 124\times 125\times 126\times 127$$

The product $$123 \times124\times125\times126 \times127$$ is a multiple of $$125$$; moreover, it also has a factor of $$2$$ three times, from $$124\left( 2\times 2\times 31 \right)$$ and from $$126\left( =2\times 63 \right)$$. Therefore it is a multiple of $$125 \times2\times2\times2= 1000$$, and so it must end in $$000$$. Alternatively, working from the options, it is easily seen that the product is a certainly an even multiple of $$5$$-so its unit digit is $$0$$.


What are the last two digits of the result of $$1\times 3\times 5\times 7\times \cdots \times 101$$?

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

The result must be a multiple of $25.$  $$a=0(\text{mod}25)$$  According to the divisibility rule of $4$, the remainder of the result divided by $4$ is $$1\times 3\times 1\times 3\times \cdots \times 1\times 3\times 1={{3}^{25}}\times 1=3(\text{mod}4)$$.  When the result is $$75(\text{mod}100)$$, it can be divisible by $25$ and have a remainder of $3$ when divided by $4.$


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

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

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 $$5$$.


The greatest common factor is smallest for which of the following pairs of numbers?

$$4$$ \& $$18$$

$$5$$ \&~$$25$$

$$6$$ \&~$$33$$

$$8$$ \&~$$35$$

The greatest common factors of the $$4$$ pairs of numbers are $$2$$, $$5$$, $$3$$, and $$1$$, respectively.


The students of a class can be divided into groups of $5$ or groups of $7$ when there is a groupwork to do. How many students at least are in that class?

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 $5$ and $7$ is $35$.


A two-digit prime number is still a prime number when its first and tenth digits are exchanged. There aresuch prime numbers.


Avril likes collecting baseball cards. The number of cards she has is divisible by $$2$$, $$3$$, and $$5$$. How many baseball cards does Avril have at least?

The total number of baseball cards is the least common multiple of $$2$$, $$3$$, and $$5$$, i.e. $$\left[ 2,3,5\right]=2\times3\times5=30$$.


$$\sqrt {3\times 12}\times \sqrt {4\times 9}=$$．

$$2\times 18$$

$$18\times 18$$

$$3\times 2$$

$$36\times 36$$


Vivian has many barbie cards. If she puts $8$ cards in each group, there will be $5$ cards left. If she puts $9$ cards in each group, there will be $3$ cards left. If she puts $10$ cards in each group, there will be $1$ card left. How many cards at least should Vivian take away, so that the remaining number of cards can be divisible by $8,$ $9,$ and $10$?

$8+8+5=9+9+3=10+10+1=21$, so after we removing $21$ cards, the number of the remaining cards can be divisible by $8, 9,$ and $10.$


The sum of the first $$5$$ \textbf{odd} positive numbers is $$25$$. What is the sum of the first $$5$$ \textbf{even} positive numbers?

Each of five positive even numbers is $$1$$ larger than the corresponding odd number in $$1$$,$$3$$, $$5$$, $$7$$ and $$9$$, which we are told have a sum of $$25$$. So the sum of the first five positive even numbers is $$25 +5 =30$$.


Cayden is drawing some words on a mural in his school. Everyday, he is only able to draw out one letter. He wants to draw:  $$I$$ $$love$$ $$Mathematics$$ $$and$$ $$English$$  If he starts on a Tuesday, and during the weekend, he is not able to draw. When will he finish the drawing?

Monday

Tuesday

Wednesday

Thursday

Friday

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

$$26\div 5 = 5R1$$  Start from Tuesday, thus, it is Tuesday


$$123+234+345$$ divided by $$4$$ has a remainder of．

The sum of each numbers remainder can be divided by $$4$$ to get $$2$$: $$123\div 4$$ R $$3$$；$$234\div 4$$ R $$2$$；$$345\div 4$$ R $$1$$; Therefore, $$\left( 3+2+1 \right)\div 4$$ R $$2$$．


The sum of the squares of the first $$20$$ positive integers is $$2870$$. What is the sum of the squares of the first $$19$$ positive integers?

The sum of the squares of the first $$20$$ positive integers is $$2870$$.  The sum of the squares of the first $$19$$ is $$2870-20^{2}=2870-400=2470$$.


What is the remainder of $$(223\times311+198\times273)\div 5$$? ．

According to the additive and multiplicative properties of remainders, the remainder of this expression equals the remainder of $$(3\times1+3\times3)\div 5=12\div5=2 \text{R} 2$$


How many numbers of the following are divisible by $3$?  $$\textasciitilde$$  $213 \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} 422\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~741\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~971\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~1197\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash~ ~ ~2937$  $\textasciitilde$  $\textasciitilde$

The sums of the digits of $422$ and $971$ are not multiple of $3$.


A positive number is called a perfect square whenever it is the square of a whole number. The first three perfect squares are $$1$$, $$4$$, and $$9$$. The $$100$$th perfect square is．

The first perfect square is $$1^{2}$$, the $$2$$nd is $$2^{2}$$, and the $$3$$rd is $$3^{2}$$.  With this pattern, the $$100$$th perfect square is $$100^{2} = 100\times100= 10000$$.


There are three whole number $A$, $B$, $C$. $A\times B=55$, $B\times C=100$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$55=5\times 11$  $100=2\times 2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=11$, $C=20$.


If I multiply four different prime numbers, the product must have  positive divisors.

$$ 8 $$

$$12 $$

$$16 $$

$$10 $$


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.

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


What is the remainder of $$(223\times311+198\times273)\div 5$$?


Bill writes down all the numbers from $1$ to $60$ inclusive. How many times does he use the digit $5$?

In tens place: there are ten $$5$$s: $$50\sim59$$  In ones place: there are six $$5$$s: $$5, 15, 25, 35, 45, 55$


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 $$16383=2^{14}-1=\left(2^{7}+1\right)\left(2^{7}-1\right) =129 \cdot 127 $$. Since $129$ is composite, $127$ is the largest prime divisible by $16383$. The sum of $127$\textquotesingle s digits is $$ 1+2+7=\text { (C) } 10 $$.


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 $5789$?

The number can be written as $$n+(n+1)=2n+1(n\geqslant 1)$$ and $$x+(x+1)+(x+2)=3x+3$$.  It must be a multiple of $3$ .  $5790$ can be divisible by both $2$ and $3$, so it is $5790-3=5787$.


Change a digit of the number $98760$ to make the new five-digit number be divisible by $250$. What is the new five-digit number?

The last three digits must be divisible by $125$, and the ones place must be divisible by $2$, so it can only be $98750$.


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

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.


$$(14\times 9\times 8)\div \left( 9\times 7\times 8 \right)=$$．

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

$$(14\times 9\times 8)\div \left( 9\times 7\times 8 \right)=14\times 9\times 8\div 9\div 7\div 8=14\div 7\times (9\div 9)\times (8\div 8)=14\div 7=2$$


A two-digit number can be divided by both $$1$$ and $$5$$. It is also an even number. How many such numbers are there?

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


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

♦ $$-\textasciitilde3$$

☺ $$\times$$ ☺

♦ $$\times$$~♦

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

♥ $$\times$$ ☺ $$=$$ ♦  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.


How many $$0$$\textquotesingle s does the product of $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$ end with?

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

There are $$6$$ factors of $$5$$ when you prime factorise~ $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$.


The largest possible sum of two different two$$-$$digit numbers is~\uline{~~~~~~~~~~}~.

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


Julie shares a bag of 70 carrots among some rabbits. Each rabbit has exactly the same number of carrots. If Julie doesn\textquotesingle t have any carrots left, how many rabbits might she have fed? ．

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


Jason has some baseball cards from the $$1920$$s. If he divides the number of cards he has by $$3$$, then he will have $$1$$ remaining cards; if he divides the number of cards he has by $$5$$, he will have $$3$$ remaining cards; if he divides the number of cards he has by $$7$$, he will have $$5$$ remaining cards. How many cards does Jason have at least?

The number of cards after adding $$2$$ is divisible by $$3$$, $$5$$, and $$7$$. Since the least common multiple of $$3$$, $$5$$, and $$7$$ is $$3\times5\times7=105$$, Jason has $$105-2=103$$ cards at least.


Two(different) numbers are selected from $$0$$，$$1$$，$$3$$，$$5$$，$$8$$ and $$9$$. How many two-digit even mumbers can be formed？


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

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


The sum of $$2016$$ integers is even. At mostof them can be odd.

The sum of any even number of odd integers is always even.


The greatest common factor of $$23$$ and $$24$$ is．

The only common factor of any two consecutive whole numbers is $$1$$.


Which of the following statements is incorrect?

$$2350$$ is a multiple of $$2$$, $$5$$ and $$10$$

$$1284$$ is a multiple of $$4$$

$$9972$$ is not a multiple of $$9$$

$$5742$$ is a multiple of $$6$$ and $$9$$

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

A: $$2350$$ is even, thus it is a multiple of $$2. 2350$$ ends with $$0$$, thus it is a multiple of $$5$$ and $$10$$;  B: $$1284$$ ends with $$84$$, and $$84$$ is a multiple of $$4$$, thus $$1284$$ is a multiple of $$4$$;  C: $$9 + 9 + 7 + 2 = 27$$ is a multiple of $$9$$, thus $$9972$$ is a multiple of $$9$$;  D: $$5 + 7 + 4 + 2 = 18$$, it is a multiple of $$9$$ and $$3$$; since it ends with $$2$$, it is also a multiple of $$2$$, and thus a multiple of $$6$$;  Hence the incorrect statement is C.


There are three whole number $A$, $B$, $C$. $A\times B=77$, $B\times C=364$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$77=7\times 11$  $364=2\times 2\times 7\times 13$  Because $B$ is the factor both number contains, $B=7$  Thus, $A=11$, $C=52$.


Which number between $$60$$ and $$80$$ is both a multiple of $$3$$ and $$8$$?~\uline{~~~~~~~~~~}~.


Divide $$2000$$ by an odd number. The remainder must be．

$$$$even

$$$$odd

$$$$prime

$$$$whole


What is the remainder when 7,999,999,999 is divided by 8 ?


Given that $$M=\overline{3abcd}$$, $$N=\overline{abcd3}$$ , and $$M-N=3177$$, what is $$\overline{abcd}$$?

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


I have equal numbers of quarters, dimes, and nickels. These coins could have a total value of any of the following $$EXCEPT$$．

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

The value of $$1$$ quarter, $$1$$ dime, and $$1$$ nickel is $$40$$￠. My coins must have a total value divisible by $$40$$, but ＄$$3.80$$ is not divisible by $$40$$.


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

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


The thousands digit of the sum of 8+88+888+8888+88888 is ．


Which of the following is not a composite number?

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


Students guess that Norb's age is $28$, $30$, $34$, $36$, $38$, and $41$. Norb says, "At least half of you guessed too low, two of your guesses are off by one, and my age is a prime number."~How old is Norb? (adapted from 2011 AMC 8 Problem, Question \#21)

If at least half the guesses are too low, the Norb\textquotesingle s age must be greater than $34$. If two of the guesses are off by one, then his age is between two guesses whose difference is $2$. It could be $29$, $35$ or $37$, but because his age is greater than $34$, it can only be $35$ or $37$. Lastly, Norb\textquotesingle s age is a prime number so the answer must be $37$.


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


What is the remainder when $$222 222 222$$ is divided by $$4$$?

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

In division by $$4$$, the last $$2$$ digits determine the remainder, use $$22\div4$$.


If the sum of two prime numbers is $$39$$, the difference between these two prime numbers will be ．

Overseas Competition->Knowledge Point->Number Theory Modules

Because $$39$$ is an odd number, one of the two numbers must be an even prime number $$2$$. The other number is $$39-2=37$$, and their difference is $$35$$.  So the answer is $$\text{B}$$.


What is the sum of the distinct prime integer divisors of $2016 ?$ (2016 AMC 8 Problem, Question \#9)

The prime factorization is $2016=2^{5} \times 3^{2} \times 7$. Since the problem is only asking us for the distinct prime factors, we have $2,3,7$. Their desired sum is then (B) 12


Which pair of numbers has a common factor greater than $$1$$?

$$4$$ and $$9$$

$$6$$ and $$27$$

$$27$$ and $$50$$

$$33$$ and $$100$$

The greatest common factors for the choices are $$1$$, $$3$$, $$1$$, $$1$$, respectively.


Suppose it is now the month of February. What month will it be $$90$$ calendar months from now?

April

May

June

July

August

$$90\div12=7R6$$  $$6$$ months afterFebruary will be August


The smallest even factor is $$2$$; $$30\div2= 15$$, the greatest odd factor.


There are two reading rooms at Think Town\textquotesingle s Library: Tulip\textquotesingle s and Lily\textquotesingle s. There are two lamps on every table in Tulip\textquotesingle s Reading Room. As for Lily\textquotesingle s Reading Room, there are three lamps on every table. Nini knows that the total number of lamps in the two reading rooms is an odd number, and the total number of tables in the two reading rooms is also an odd number. Which reading room has an odd number of tables?

Tulip\textquotesingle s Reading Room

Lily\textquotesingle s Reading Room

Both rooms

None of the rooms

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

There are two lamps on every table in Tulip\textquotesingle s Reading Room, so the number of lamps in it is an even number. The total number of lamps in the two rooms is odd, and thus the number of lamps in Lily\textquotesingle s Reading Room is an odd number. Since there are three lamps on every table in Lily\textquotesingle s Reading Room, the number of tables in it is also an odd number. However, the total number of tables in the two rooms is odd, so the number of tables in Tulip\textquotesingle s is even. Therefore, the answer is $$\text{B}$$.


Any number that is divisible by both $$12$$ and $$5$$ must also be divisible by．

A number divisible by $$12$$ \& $$5$$ is divisible by a product of their factors, such as $$3\times5$$.


Linda wrote down a natural number. When she divided the number by $$7$$, the remainder was $$5$$. What is the remainder when twice that number is divided by $$7$$? ．

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

The remainder of $$A$$ $$\div7$$ is $$5$$, and $$2A=A+A$$. Therefore the remainder of $$2A\div7$$ is $$5+5=10$$. $$10= 7+3$$, therefore the remainder is $$3$$.


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.


How many numbers listed below are positive?  $$23$$, $$32$$, $$-56$$, $$-98$$, $$7$$


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

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

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.


Pip swam $$3$$ back and forth in the lane for a total of $$156$$ metres, how long is the lane in this pool?

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

one back and forth $$156\div3=52$$metres  length of the lane$$52\div2=26$$metres


The sum of three different positive even numbers could be．

The sum must be even, and it could be $$2+4+8=14$$.


Della has a box of ping pong balls. Each time she counts the balls $8$ by $8$, $10$ by $10$, or $12$ by $12$, there are always $3$ balls left. How many ping pong balls are there in the box at least?

If the ping pong balls in the box are reduced by $3$, then there will be no extra balls when you count them $8$ by $8$, $10$ by $10$, or $12$ by $12$. It means that the number of the ping pong balls is a common multiple of $8$, $10$, and $12$ after being reduced by $3$. If you want to know how many ping pong balls there are at least, you can first find the least common multiple of $8$, $10$ and $12$, and then add $3$ to get the answer.  $$\begin{array}{l} {2\left\textbar{} \underline{\textasciitilde8\textasciitilde\textasciitilde10\textasciitilde\textasciitilde12}\right. }\textbackslash\textbackslash{\textasciitilde2\left\textbar{} \underline{4\textasciitilde\textasciitilde\textasciitilde5\textasciitilde\textasciitilde\textasciitilde6}\right. }\textbackslash\textbackslash{\textasciitilde\textasciitilde\textasciitilde\textasciitilde\textasciitilde2\textasciitilde\textasciitilde\textasciitilde5\textasciitilde\textasciitilde\textasciitilde3} \end{array}$$  $2\times2\times2\times5\times3=120$  $120+3=123$


How many 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.


There are some flowers along the corridor, arranged in the following order: $3$ red flowers, $2$ yellow flowers, $2$ pink flowers$\cdots$  If there are $100$ flowers altogether, how many red flowers are there altogether? ．


There are three whole numbers $A$, $B$ and $C$ ($B\neq1$). $A\times B=55$, $B\times C=100$. $A+B+C=$~\uline{~~~~~~~~~~}~.

$55=5\times 11$  $100=2\times 2\times 5\times 5$  Because $B$ is the factor both number contains, $B=5$  Thus, $A=11$, $C=20$.


In how many ways can we represent the number $$2003$$ as a sum of two prime numbers? ($$2003$$ Math Kangaroo Problems, Level $$7-8$$, Question \#$$15$$)

Such a representation is impossible.

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

If so, $2$ must be one of the prime numbers, and the other must be $2003-2=2001$, which is not prime. Thus, such a representation is impossible.


How many of the numbers $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, and $$9$$ can be written as a sum of exactly two of the other numbers?


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

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

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


Which of the following is an even number?

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


A three-digit number can be written as $\overline{6\square4}$~and this three-digit number is divisble by $4$. How many different digits can we fill in this square?

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


Jason has some baseball cards from the $$1920$$s. If he divides the number of cards he has by $$3$$, the remainder is $1$; if he divides the number of cards he has by $$5$$, the remainder is $3$; if he divides the number of cards he has by $$7$$, the remainder is $5$. Suppose Jason has at least $k$ cards, the sum of digits of $k$ is~\uline{~~~~~~~~~~}~.

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

The number of cards after adding $$2$$ is divisible by $$3$$, $$5$$, and $$7$$. Since the least common multiple of $$3$$, $$5$$, and $$7$$ is $$3\times5\times7=105$$, Jason has $$105-2=103$$ cards at least.


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 $$15+4=19$$.


Every positive integer is congruent modulo $9$ to the sum of its decimal digits.  Now, let $S(n)$ equal the sum of the digits of positive integer $n$. For example, $S(1507)=13$. For a particular positive integer $n, S(n)=1274$. Which of the following could be the value of $S(n+1)$? (Adapted From 2017 AMC 12A Problems, Question \#18)

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

Note that $n \equiv S(n) \bmod 9$, so $S(n+1)-S(n) \equiv n+1-n=1 \bmod 9$. So, since $S(n)=1274 \equiv 5 \bmod 9 $, we have that $S(n+1) \equiv 6 \bmod 9$. Then, only one of the answer choices is congruent to $6 \bmod 9$, which is $(D)=1239$.


How many different ways are there to form a $3-$digit number without repeating digits, using $1, 2, 3, 4,$ and $0$?

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


Arthur writes down three two-digit integers. One is square, one is prime and one is triangular.  He 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 $$41$$.


Each of the following products is an even number except．

The product is even $$except$$ when both numbers you multiply are odd.


Of the following, which has an odd quotient when divided by $$2$$?

$$678 678678678678$$

$$876876876 876876$$

To check for an odd quotient, divide only the last two digits by $$2$$.


Three different numbers are chosen from the numbers 3, 5, 6 and 8.  They are then added together.  Which of these statements is/are correct?  \textbf{1} The total cannot be a multiple of 8.  \textbf{2} The total can be a multiple of 3.  \textbf{3} The total is always odd.

none of them

statement 1 only

statement 2 only

statement 3 only

statements 2 and 3 only


What is the angle between the hour hand and the minute hand at seven o\textquotesingle clock? ．

$50^{}\circ $

$120^{}\circ $

$135^{}\circ $

$150^{}\circ $

$165^{}\circ $

Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Reading the Clock

The smaller angle is $\frac 5{12}$ of a full circle.  A full circle has $360$ degrees, so the angle is $\frac 5{12}\times 360^{}\circ =150^{}\circ $.  So, the answer is $\rm D$.


Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris\textquotesingle s. What is the expected number of balls that occupy their original positions after these two successive transpositions?

Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability

After the first swap, we do casework on the next swap. Case 1: Silva swaps the two balls that were just swapped There is only one way for Silva to do this, and it leaves 5 balls occupying their original position. Case 2: Silva swaps one ball that has just been swapped with one that hasn\textquotesingle t swapped There are two ways for Silva to do this, and it leaves 2 balls occupying their original positions. Case 3 : Silva swaps two balls that have not been swapped There are two ways for Silva to do this, and it leaves 1 ball occupying their original positions. Our answer is the average of all 5 possible swaps, so we get $$ \frac{5+2 \cdot 2+2 \cdot 1}{5}=\frac{11}{5}=(\text { D) } 2.2 $$


A robot is facing south-east.  It makes 58 quarter-turns clockwise, then 93 quarter-turns anti-clockwise.  In which direction is the robot now facing?

north

north-east

north-west

south-east

south-west

Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations

58 quarter-turn clockwise, 93 quarter-turn anti-clockwise, is the same thing as 35 and 1/4 quarter-turn counterclockwise, 35*1/4=8 turns more than 3/4, so it is facing southwest.


$$2006$$ students participated in a survey. The survey stated that $$1500$$ of them participated in the Math Kangaroo contest, and $$1200$$ of them participated in an English Language contest. Out of the students who participated in the survey, how many participated in both contests if it is known that $$6$$ people did not take part in either of the competitions? ($$2006$$ Math Kangaroo Problems, Level $$7-8$$, Question \#$$5$$)

Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets


In how many ways can the letters in $BEEKBBPERPP$ be rearranged so that two or more $E$s do not appear together?

Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations

There are $3$ $E$s in total now with other $8$ letters remaining. But pay attention to $B$ and $P$: there are $3$ $B$s and $3$ $P$ here.  There are $\_8P\_5 \div \_3P\_3$ ways for us to arrange the $8$ letters\textquotesingle{} positions. Then, we can put the $3$ $E$s in the $9$ intervals.  So the answer is $\_8P\_5 \div \_3P\_3 \times \_9C\_3=94080$.


I have $$11$$ pieces of candy in each of three baskets. From each basket I take out one piece of candy in the following order: from the left, from the middle, from the right, from the middle, from the left, from the middle, from the right, and so on. What is the largest number of pieces of candy left in one of the baskets when the middle basket is empty? (2001 Math Kangaroo Problem, Level 3-4, Question \#23)

Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Operational Problem

Start from either the left or the right and follow the steps until the basket in the middle has $$0$$ candy. And you can see that one side has $$5$$ pieces and the other side has $$6$$ pieces. So the largest number of pieces of candy is $$6$$.


A positive integer divisor of $12 !$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$?~ (2020 AMC 10A Problem, Question \#15)

The prime factorization of $12 !$ is $2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7 \cdot 11$. This yields a total of $11 \cdot 6 \cdot 3 \cdot 2 \cdot 2$ divisors of $12 !$. In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization.  Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in $12 !$. Thus, there are $6 \cdot 3 \cdot 2$ perfect squares. (For $2$ , you can choose $0,2,4,6,8,$ or $10$, etc.  The probability that the divisor chosen is a perfect square is $$ \frac{6 \cdot 3 \cdot 2}{11 \cdot 6 \cdot 3 \cdot 2 \cdot 2}=\frac{1}{22} \Longrightarrow \frac{m}{n}=\frac{1}{22} \Longrightarrow m+n=1+22=\text { (E) } 23 $$.


Abe has $1$ green and $1$ red jelly beans in his hand. Bob has $1$ green and $2$ yellow jelly beans in his hand. Each randomly picks a jelly bean to show to the other. What is the probability that the colours match? ．

The probability that both show a green bean is $\dfrac{1}{2}\cdot \dfrac{1}{3}=\dfrac{1}{6}$.


If the sum of $$9$$ consecutive odd integers is $$1935$$, what is the sum of the next $$9$$ consecutive odd integers?

Each number in the second sum is $$18$$ greater than the corresponding number in the first sum. Thus the second sum is $$1935 +18 \times9 = 2097$$.


A box contains $11$ cards, numbered from $1$ to $11$. One card is selected randomly from the box. What is the probability that the number on the selected card is greater than $7$? (adapted from 2017 AMC 8 Problem, Question \#10)

There are $4$ numbers greater than $7$. Thus, the probability is $\frac4{11}$.


Three kids line up to play games. In how many different ways can they form the line?

$3\times 2\times 1=6$  There are six different ways for three kids to line up.


A city has a bicycle hire scheme where it is possible to hire a bicycle for short journeys.  Last year I hired a bicycle $$60$$ times and rode for $$13$$ hours altogether.  For how long on average did I hire the bicycle on each ride?

$$13$$ minutes

$$23$$ minutes

$$39$$ minutes

$$47$$ minutes

$$73$$ minutes

Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Using Formulas

The rider travels for $$13$$ hours over $$60$$ rides, which is anaverage time of $$\frac{13}{60}$$ of an hour per ride, hence $$13$$ minutes.


A $$3$$-digit integer is called a \textquotesingle V-number\textquotesingle{} if the digits go \textquotesingle high-low-high\textquotesingle{} $$-$$ that is, if the tens digit is smaller than both the hundreds digit and the units (or \textquotesingle ones\textquotesingle) digit.  How many $$3$$-digit \textquotesingle V-numbers\textquotesingle{} are there?

Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers

The smallest \textquotesingle V-number\textquotesingle{} is $$101$$ and the largest \textquotesingle V-mumber\textquotesingle{} is $$989$$.  Consider the tens digits. The smallest tens digit is $$0$$ and the largest tens digit is $$8$$.  If the tens digit is $$0$$, the hundreds digit can be $$1$$ to $$9$$, and the units digit can be $$1$$ to $$9$$, giving $$9 \times 9$$ possible \textquotesingle V-numbers\textquotesingle.  If the tens digit is $$1$$, then the hundreds digit can be $$2$$ to $$9$$ and the units digit can be $$2$$ to $$9$$, giving $$8 \times 8$$ possible \textquotesingle V-numbers\textquotesingle.  If the tens digit is $$d$$, where $$d$$ can be any digit from $$0$$ to $$8$$, the hundreds digit can be $$(d + 1)$$ to $$9$$ and the units digit can be $$(d + 1)$$ to $$9$$, giving $$(9-d)\times (9-d)$$ possible \textquotesingle V-numbers\textquotesingle.  The greatest value of $$d$$ is $$8$$. In this case, the hundreds digit can only be $$9$$ and the units digit can only be $$9$$, which gives just $$1 \times 1$$ possibilities.  This gives the total number of possible \textquotesingle V-numbers\textquotesingle{} to be $$9\times9+8\times8+\cdots +1\times1 = 285$$, which is the sum of the squares from $$1$$ to $$9$$ inclusive.


On $2021$ December $31$\textsuperscript{st}, Lucas, Jeremy, and Irene visited their grandpa together. Then Lucas visited him every $4$ days, Jeremy visited him every $5$ days, and Irene visited him every $6$ days. In the first three months of $2022$, how many days could the grandpa be visited by at least one person?

Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle


Harriet tells Topaz that she is thinking of three positive integers, not necessarily all different. She tells her that the product of her three integers is $$36$$. She also tells her the sum of her three integers. However, Topaz still cannot work out what the three integers are. What is the sum of Harriet\textquotesingle s three integers?

Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration

The possible groups of three integers with product $$36$$ are $$(1,1,36)$$, $$(1,2,18)$$, $$(1,3,12)$$, $$(1,4,9)$$, $$(1,6,6)$$, $$(2,2,9)$$, $$(2,3,6)$$ and $$(3,3,4)$$ with sums $$38$$, $$21$$, $$16$$, $$14$$, $$13$$, $$13$$, $$11$$ and $$10$$ respectively. The only value for the sum that occurs twice is $$13$$. Hence, since Topaz does not know what the three integers chosen are, the sum of Harriet\textquotesingle s three integers is $$13$$.


Find the number of positive integers from $100$ to $300$ which is divisible by $6$, $8$ and $10$.

None of the above


Jack, Sarah, and Jimmy participated in a maths competition.  Jack says: "I won the competition."  Sarah says: "I didn\textquotesingle t win the competition."  Jimmy says: "Jack didn\textquotesingle t win the competition."  Only one of them told the truth. Who won the maths competition? ．

$$$$Jack

Sarah

Jimmy

Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning using Hypothesis

We can spot that Jack\textquotesingle s statement and Jimmy\textquotesingle s statement contradict each other, so one of them is telling the truth. Therefore, Sarah tells a lie.


Manjari\textquotesingle s average score on six tests is $$82$$. Her average score on the $$7^{}\text{th}$$ and $$8^{}\text{th}$$ tests is $$98$$. What is her average score on all eight tests? ．

Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)

Total score on the first six tests is $$6\times82=492$$. Her total score on all eight tests is $$492+2\times98=688$$, and her average score is $$688\div8=86$$.


If three couples stand in a row for taking a photo, how many ways are there for only one couple to stand next to each other?


There is a ball in a box and three kids are guessing what colour it is.  \textbf{Val says: "The ball is white."}  \textbf{John says: "The ball is blue."}  \textbf{Elvis says: "I agree with Val."}  They open the box and find only one of them guessed right.  What colour is the ball?

white

blue

Uncertain

We can spot that Val\textquotesingle s guess and Elvis\textquotesingle{} guess are the same, so both of them must be wrong. Therefore, John guessed it right.


Ming, Fanny and other $$6$$ classmates sit at a row. Ming and Fanny cannot sit at $$2$$ ends at the same time. How many sitting arrangements are there?

Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication


Professor Chang has ten different language books lined up on a bookshelf: three Arabic, three German, and four Spanish. How many ways are there to arrange the ten books on the shelf keeping the Arabic books together and keeping the Spanish books together? (Adapted from $2018$ AMC $8$ Problem, Question \#$16$)

Since the three Arabic books and four Spanish books have to be kept together, respectively, we can treat them both as just one book. That means we\textquotesingle re trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $\_5P\_5$. Now we multiply this product by $\_3P\_3\times \_4P\_4$~because there are $\_3P\_3$~ways to arrange just three Arabic books, and $\_4P\_4$~ways to arrange just four Spanish books. Multiplying all these together, we have the answer $D$.


Marcie is $4$ years old, and Sara is $6$ years old this year. $8$ years later, what is the sum of their ages?

Overseas Competition->Knowledge Point->Counting Modules


Annie has four cards of different colors. She writes letter $B$ on the red card and blue card, and writes letter $O$ on the yellow card and green card. Now Annie puts the four cards in a box. Bob is going to draw three of them from the box randomly. How many different possible results can Bob get? Among them, how many can Bob get his name?

$6$; $3$

$4$; $3$

$6$; $2$

$4$; $2$

$6$; $1$

There are $4$ ways to choose three from four. Two $B$s with one $O$ can form Bob\textquotesingle s name. Thus, $2$ of the $4$ ways can get Bob\textquotesingle s name.


Among the natural numbers from $1$ to $600$, how many numbers are multiples of $3$ or $5$?

There are $200$ multiples of $3$ and there are $120$ multiples of of $5$. There are also $40$ multiples of $15$. By the Inclusion Exclusion Principle, there are $200+120-40=280$ numbers which are multiples of $3$ or $5$.


Eddie is ordering lunch at a fast food restaurant that has sandwiches and burgers on the lunch menu, along with coffee, milk, and tea as drink options. If Eddie chooses one food item and one drink item from the lunch menu, he hasdifferent ways to order lunch.


There are $$17$$ balls in a bag. Each ball has a number from $$1$$ to $$17$$ on it. We randomly pick a ball from the bag. What is the smallest number of balls we have to pick in order to be sure that we have at least one pair of balls with a sum equal to $$18$$? ($$2005$$ Math Kangaroo Problem, Level $$9-10$$, Question $$ \textbackslash\# $$$$15$$

Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Worst Case in Pigeonhole Principle Problems

Among these numbers, there are $$8$$ pairs of numbers can get the sum of $$18$$($$1+17=2+16=3+15=4+14=5+13=6+12=7+11=$$$$8+10$$), and $$9$$ is useless. So in the worst case, after we choose $$9$$, we need $$8+1=9$$ more numbers to make sure a pair appears.  Thus, the answer is $$1+8+1=10$$.  Copyrighted material used with permission from Math Kangaroo in USA, NFP Inc.


$$\textgreater$$

$$\textless$$

Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication

Isolate the decimals and whole number parts


A bus was supposed to arrive at Think Station at $$5:30$$ PM, but it arrived $45$ minutes earlier by changing another route because of a traffic accident. When did the bus reach Think Station?

$5:00$ PM

$4:30$ PM

$4:00$ PM

$5:30$ PM

$4:45$ PM

Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation


Miruna had to multiply two $$2$$-digit numbers together, but she accidentally reversed the digits of both of them before multiplying and reached the answer $$209$$. Which of the following answers should she have obtained?

The prime factors of $$209$$ are $$11$$ and $$19$$, so these must have been the reversed numbers that Miruna multiplied. The correct multiplication was $$11 \times 91=1001$$.


My average score on $$8$$ math tests is $$90$$. If my average score on the first $$5$$ tests was $$87$$, what was my average score on the last $$3$$ tests?

I scored a total of $$720$$ on all $$8$$ tests. The total of $$435$$ on the first $$5$$ tests leaves a total of $$285$$ for the last $$3$$ tests, so the average is $$285\div3 = 95$$.


Tiffany wants to pack $$9$$ shirts into several bags. There are at least $$2$$ shirts in each bag, and the number of shirts in each bag should be different. There are~\uline{~~~~~~~~~~}~ways to pack the shirts.


A fair $6$-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number? (2011 AMC 8 Problems, Question \#18)

Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability

There are $6\cdot 6=36$ ways to roll the two dice, and $6$ of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are $\dfrac{30}{2}=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is $\dfrac{15+6}{36}=\dfrac{21}{36}=\boxed {\left (\text{D}\right )\dfrac{7}{12}}$.


Sophia's average score on six tests is $$82$$. Her average scores on the $$7^{}\text{th}$$ and $$8^{}\text{th}$$ tests is $$98$$. What is her average score on all eight tests? ．

Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability

Sophia\textquotesingle s total score on the first six tests is $$6\times82=492$$. Her total score on all eight tests is $$492+2\times98=688$$, and her average score is $$688\div8=86$$.


Eddie finished reading a story book last week. He read an average of $19$ pages per day in the first six days of the week and $26$ pages in the last day. How many pages on average did Eddie read per day?


Amy picks some number cards as shown below:  $13, 20, 14, 15, 19, 20, 20, 19, 19, 15, 19, 19, 20, 13, 15$.  What is the difference between their mode and median?

She picks two $13$s, one $14$, three $15$s, five $19$s, and four $20$s.  The mode is $19.$  The median is $19.$  Thus, their difference should be $0.$


Maria arrived at the market at $$9:35$$. She spent $$1$$ hour $30$ minutes at the market, and it took her an hour to arrive home. When did she arrive home?

Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations


Granny has $$10$$ grandchildren. Alice is the oldest. One day, Granny notices that her grandchildren all have different ages. If the sum of her grandchildren\textquotesingle s ages is $$180$$, what is the youngest Alice can be? (2014 Math Kangaroo Problem, Level 5-6, Question \#30)

Try to consider the problem of the youngest Alice in such way: make each grandchild has a similar age as possible, and the age of each grandchild should be different, that is, $$1 + 2 + 3 +\cdots 9 + 10 = 55$$; then $$180 - 55 = 125$$, $$125 \div 10 = 12R5$$, and $$5$$ is left. If every child\textquotesingle s age is added by $$12$$, then $$5$$ is left. Give the extra year to each of the $$5$$ oldest children. In doing so, the minimum age of Alice is $$10+12+1=23$$.


Throw two dice of the same quality and size. The six sides of each die are marked with number of dots from $$1$$ to $$6$$, respectively. Among the following options,~\uline{~~~~~~~~~~}~is an impossible event.

The sum of dots is $$12$$.

The sum of dots is smaller than $$3$$.

The sum of dots is larger than $$4$$ but smaller than $$8$$.

The sum of dots is $$13$$.

The maximum sum is $$6+6=12$$, so "the sum of dots is $$13$$" is an impossible event.  So $$\text{D}$$ is the answer.


There are  different ways for a librarian, lending six different books to three students, given that each student gets only a book.

The first student has six choices of books; the second has five; and the third has four. By the Rule of product, there is a total of $$6\times5\times4=120$$ways.


What is the smallest possible sum of two positive integers whose product is $$240$$？

Since the product of the two positive integers is $$240$$, the possible pairs of integers are $$\left( 1,240 \right)$$, $$\left( 2,120 \right)$$, $$\left( 3,80 \right)$$, $$\left( 4,60 \right)$$, $$\left( 5,48 \right)$$, $$\left( 6,40 \right)$$, $$\left( 8,30 \right)$$, $$\left( 10,24 \right)$$, $$\left( 12,20 \right)$$ and $$\left( 15,16 \right)$$. The respective sums of these pairs are $$241$$, $$122$$, $$83$$, $$64$$, $$53$$, $$46$$, $$38$$, $$34$$, $$32$$ and $$31$$. Of these, the smallest value is $$31$$.


A three-digit integer contains one of each of the digits $1,3$ , and $5$ . What is the probability that the integer is divisible by $5$ ? (2009 AMC 8 Problem, Question \#13)

The three digit numbers are $135,153,351,315,513,531$. The numbers that end in 5 are divisible are 5 , and the probability of choosing those numbers is (B) $\frac{1}{3}$.


Timi has $8$ paintings: $3$ of them are drawing landscape, and $5$ of them are drawing figure. Among the $5$ figure paintings, there are $3$ drawing the whole family of Timi, and the other $2$ are drawing himself. Now, Timi wants to put those painting in a line. The $3$ landscape paintings cannot be adjacent. How many ways can he do this?


Pip and Bud are playing a game. Each time Pip has to flip a £$$1$$ coin. If Pip gets a head, Bud will pay him £$$1$$. If Pip gets a tail, Bud will take one of her coins. After they played $$50$$ rounds, how much would you expect Pip to gain or lose?

Gaining £$$50$$

Losing £$$50$$

Gaining £$$25$$

Losing £$$25$$

Break Even


A cup costs ￡$$8$$. Which of the following payment is not correct?．

One £5 note and three~£1 coins

Eight £1 coins

One £5 note and four 50p coins


Molly, Dolly, Sally, Elly and Kelly are sitting on a park bench.  Molly is not sitting on the far right and Dolly is not sitting on the far left.  Sally is not sitting at either end.  Kelly is not sitting next to Sally and Sally is not sitting next to Dolly.  Elly is sitting to the right of Dolly but not necessarily next to her.  Who is sitting at the far right end?

Molly

Dolly

Sally

Kelly

Elly

Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Conditions

The question tells us that Sally is not sitting at either end. This leaves three possible positions for Sally, which we will call positions $$2$$, $$3$$ and $$4$$ from the left-hand end. Were Sally to sit in place $$2$$, neither Dolly nor Kelly could sit in places $$1$$ or $$3$$ as they cannot sit next to Sally and, since Elly must sit to the right of Dolly, there would be three people to fit into places $$4$$ and $$5$$ which is impossible. Similarly, were Sally to sit in place $$3$$, Dolly could not sit in place $$2$$ or $$4$$ and the question also tells us she cannot sit in place $$1$$ so Dolly would have to sit in place $$5$$ making it impossible for Elly to sit to the right of Dolly. However, were Sally to sit in place $$4$$, Dolly could sit in place $$2$$, Kelly in place $$1$$, Molly (who cannot sit in place $$5$$) in place $$3$$ leaving Elly to sit in place $$5$$ at the right-hand end.


Only $$1$$ of the $$3$$ boys Abel, Ben and Cain can swim. Abel says, "I can swim." Ben says, "I cannot swim." Cain says, "Abel cannot swim." Only $$1$$ boy is telling the truth. Who can swim? ．

Abel

Ben

Cain

Only $$1$$ of the $$3$$ boys can swim. Only $$1$$ of the $$3$$ boys is telling the truth!  Since Abel and Cain contradict each other, there must be one who is telling the truth!  If Abel is true, Ben is lying and Ben can swim. Then we have $$2$$ boys (Abel and Ben)  who can swim. Contradiction.  Hence, Cain is true. Abel is lying and so is Ben. Then, Ben can swim. Abel cannot swim and we are not sure if Cain can swim.


How many $3$-digit positive integers have digits whose product equals $24$ ? (2009 AMC 8 Problem, Question \#16)

With the digits listed from least to greatest, the $3$-digit integers are $138,146,226,234.$ $226$ can be arranged in $3$ ways, and the other three can be arranged in $6$ ways. There are $3+6(3)=(\mathbf{D}) 21$ 3-digit positive integers.


Summer has $100$ number cards each with a different number from $0$ to $99$. What is the probability that when she chooses two cards randomly, the sum of those two is an even number?

There are $50$ even number cards and $50$ odd number cards here. If she chooses odd+odd or even+even, she will get an even sum. But if she chooses odd+even or even+odd, she will get an odd sum. So each of the two probabilities is equal.


Karl, Lim, Mary and Navin are each competing in a different sporting event at the Olympic Games - basketball, swimming, taekwondo and women\textquotesingle s rhythmic gymnastics. Karl and Navin do not know how to swim. Mary is the only female. Karl\textquotesingle s event does not require a ball. Which event is Navin competing in?

Basketball

Swimming

Taekwondo

Rhythmic gymnastics

Since \textbf{Mary} is the only female, she must have participated in the women\textquotesingle s\textbf{~Rhythmic gymnastics} as the rest are males.  Since Karl and Navin do not know how to swim, \textbf{Lim} is the only male left and thus he participated in the S\textbf{wimming} event.  And since \textbf{Karl}\textquotesingle s event does not require a ball, he must have participated in \textbf{Taekwondo}.  We can thus conclude that \textbf{Navin}~participated in \textbf{Basketball}.


The average of the first ten odd whole numbers is．