Una rolls $6$ standard $6$-sided dice simultaneously and calculates the product of the $6$ numbers obtained. What is the probability that the product is divisible by $4$?

We will use complementary counting to find the probability that the product is not divisible by $4$ . Then, we can find the probability that we want by subtracting this from $1$.  We split this into two cases.  Case 1: The product is not divisible by $2$. We need every number to be odd, and since the chance we roll an odd number is $\frac{1}{2}$, our probability is $\left(\frac{1}{2}\right)^{6}=\frac{1}{64}$.  Case 2: The product is divisible by $2$, but not by $4$. We need 5 numbers to be odd, and one to be divisible by $2$, but not by $4$. There is a $\frac{1}{2}$ chance that an odd number is rolled, a $\frac{1}{3}$ chance that we roll a number satisfying the second condition (only $2$ and $6$ work), and $6$ ways to choose the order in which the even number appears. Our probability is $\left(\frac{1}{2}\right)^{5}\left(\frac{1}{3}\right) \cdot 6=\frac{1}{16}$. Therefore, the probability the product is not divisible by 4 is $\frac{1}{64}+\frac{1}{16}=\frac{5}{64}$. Our answer is $1-\frac{5}{64}=$ (C) $\frac{59}{64}$.


Eight students from Think Academy School take a Mathematics test. Unfortunately, none of the students wrote his/her name on the test paper. As a result, the tests are handed back to the students at random. In how many ways can exactly $5$ of the $8$ students get the correct test back?

None of the above

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

If exactly $5$ pupils get the correct test, then exactly $3$ pupils must get the wrong test.  No. of ways to choose $5$ pupils to get the correct test is $$\frac{8 \times 7 \times 6 \times 5 \times 4}{5 \times 4 \times 3 \times 2 \times 1}-56.$$  To make sure that the other $3$ pupils get the wrong tests, the correct number is $2$.  Hence, the total no, of ways $=56 \times2=112$.


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

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

There are $5$ $E$s in total, which have $4$ intervals leaving for the other $4$ letters. Thus, the answer is $\_4P\_4=24$.


Two tiles numbered $5$ and $6$ are turned face down, respectively. One tile is turned up at random, and throw a die to get a number from $1$ to $6$. What is the probability that the product of the numbers on the tile and the die is smaller than $6$?

$\frac {1}{2}$

$\frac {1}{3}$

$\frac {1}{4}$

$\frac {1}{6}$

$\frac {1}{12}$

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

There are $2\times6=12$ different combinations. The product of two numbers is smaller than $6$ will be $5\times1$. Thus, the probability is $\frac 1{12}$.


The mean, median, and unique mode of the positive integers $3,4,5,6,6,7$, and $x$ are all equal. What is the value of $x$ ? (2012 AMC8, Question \#11)

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


Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

Overseas Competition->Knowledge Point->Counting Modules

We will use complementary counting. First, the frog can go left with probability $\frac{1}{4}$. We observe symmetry, so our final answer will be multiplied by $4$ for the $4$ directions, and since $4 \cdot \frac{1}{4}=1$, we will ignore the leading probability. From the left, she either goes left to another edge $\left(\frac{1}{4}\right)$ or back to the center $\left(\frac{1}{4}\right)$. Time for some casework.  Case 1: She goes back to the center. Now, she can go in any $4$ directions, and then has $2$ options from that edge. This gives $\frac{1}{2}$.  Case 2: She goes to another edge (rightmost).  Subcase 1: She goes back to the left edge. She now has $2$ places to go, giving $\frac{1}{2}$  Subcase 2: She goes to the center. Now any move works. $\frac{1}{4} \cdot \frac{1}{2}+\frac{1}{4} \cdot 1=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$ for this case.  She goes back to the center in Case 1 with probability $\frac{1}{4}$, and to the right edge with probability $\frac{1}{4}$ So, our answer is $\frac{1}{4} \cdot \frac{1}{2}+\frac{1}{4} \cdot \frac{3}{8}=\frac{1}{4}\left(\frac{1}{2}+\frac{3}{8}\right)=\frac{1}{4} \cdot \frac{7}{8}=\frac{7}{32}$ But, don\textquotesingle t forget complementary counting. So, we get $1-\frac{7}{32}=\frac{25}{32} \Longrightarrow D$.


None of the above

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


How many whole numbers between $$1$$ and $$500$$ are divisible by $$6$$ but are not divisible by $$8$$?

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

Every $$6$$th number $$\left( 83 \right.$$ of them$$\left. {} \right)$$ is divisible by $$6$$. Every $$24$$th number $$\left( 20 \right.$$ of them$$\left. {} \right)$$ is divisible by $$6$$ and $$8$$, and $$83-20 = 63$$.


There is a ball in a box and three kids are guessing what colour it is.  Val says: "The ball is white."  John says: "The ball is blue."  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

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

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


A train was supposed to arrive at a $$MRT$$ station at $$5:30 \rm pm.$$ But the train arrived half an hour early because of a traffic accident and changed another line. What time did the train reach the $$MRT$$ station? (adapted from 2008 Math kangaroo Problems, Level 3-4 , Question \#6）

$5:00$ pm

$4:30$ pm

$4:00$ pm

$5:30$ pm

$4:40$ pm

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


How many multiples of $6$ are there between $14$ and $100$?

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


Rose is $$150\rm cm$$ tall. Rose\textquotesingle s older brother Quentin is $$10 \rm cm$$ taller than Rose is. Rose\textquotesingle s younger brother Sam is $$4 \rm cm$$ shorter than Rose is. What is the average of the heights of Rose, Sam, and Quentin?

$$148\rm cm$$

$$150\rm cm$$

$$152\rm cm$$

$$154\rm cm$$

Rose is $$150\rm cm$$ tall. Quentin is $$10 \rm cm$$ taller than Rose, so Quentin is $$160 \rm cm$$ tall. Sam is $$4 \rm cm$$ shorter than Rose, so Sam is $$146 \rm cm$$ tall. Their average height is $$ (150+160+146)\div3 =152 \rm cm$$.


There are $$24$$ four-digit numbers which use each of the digits $$3$$, $$5$$, $$6$$ and $$9$$ once only.  When all of these $$24$$ four-digit numbers are put in order from smallest to largest, which one is in \emph{eighth} position?

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


A dataset of $9$ numbers has an average of $72$. After removing one of the numbers, the average of the remaining numbers becomes $78$. The number that gets removed is~\uline{~~~~~~~~~~}~.


Joann has a robot toy that can walk. The robot can only walk in straight line, and each step can only walk forward for $$1$$ cm or $$3$$ cm. If the robot walks $$4$$ steps forward, it can travel~\uline{~~~~~~~~~~}~different distances.


The numbers $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, $$9$$, $$10$$, $$11$$, and $$12$$ are arranged in $3$ columns of $4$ numbers eadh so that the sum of the numbers in each column is the same. The sum of the numbers in each column is．

It\textquotesingle s just like a magic square! The sum of all $$12$$ numbers is $$78$$. Hence, the answer is $$78\div3 = 26$$.


Math Town is hosting a triathlon competition (a combination of swimming, cycling, and running). The competition starts at $$9:30$$ A.M.. It takes the winner $$25$$ minutes to swim, $$1$$ hour $$10$$ minutes to cycle, and $$35$$ minutes to run. At what time does the winner complete the competition?

$$11:30$$ A.M.

$$10:40$$ A.M.

$$11:40$$ A.M.

$$11:50$$ A.M.

$$11:40$$ P.M.

It takes $25+35=60$ min, which is equal to $1$ hour to swim and run.  It takes $1$ h + $1$ h $10$ min = $2$ h $10$ min in total.  Thus, the end time is $9:30+$$2$ h $10$ min$=11:40$.


Joann and Sana are practicing writing stories. Joann writes $6$ stories each day, and she writes $2$ more than that of Sana each day. How many stories do they write in total in one week?


A win, a loss and a draw are three outcomes of a football game: a win scores $$2$$ points, a draw scores $$1$$ point for each of two teams and a loss scores $$0$$ points. Now, $15$ teams run a single round-robin tournament. How many scores in total will all the $15$ teams get?


On a fine morning, a rabbit, a dog, a cat, and a duck went out to look for food.  The rabbit said: "If I get food, the dog will also get food."  The dog said: "If I get food, the cat will also get food."  The cat said: "If I get food, the duck will also get food."  That evening, they found that all of them were telling the truth but only two of them did get food.~\uline{~~~~~~~~~~}~and~\uline{~~~~~~~~~~}~didn\textquotesingle t get any food.

The rabbit; the dog

The dog; the cat

The cat; the duck

The cat; the rabbit

We can infer that if the rabbit gets food, then all of the other three would get food; if the dog gets food, then both of the cat and duck would get food. Therefore, only when the rabbit, and the dog don\textquotesingle t get food, the cat and the duck would get food.


Seattle is $3$ hours behind New York. For example, when Seattle is $6$ AM, New York is $9$ AM. One day, Rosie took a flight from Seattle to New York. The flight took $$5$$ hours to reach New York, and the time was $9:40$ PM in New York when Rosie arrived. What was the time in Seattle when Rosie departed?

$3:40$ PM

$3:20$ PM

$2:40$ PM

$1:40$ PM

$1:20$ PM

It\textquotesingle s $9:40$ PM in New York, which means it\textquotesingle s $6:40$ PM in Seattle.  The plane departed at $6:40$-$5$ hours=$1:40$ PM.


Fiona spent $$1$$ h $45$ min at the library. At the library, she spent $$30$$ min looking for reading materials. She spent the rest of the time reading them. How many minutes did Fiona spend reading? (adapted from 2011 Math kangaroo Problems, Level 3-4 , Question \#4)

$$45$$ min

$$35$$ min

$$75$$ min

$$55$$ min

$$65$$ min

Time spent reading  $$\rm =1h45min-30min$$  $$\rm =1h15min$$  $$\rm =60min+ 15min$$  $$\rm =75min$$.


Four friends, Edwin, Fred, Gary and Howard, were playing together when one of them broke a vase. The teacher asked,~"Who is the culprit?"  \textbf{Both Edwin and Howard said, "Not me."}  \textbf{Fred said, "Howard broke the vase."}  \textbf{Gary said, "Fred is the culprit."}  If only one of four boys was lying, who broke the vase?

Edwin

Howard

Fred

Gary

Either Fred or Howard must be lying since what they said is conflicting. Since only one person was lying, Gary was telling the truth, which means Fred broke the vase.


Teacher Anderson has pens with three different colours, $$12$$ black, $$8$$ red and $$10$$ green. He put all of these pens in his pencil case. If he close his eyes and randomly pick a pen, at least how many pen he must pick to guarantee he get a red pen?．

To guarantee he get a red pen, the worse cases scenario is he picked all the green and black pens. Which is $$12+10=22$$. Then, the next pen must be the red. Thus, $$22+1=23$$.


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

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 $$688\div8=86$$.


In my suitcase I have $$5$$ sweaters and $$6$$ pairs of pants. If I make an outfit of a sweater and a pair of pants, how many different outfits can I select?

I have $$5$$ sweaters and $$6$$ pairs of pants. For each sweater, there are $$6$$ pairs of pants with which that sweater can be paired. There are $$5$$ sweaters, so there are $$5\times6=30$$ different possible outfits.


The time on a $$12$$-hour circular clock is $$11:00$$ A.M. When the \emph{minute} hand goes around $$3$$ times, the time will be．

$$11:03$$ A.M.

$$11:30$$ A.M.

$$1:00$$ P.M.

$$2:00$$ P.M.

It takes $$3$$ hours for the \emph{minute} hand to go around $$3$$ times, so the time will be $$2$$ P.M.


How many ways are there to rearrange the letters in the word \textquotesingle BEAUTY\textquotesingle{} if the vowels are never together?


Four friends, Edwin, Fred, Gary and Howard, were playing together  when one of them broke a vase.  The teacher asked: "Who is the culprit?"  Both Edwin and Howard said, "Not me."  Fred said, "Howard broke the vase."  Gary said, "Fred is the culprit."  If only one of four boys was lying, who broke the vase?

Edwin

Fred

Gary

Howard

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

Either Fred or Howard must be lying since what they said did not tally.  Since only one person was lying, Gary was telling the truth i.e, Fred broke the vase.


Eight pupils stand in a row to take a photo. Four of them insist on standing together. How many different ways are there to arrange them?


Two different numbers are randomly selected from the set $-2, -1, 0, 3, 4, 5$~and multiplied together. What is the probability that the product is $0$? ($2016$ AMC $8$ Problem, Question \#$13$)

The product can only be $0$ if one of the numbers is $0$. Once we choose $0$, there are $5$ ways of choosing the second number, and there are $15$ ways of choosing $2$ numbers randomly. Thus $\frac{5}{15} = \frac{1}{3}$. The answer is $D$.


Four students asked their teacher, Mr. Carter, to line up with them to take a picture.  ①If Mr. Carter does not want to stand on either ends, how many different ways can they line up for the picture?  ②If Mr. Carter insists on standing on one of the $2$ ends, how many different ways can they line up for the picture?

$72$ , $24$

$96$ , $24$

$72$ , $48$

$96$ , $48$

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

①$$3\times 4\times 3\times 2\times 1=72$$ ,  ②$$2\times 4\times 3\times 2\times 1=48$$ .


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

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.


The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number? ．

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

There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order). We can solve for how many ways the $2$ numbers add up to an odd number and subtract the answer from $1$.  How to solve the problem: The probability of getting an odd number first is $\dfrac{4}{6}=\dfrac{2}{3}$. In order to make the sum odd, we must select an even number next. The probability of getting an even number is $\dfrac{2}{6}=\dfrac{1}{3}$. Now we multiply the two fractions: $\dfrac{2}{3}\times\dfrac{1}{3}=\dfrac{2}{9}$. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except switched, and by the Communitive Property of Multiplication, it does not matter if the equations are switched. Thus we do $\dfrac{2}{9}\times2=\dfrac{4}{9}$. This is the probability of getting an odd-number sum. In order to get the probability of getting an even number we do $1-\dfrac{4}{9}=\left (\text{C}\right )\dfrac{5}{9}$.


Eight pupils from Victory Primary School take a Mathematics test, but none of the pupils wrote his/her name on the test. The tests are therefore handed back to the pupils at random. In how many ways can exactly $5$ of the $8$ pupils get the correct test back?

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

If exactly $5$ pupils get the correct test, then exactly $3$ pupils must get the wrong test.  No. of ways to choose $5$ pupils to get the correct test is $$\frac{8 \times 7 \times 6 \times 5 \times 4}{5 \times 4 \times 3 \times 2 \times 1}-56.$$  To make sure that the other $3$ pupils get the wrong tests, the correct number is $2$.  Hence, the total no, of ways $=56 \times2=112$.


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 $\rm D$.


A fair $6-$sided die is rolled twice. What is the probability that the sum of the two rolls\textquotesingle{} outcomes is a perfect square?

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


How many two-digit numbers are there where the ones digit is greater than the tens~ digit?．


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

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 $\_8P\_5 \div \_3P\_3 \times \_9C\_3=94080$.


$$150$$ students went to a $$\text{CCA}$$ fair and all of them at least tried out one activity. $$85$$ students tried out Sports activity. $$77$$ students tried out Art activity. How many students tried out both Sport activity and Art activities?

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


Maria went to the market at $$9:35$$ a.m. She spent $$1.5$$ hours at the market, and she took an hour to reach home. What time did she reach home?

$11:05$ am.

$12:05$ am.

$11:45$ am.

$11:55$ am.

$12:00$ am.


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

The average is the middle number, $$5$$.


Pat and Lee counted leaves on two plants. Pat\textquotesingle s got a $$1$$-digit number. Lee got a $$3$$-digit number. If the difference between their numbers was $$91$$, what was the sum of their numbers?

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

Pat and Lee counted leaves on two plants. Pat\textquotesingle s got a $$1$$-digit number. Lee got a $$3$$-digit number. If the dīfference of the numbers was $$91$$, the numbers were $$100$$ and $$9$$, and the sum is $$109$$.


How many whole numbers between $$1$$ and $$1000$$ do not contain the digit $$1$$? ($2009$ AMC $8$ Problems, Question \#$22$)

One-digit number: $9-1=8$  Two-digit number: $8\times9=72$  Three-digit number: $8\times9\times9=648$  So, there are $8+72+648=728$ numbers that meet the requirements.


Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles->Number Puzzles (horizontal forms)


Three girls and two boys were dancing. They danced in pairs so that each girl danced with each boy for exactly one minute. At any time, there was only one pair on the dance floor. For how many minutes did they dance? (2021 Math Kangaroo Problem, Level 1-2, Question \#21)

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

In total, there are $2+2+2=6$ different pairs.


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


A cuboid has a length of $2\textasciitilde cm$, breadth of $4\textasciitilde cm$ and a height of $8\textasciitilde cm$. A cube has the same volume. Determine the side length of the cube.

$3\textasciitilde cm$

$4\textasciitilde cm$

$5\textasciitilde cm$

$6\textasciitilde cm$

Volume of cube $=2\times4\times8=64\textasciitilde cm^{3}$  Side length of cube $=$$\sqrt[3]{64}$$=4\textasciitilde cm$


In Carl\textquotesingle s pencil case there are nine pencils. At least one of the pencils is blue. In any group of four pencils, at least two have the same colour. In any group of five pencils, at most three have the same colour. How many pencils are blue?

More information needed

The information that in any group of four pencils, at least two have the same colour, tells us that there at most three different coloured pencils in Carl\textquotesingle s pencil case. The information that in any group of five pencils, at most three have the same colour, tells us that there are at most three pencils of any single colour in the pencil case. Hence there are three pencils of each of the three different colours and so Carl\textquotesingle s pencil case contains three blue pencils.


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 and one drink from the lunch menu, he hasdifferent ways to order lunch.


Suzie flips a fair coin $6$ times. The probability that Suzie flips $3$ heads in a row but not $4$ heads in a row is given by $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. $m+n=$~\uline{~~~~~~~~~~}~.  $\textasciitilde$

Overseas Competition->Knowledge Point->Counting Modules

Consider the distribution of the three consecutive heads. We have four cases:  1110xx, 01110x, x01110, xx0111.  Since case 1 and case4, case 2 and case 3 are the same, we only need to think about case 1 and case 2.  The probability of case 1: $(\dfrac{1}{2})^{4}=\frac{1}{16}$.  The probability of case 2: $(\dfrac{1}{2})^{5}=\frac{1}{32}$.  The sum of the probabilities of 4 cases is $2\times(\frac{1}{16}+\frac{1}{32})=\frac{3}{16}$.


In Tim\textquotesingle s class, there are $$20$$ students who can swim, $$25$$ students who can play basketball, and $$10$$ students who can do both. If everyone in the class plays at least one sport, how many students are there in Tim\textquotesingle s class?~\uline{~~~~~~~~~~}~.

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


Dexter sprinted for the first $600\textasciitilde m$ of a race in $160\textasciitilde s$ and jogged the remaining $\frac{2}{5}$ of the race in $240\textasciitilde s$. What was his average speed?

$2.5\textasciitilde m/s$

$5\textasciitilde m/s$

$2\textasciitilde m/s$

$0.4\textasciitilde m/s$

Average speed $=\frac{Total distance}{Total Time}=\frac{1000\textasciitilde m}{400\textasciitilde s}=2.5\textasciitilde m/s$


April wants to split $$6$$ identical sweets into $$2$$ identical containers. How many different ways can she do so if a container can be empty?


Tiffany uses $3$ identical red gems, $6$ identical blue gems, and $9$ identical pearls to make a necklace. If any two gems cannot be adjacent, how many ways does she have to make the necklace? (After flipping or rotating, if two neckleces can be the same, then count them as $1$ way.)

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

The $9$ pearls should be put in each of the two gems. Thus, just consider the combination of gems:  if all red gems are together, she has $1$ way;  if two red gems are together, she has $3$ ways;  if all the red gems are not together, the blue gems can be put $1-2-3$, $1-1-4$ or $2-2-2$, so there are $3$ ways.  There are $7$ ways.


The product of two whole numbers is $$30$$. What is the least possible value of their sum?

Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Problems of Extreme Value with Fixed Products

The product of two whole numbers is $$30$$. If the numbers are $$5$$ and $$6$$, their sum is $$5+6=11$$.


Amos is taller than Eugene.  Leo is shorter than James but taller than Eugene.  James is shorter than Amos.  is the tallest andis the shortest.

Amos, James

James, Eugene

James, Leo

Amos, Eugene

From clue $$2$$, James is taller than Leo and Leo is taller than Eugene.  From clue $$3$$, Amos is taller than James.  Rank from tallest to shortest: \textbf{Amos}, James, Leo, \textbf{Eugene}.


A meal was priced at $\textbackslash$80$. The shop offered Jane the meal at a $40\textbackslash\%$ discount. After service charges of $10\textbackslash\%$, how much did Jane pay for her meal?

Price after discount $=\textbackslash$80\times(100\textbackslash\%-40\textbackslash\%)=\textbackslash$48$  Price after service charges $=\textbackslash$48\times110\textbackslash\%=\textbackslash$52.80$


Amy, Bill and Celine are friends with different ages. Only one of the following statements is true.  $$\rm I$$. Bill is the oldest.  $$\rm II$$. Amy is not the oldest.  $$\rm III$$. Celine is not the youngest.  Rank the friends from oldest to youngest.

Bill, Amy, Celine

Amy, Bill, Celine

Celine, Amy, Bill

Celine, Bill, Amy

Amy, Celine, Bill

If Bill is the oldest, then Amy is not the oldest, and both statements $$\rm I$$ and $$\rm II$$ are true, so statement $$\rm I$$ is not the true one.  If Amy is not the oldest, and we know Bill cannot be the oldest, then Celine is the oldest. This would mean she is not the youngest, and both statements $$\rm II$$ and $$\rm III$$ are true, so statement $$\rm II$$ is not the true one.  Therefore, statement $$\rm III$$ is the true statement, and both $$\rm I$$ and $$\rm II$$ are false. From this, Amy is the oldest, Celine is in the middle, and Bill is the youngest.


Overseas In-curriculum->Knowledge Point->Knowing Numbers->Decimals->Converting Between Fractions and Decimals->Converting Fractions to Decimals


How many minutes is it from $$22:45$$ today to $$00:35$$ tomorrow?

$$90 $$

$$ 100 $$

$$ 110 $$

$$120 $$

It is $$75$$ minutes from $$22:45$$ to midnight and then another $$35$$ minutes from midnight until $$00:35$$. So the required number of minutes is $$75 + 35 = 110$$.


How many of the following statements are correct?  Statement $1$: The probability of a certain event to happen is $$1$$.  Statement $2$: Indefinite events include impossible events.  Statement $3$: The probability of an impossible event to happen is $$0$$.  Statement $4$: The probability of an indefinite event to happen is between $$0$$ and $$1$$.

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

Impossible events are definite events.  $$\text{Statement 2}$$ is wrong.  $$\text{Statement 1}$$, $$\text{Statement 3}$$, and $$\text{Statement 4}$$ are right.  Thus, the answer is $$\text{D}$$.


There are $$38$$ students in the class. Each student is in either math club or music club or both. Given that among students in this class, $$26$$ students are in the math club, and $$18$$ students are in the music club, there are~\uline{~~~~~~~~~~}~students in the math club only and not in the music club.


What is the sum of the smallest $4$-digit number and the largest $1$-digit number?


There are some pieces of candy on a table. You are challenged by your friend to play the following game: The two players take turns taking some candy. Every turn, you can take away either $$1$$, $$2$$, $$3$$, $$4$$ or $$5$$ pieces from the table. The person who takes away the final piece of candy from the table wins. If you go second, how many pieces of candy should be on the table before the game starts such that you can ensure victory?

Only $$18$$ is one of the multiples of $$5+1$$, and the second player can ensure victory.


How many of the following statements are current?  Statement $1$: The probability of an indefinite event to happen is between $$0$$ and $$1$$.  Statement $2$: The probability of an impossible event to happen is $$0$$.  Statement $3$: The probability of a certain event to happen is $$1$$.  Statement $4$: Indefinite events include impossible events.

Impossible events are definite events.  $$\text{Statement 4}$$ is wrong.  $$\text{Statement 1}$$, $$\text{Statement 2}$$, and $$\text{Statement 3}$$ are right.  Thus, the answer is $$\text{D}$$.


Between them, the two four-digit integers $$M$$ and $$N$$ contain all ten digits from $$1$$ to $$8$$.  What is the least possible difference between $$M$$ and $$N$$?

Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value in Enumeration Problems

The first digits of the two numbers will need to be as close as possible to each other. Since they cannot be equal, they will have to differ by $$1$$; say they are $$n$$ and $$n + 1$$. The difference between the two numbers will then be minimised by making the four digits after $$n + 1$$ as small as possible and the four digits after $$n$$ as large as possible. The smallest four-digit number available is $$123$$ and the largest is $$876$$. So we need to make $$n= 4$$ and then the required diference is $$5123 - 4876 = 247$$.


Eddie reads at least one chapter of a book with $$16$$ chapters each day. If he is asked to read a different number of chapters each day, this book can be finished reading for at most~\uline{~~~~~~~~~~}~days.

Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value with Fixed Sums


A chess singles tournament had $10$ players. Each player played with every other player only once. $2$ points are earned for winning a game, $0$ points are earned for losing a game, and $1$ point is earned by each player in a tie. After the tournament, the judge finds that the sum of these $10$ players\textquotesingle{} points is~\uline{~~~~~~~~~~}~.

There are $$10$$ players, so there are $$10\times9\div2=45$$ games in total. The sum of the points of two players in each game must be $2$. Thus, the sum of all points earned by $10$ players in $45$ games is $2\times45=90$.


Hilla can make $9$ burgers every hour. In a day, she makes burgers for $6$ hours. How many burgers can she make?

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


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

Abel

Ben

Charles

more information needed

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.


Abby, Bret, Carl, and Dana are sitting in a row of four seats numbered $$1$$ to $$4$$. Joe looks at them and says:  "Bret is next to Carl."  "Abby is between Bret and Carl."  However, all of Joe\textquotesingle s statements are false. Bret is actually sitting in the seat $$3$$. Who is sitting in seat the $$2$$? ．

Abby

Bret

Carl

Dana

We know that Carl does not sit next to Bret, so he must sit in seat $$1$$. Since Abby is not between Bret and Carl, she must sit in seat $$4$$. Finally, Dana has to take the last seat available, which is $$2$$.


In numbers 30 to 90, there are some two-digit numbers in which the sum of the tens digit and the ones digit is 13. How many of such numbers are there?~\uline{~~~~~~~~~~}~

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


A top hat contains $$3$$ red chips and $$2$$ green chips. Chips are drawn randomly, one at a time without replacement, until all $$3$$ of the reds are drawn or until both green chips are drawn. What is the probability that the $$3$$ reds are drawn?

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

There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out $3$ red chips, $3$ red chips and $1$ green chip, $2$ green chips, $2$ green chips and $1$ red chip, and $2$ green chips and $2$ red chips. Because order is important in this problem, there are $1+4+1+3+6=15$ ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which $15-5=10$. Out of the $10$ ways to end the game, $4$ of them ends with a green chip. The answer is $\frac{4}{10}=\frac{2}{5}$, or $(\mathbf{B}) \frac{2}{5}$.


Ali, Barb, and Cal were all born on April $$1$$, in different years.  This coming April $$1$$, if I add all their ages together, I\textquotesingle ll get $$9$$.  On that day, Ali\textquotesingle s age could not be．

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

Ali, Barb, and Cal \emph{were} all born on April $$1$$, in \emph{different} years.  This coming Apr $$1$$, if I add all their ages, I\textquotesingle ll get $$9$$. Since the youngest possible ages are $$1$$ and $$2$$, the oldest possible age is $$6$$.


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? ．

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 $$688\div8=86$$.


There are $$10$$ boys in Pat\textquotesingle s math class. If there are twice as many girls as boys in the class, how many girls are there in the class?


There are three boxes labeled $$A$$, $$B$$, and $$C$$. Box $$A$$ contains $$1$$ red ball and $$4$$ white balls. Box $$B$$ contains $$2$$ red balls and $$3$$ white balls. Box $$C$$ contains $$3$$ red balls. All the balls are the same except for the color. Oscar randomly selects one box and then randomly selects one ball from the selected box. What is the probability that he selects a red ball?


Matt had to deliver flyers to all houses numbered from 25 to 57. How many houses got th flyers?


Which of the following expression doesn\textquotesingle t mean the sum of 6 + 6 + 6 + 6 + 6?

$$5\times 6$$

By the meaning of multiplication, $$5\times 6$$ means the sum of five $$6$$\textquotesingle s.


How many thousands of seconds are there in $$365$$ days?

$$365$$ days have $$365\times24\times60\times60\div1000=31536$$ thousand seconds．


There is a box contains $7$ chips numbered $1$, $2$, $3$, $4$, $5$, $6$, and $7$. A chip is drawn randomly from the box. What is the probability that the number on the chip is an even number? (adapted from 2015 AMC 8 Problem, Question \#7)

$2$, $4$, and $6$ are even numbers. Thus, the probability is $\frac37$.


Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?

all $4$ are boys

all $4$ are girls

$2$ are girls and $2$ are boys

$3$ are of one gender and $1$ is of the other gender

all of these outcomes are equally likely

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


Which of the following statements is a certain event?

$3$ is a factor of $98765$.

The sum of the interior angles of a triangle is $180$°.

The age difference between Michael and Candy will increase in $10$ years.

Choose a $3$-digit number and it can be divided by $9$ without any remainder.

$A$ is an impossible event. $C$ is an impossible event. $D$ is a random event.


Eddie reads at least one chapter of a $$15-$$chapter book each day. If he is asked to read a different number of chapters each day, this book can be read for at most~\uline{~~~~~~~~~~}~days.


Seven girls and three boys are standing in a line randomly. What is the probability that three boys are next to each other?


None of the above


A candy store sells some assorted candies. Each bag of assorted candies is made up of $2$ kilograms of toffee and $3$ kilograms of fruit drops. The price of toffee is $\textbackslash$6$ per kg, and the price of fruit drops is $\textbackslash$1$ per kg. How much is the cost of the assorted candies per kilogram?

The assorted candy weighs $2+3=5$ kg and is priced at $2\times6+3\times1=\textbackslash$15$ in total. Therefore, each kilogram of the assorted candy is priced at $15\div5=3$ dollars.


How many positive integers from $1$ to $100$ do not have $2,3$ or $5$ as its factors?

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


Elson finished reading a storybook last week. He read an average of $19$ pages per day for the first four days of the week. He read an average of $25$ pages per day from Friday to Saturday. He did not read on Sunday. How many pages on average did Elson read per day throughout the entire week?


There are $6$ more red fish than yellow fish in Joann\textquotesingle s aquarium. Joann buys $8$ new red fish and $3$ new yellow fish. How many more red fish than yellow fish are in the aquarium now?


Three children have $$10$$ balloons in total, and each of them has a different number of balloons. Joann has the most balloons, and at least how many balloons should she have?

Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Splitting Whole Numbers


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.


There are $$20$$ red balls, $$2$$ black balls, and 1 white ball in a cloth bag. They are of identical shape, size and quality except for color. Take out 1 ball at random. Among the following statements, which one is true?

The probability of taking out a black ball is the smallest.

It\textquotesingle s impossible to take out a white ball.

The probability of taking out a red ball is larger.

A red ball is surely to be taken out.

The number of red balls in the bag is the most compared with other colored balls. If we take out one ball at random, the probability of taking out a red ball is larger. So $$\text{C}$$ is the answer.


Joshua took a $$3$$ hours $$20$$ minutes train ride from Town $$X$$ to Town $$Y$$. The train departed at $$7:55$$ am, but stopped for $30$ minutes due to heavy rain during his trip. When did Joshua arrive in Town $$Y$$? (2009 Math kangaroo Problems, Level 3-4 , Question \#8)

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


A whole number has two digits. The product of the digits of this number is 15. The sum of digits of this number is:


Abe holds $1$ green and $1$ red jelly bean in his hand. Bob holds $1$ green, $1$ yellow, and $2$ red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match? ($2013$ AMC $8$ Problem, Question \# $14$)

The probability of both show a green bean is $\dfrac{1}{2}\times \dfrac{1}{4}=\dfrac{1}{8}$. The probability of both show a red bean is $\dfrac{1}{2}\times \dfrac{2}{4}=\dfrac{1}{4}$. Therefore, the probability is $\dfrac{1}{4}+\dfrac{1}{8}=\boxed {\left (\text{C}\right )\dfrac{3}{8}}$.


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


How many whole numbers between $$5000$$ and $$6000$$ consist of four different digits that decrease from left to right?

Overseas Competition->Knowledge Point->Counting Modules->Finding the Shortest Path by Number Notation->Finding the Shortest Path by Number Notation (with specific points or areas)


$$151$$ traffic ambassadors lined up neatly in a straight line (at equal distances from each other) on a highway to promote traffic laws. After completing their task, at which position of the highway should they assemble in order to minimise their total walking distance from their respective positions to the assembly area? ．

$$75^{}\text{th}$$ traffic ambassador\textquotesingle s position

Between the $$75^{}\text{th}$$ and $$76^{}\text{th}$$ traffic ambassadors\textquotesingle~positions

Any position on the highway

Overseas Competition->Knowledge Point->Counting Modules

There are an odd number of positions, the assembly area should be in the mid-point, $$\left( 151+1 \right)\div 2=76$$, that is, the $$76^{}\text{th}$$ traffic ambassador\textquotesingle s position


One morning, a rabbit, a dog, a cat, and a duck went looking for food outside.  The rabbit says: "If I get food, the dog will also get food."  The dog says: "If I get food, the cat will also get food."  The cat says: "If I get food, the duck will also get food."  In the evening, they find that all of them tell the truth but only two of them get food.~\uline{~~~~~~~~~~}~and~\uline{~~~~~~~~~~}~don\textquotesingle t get food.

The rabbit; the dog

The dog; the cat

The cat; the duck

The cat; the rabbit

We can infer that if the rabbit gets food, then all of the other three would get food; if the dog gets food, then both of the cat and duck would get food. Therefore, only when the rabbit and the dog don\textquotesingle t get food, the cat and the duck would get food.


Sam\textquotesingle s exam was meant to begin at $$13:45$$ but started $$17$$ minutes late. At what time did it begin?

Seventeen minutes after $$13:45$$ is $$14:02$$.


There are $$24$$ four-digit numbers which is formed using each of the digits $$3$$, $$4$$, $$5$$ and $$6$$ once only. When all of these $$24$$ four-digit numbers are put in order from smallest to largest, which one is in the seventh position?


Jack enters the classroom and sees all the $6$ classmates in the classroom. How many students are there in the classroom now?

It cannot be determined.


In how many different ways can six identical coins be distributed among Al, Bo, and Carl so that each gets at least $$1$$ oin?


If $$a⊕b= \frac{1}{ \dfrac{1}{a}+ \dfrac{1}{b}}$$, then what is the value of $$ (1\times2)⊕(2\times3)⊕(3 \times4)⊕\ldots⊕(2013\times2014)$$?

$$\frac {2013}{2014}$$

None of the above


What is the angle between the hour hand and the minute hand at seven o\textquotesingle clock?~ ~ ．(Only consider angles less than 180^{}\circ)

$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 $\rm D$.


If we throw two $6-$sided dice of the same shape and size, which statements among the following options is an impossible event?

The sum of dots is $$12$$.

The product is a prime number.

The sum of dots is larger than $$1$$ but smaller than $$3$$.

The product of is $40$.


The time twelve thousand and twelve hours after $$7$$ A.M. is．

$$1$$ A.M.

$$7$$ P.M.

$$7$$ A.M.

$$7$$ P.M.

Divide $$12012\div24$$ to get remainder $$12$$. $$12$$ hours after $$7$$ A.M. is $$7$$ P.M.


There are $$18$$ boys in Pat\textquotesingle s math class. If there are twice as many girls as boys in the class, how many girls are there in the class?


Kevin has 3 regular dice. Each dice has numbers from 1 to 6. Which of the following could not be the sum of the numbers on top of the 3 dice?

All the above numbers are possible sum


Veronica rolls three six-sided dice. The product of the three numbers is $$90$$.  What is the sum of the three numbers?

more information is needed

Given that the numbers on a die are $$1$$, $$2$$, $$3$$, $$4$$, $$5$$ and $$6$$, to achieve a product of $$90$$, one of the numbers must be $$5$$ and the other two must have a product of $$18$$. Thus the numbers can only be $$3$$, $$5$$ and $$6$$ , whose sum is $$14$$.


Mike and Sara are looking for a place to eat lunch. They know there are nine Chinese restaurants, three Mexican restaurants, and two fast food restaurants nearby. How many different choices do they have in total to eat one meal?

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

They can only choose one place, so it can only be Chinese, Mexican, or fast food. Therefore, we can add each one up to get $$9+3+2 = 14$$.  ~


Mary had equal number of green, yellow and red tokens. She used some of those tokens to make a pile. You can see all used tokens in the figure. She still has five tokens which are not on the pile. How many yellow tokens did she have have at the beginning?  insert pic

Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Counting Solid Figures

The used tokens are 5 red tokens at the bottom, 4 green tokens and 4 yellow tokens. Now we distribute the unused 5 tokens to the 3 colors such that each color has the same number of tokens.  Firstly, we need 1 more green token and 1 more yellow token to make all colors equal in number of tokens.  We are left with 3 tokens, and we will equally distribute these tokens to the colors, i.e. each color will have 1 more token.  Hence, the unused tokens are 1 red, 2 green and 2 yellow. There are 4+2 = 6 yellow tokens in total.


Josip has 4 toys - a car, a doll, a ball and a ship. He wants to put them on a line on a shelf. The ship has to be next to the car and the doll has to be next to the car. In how many ways can he arrange them so all the conditions would be fulfilled?

The car is in the middle of the ship and the doll. There are 2 ways to arrange the three toys, i.e. door-call-ship and ship-car-doll.  The last toy, the ball, can be placed on either of the 2 ends.  Therefore, there are 2 x 2 = 4 ways to arrange the toys.


In a toy store, cars are available in 5 different colours: blue, white, yellow, black and red. A car has either 2 or 4 doors. How many different version of the car are available?

None of the above


A box contains $5$ red balls and $3$ white balls that are identical in all aspects except color. One ball is drawn at random from the box and then replaced. The box is then thoroughly shaken so that the balls are arranged at random again and a second ball is drawn randomly from the box. What is the probability of drawing white ball on both draws?

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


There are $6$ stairs, and you can walk for one or three staris each time. There are~\uline{~~~~~~~~~~}~different ways to walk the $6$ stairs.


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 $$18$$?

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.


The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number? (2019 AMC 8 Problems, Question \#18)

We have a 2 dice with 2 evens and 4 odds on both dice. For the sum to be even, the 2 rolls can be 2 odds or 2 evens.  Ways to roll 2 odds (Case 1 ): The total number of ways to obtain 2 odds on 2 rolls is $4 * 4=16$, as there are 4 possible odds on the first roll and 4 possible odds on the second roll.  Ways to roll 2 evens (Case 2 ): Similarly, we have $2 * 2=4$ ways to obtain 2 evens. Probability is $\frac{20}{36}=\frac{5}{9}$.


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

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

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


The numbers $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, $$9$$, $$10$$, $$11$$, and $$12$$ are arranged in $3$ columns of $4$ numbers each so that the sum of the numbers in each column is the same. The sum of the numbers in each column is．

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

It\textquotesingle s just like a magic square! The sum of all $$12$$ numbers is $$78$$. Hence, the answer is $$78\div3 = 26$$.


Mother has $$12$$ identical peaches. She wants to place them into $$3$$ identical baskets. How many ways can she place the peaches if the number of peaches in each basket cannot be zero and must be different from one other?

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

Order does not matter as peaches and baskets are identical. But we need to have different number of peaches in each basket and none of the baskets can be empty.  $$12=1+2+9$$  $$12=1+3+8$$  $$12=1+4+7$$  $$12=1+5+6$$  $$12=2+3+7$$  $$12=2+4+6$$  $$12=3+4+5$$  There are $$7$$ ways she can place the peaches.


Fill in the blank:~\uline{~~~~~~~~~~}~is $2$ tens $8$ ones less than $5$ tens $5$ ones.

None of the above


Una rolls 6 standard 6 -sided dice simultaneously and calculates the product of the 6 numbers obtained. What is the probability that the product is divisible by 4 ?

We will use complementary counting to find the probability that the product is not divisible by 4 . Then, we can find the probability that we want by subtracting this from 1 . We split this into 2 cases. Case 1: The product is not divisible by 2. We need every number to be odd, and since the chance we roll an odd number is $\frac{1}{2}$, our probability is $\left(\frac{1}{2}\right)^{6}=\frac{1}{64}$ Case 2: The product is divisible by 2 , but not by 4 . We need 5 numbers to be odd, and one to be divisible by 2 , but not by 4 . There is a $\frac{1}{2}$ chance that an odd number is rolled, a $\frac{1}{3}$ chance that we roll a number satisfying the second condition (only 2 and 6 work), and 6 ways to choose the order in which the even number appears. Our probability is $\left(\frac{1}{2}\right)^{5}\left(\frac{1}{3}\right) \cdot 6=\frac{1}{16}$. Therefore, the probability the product is not divisible by 4 is $\frac{1}{64}+\frac{1}{16}=\frac{5}{64}$. Our answer is $1-\frac{5}{64}=$ (C) $\frac{59}{64}$.


A data set consists of~~(not distinct) positive integers: 1,7,5,2,5 and x. The average (arithmetic mean) of the~~numbers equals a value in the data set. What is the sum of all positive values of x?


If the average of two numbers is $$7$$, the numbers could be．

$$3$$ and $$4$$

$$1$$ and $$8$$

$$2$$ and $$14$$

$$6$$ and $$8$$

If the average of two numbers is $$7$$, their sum is $$2\times7 = 14$$.


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?

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 $$ \frac{5+2 \cdot 2+2 \cdot 1}{5}=\frac{11}{5}=2.2$$.


The average mass of $$2$$ bags of flour is $$1.72 \text{kg}$$. The mass of one of the bag of flour is $$1.68 \text{kg}$$. What is the mass of the other bag of flour?


Two different numbers are randomly selected from the set $-5, -3, -1, 3, 5$~and multiplied together. What is the probability that the product is a negative number?~


Jack built a cube using 27 small cubes which are colored either black or white (see figure). No two of the small cubes which are colored in the same color have a common face. How many white cubes did Jack use?


I donate a ＄$$100$$ bill, $$2$$＄$$50$$ bills, $$3$$＄$$20$$ bills, $$4$$＄$$10$$ bills, and $$5$$＄$$5$$ bills. If $$5$$ people divide my money equally, each person receives．


Find the number of two-digit positive integers whose digits total $7$. (2004 AMC 8 Problem, Question \#8)

The numbers are $16,25,34,43,52,61,70$ which gives us a total of $(\text{B}) 7$.


Rabbit Borya likes cabbages and carrots very much. In a day he eats either 9 carrots, or 2 cabbages, or 1 cabbage and 4 carrots. During one week Borya has eaten 30 carrots. How mnay cabbages has he eaten during this week?

Borya ate 30 carrots in a week, which is a number divisible by 3.  On the days that he ate 9 carrots, the total number of carrots is also divisible by 3.  Hence, the total number of carrots he ate on days that he ate 4 carrots must also be divisible by 3, meaning the number of days he ate 4 carrots is divisible by 3.  If there were no days he ate 4 carrots, he ate 9 carrots in 30/9 days, which is impossible. If there were 6 days he ate 4 carrots, he was left with only 30-6 x 4 = 6 carrots to eat on other days, which is also impossible.  therefore, he ate 4 carrots on 3 days, meaning there are \_\_\_ 2 days he ate 9 carrots. There were 3 days Borya ate 1 cabbage, and 7 \_ days he ate 2 cabbages.  He ate a total of 3 + 2~ x 2 = 7 cabbages in a week.


What is the average (mean) of all $$5-$$digit numbers that can be formed by using each of the digits $$1$$, $$3$$, $$5$$, $$7$$, and $$8$$ exactly once? ($$2005$$ AMC $$10B$$ Problem, Question \#$$20$$)

Method $$1$$: We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are $$4!=24$$ ways to arrange the other numbers, so each number appears in each spot $$24$$ times. Therefore, the sum of all such numbers is $$24\times (1+3+5+7+8)\times (11111)=24\times24\times11111=6399936$$.  Since there are $$5!=120$$ such numbers, we divide $$6399936\div120$$~ to get $$53332.8$$.  Method $$2$$: We can first solve for the mean for the digits $$1$$, $$3$$, $$5$$, $$7$$, and $$9$$ since each is $$2$$ away from each other. The mean of the numbers than can be solved using these digits is $$55555$$. The total amount of numbers that can be formed using these digits is $$5!=120$$. The sum of these numbers is $$55555(120)=6666600$$. Now we can find out the total value that was gained by replacing the $$8$$ with a $$9$$. We can start by calculating the gain when the $$8$$ was in the ones digit. Since there are $$4!=24$$ numbers with the $$8$$ in the ones digit and $$1$$ was gain from each of them, $$24$$ is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $$24+240+2400+24000+240000=266664$$ as the total amount that was gained. Subtract this amount from the sum of the digits using the~ $$9$$ instead of the $$8$$ to get $$6666600-266664=6399936$$. Finally, we divide this by $$120$$ to get the average $$\frac{6399936}{120}=53332.8$$.  Method $$3$$: The average value of the digits is $$\frac{(1 + 3 + 5 + 7 + 8)}{5} = 4.8$$. Values occur in every place so $$4.8 \times 11111 = 53332.8$$.


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?

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


How many digits are there from $8$ to $78$?


There are $$124$$ teams participating in a volleyball match held in the Boston. Using the single elimination method, how many games will be played in total?

Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Sports Competition

Single elimination tournament: $$124-1=123$$ games


SASMO 2016 P2 Q6  Uncle John has a farm. His wife and his two sons are staying with him in the farm. They raise 10 cows and 20 chickens. How many total legs are there in the farm?

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


Abby, Bret, Carl, and Dana are seated in a row of four seats numbered $$1$$ to $$4$$. Joe looks at them and says:  ``Bret is next to Carl."  "Abby is between Bret and Carl."  However, all of Joe\textquotesingle s statements are false. Bret is actually sitting in seat $$3$$. Who is sitting in seat $$2$$?

Abby

Bret

Carl

Dana

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

We know that Carl does not sit next to Bret, so he must sit in seat $$1$$. Since Abby is not between Bret and Carl, she must sit in seat $$4$$. Finally, Dana has to take the last seat available, which is $$2$$.


Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132$ . (2016 AMC 8 Problem, Question \#11)

We can write the two digit number in the form of $10 a+b$; reverse of $10 a+b$ is $10 b+a$. The sum of those numbers is: $$ \begin{gathered} (10 a+b)+(10 b+a)=132 \textbackslash\textbackslash{} 11 a+11 b=132 \textbackslash\textbackslash{} a+b=12 \end{gathered} $$ We can use brute force to find order pairs $(a, b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than 10 . Thus our ordered pairs are $(3,9) ;(4,8) ;(5,7) ;(6,6) ;(7,5) ;(8,4) ;(9,3)$ or $(\text{B}) 7$ ordered pairs.


Ms. Osborne asks each student in her class to draw a rectangle with integral side lengths and a perimeter of $$50$$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?

Overseas Competition->Knowledge Point->Geometry Modules->Objects with Straight Sides->Knowing Graphs

As we know, the sum of the length and width is $$25$$. The largest area is $$13\times12=156$$ and the smallest area is $$24\times1=24$$, so the difference is $$156-24=132$$.


A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won $$4$$ games, Ines won $$3$$ games, Janet won $$2$$ games, Kendra won $$2$$ games and Lara won $2$ games, how many games did Monica (the sixth player) win? ($$2006 \text{ AMC } 8$$ Problem, Question \#$$20$$)

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

There are $$6$$ players, so there are $$6\times5\div2=15$$ games in total. By now, $$4+3+2+2+2=13$$ games have been finished (there is one winner in each game), so Monica needs to win $$15-13=2$$ games. Therefore, the answer is $$\rm C$$.


Each of $$300$$ students belongs to exactly $$2$$ of the $$5$$ school clubs. What is the average number of students in each club?

Since $$300$$ students belong to $$2$$ clubs each, there are $$600$$ club memberships. The average membership of the $$5$$ clubs is $$600\div 5 = 120$$ students.


How many terms are there in the arithmetic sequence~$2,\textbackslash{} 5,\textbackslash{} 8,\textbackslash{} 11,\textbackslash{} 14\cdots \textbackslash{} 95,\textbackslash{} 98$$?$terms.


Three friends, Ms Raja, Ms Omar and Ms Beatty all live in the same street. They are a doctor, an engineer and a musician in some order. The youngest one, the doctor, does not have a brother. Ms Beatty is older than the engineer and is married to Ms Omar\textquotesingle s brother. What are the names, in order, of the doctor and the engineer?

Raja and Omar

Omar and Beatty

Beatty and Omar

Raja and Beatty

Omar and Raja

The doctor is the youngest and does not have a brother. Since Ms Omar has a brother and Ms Beatty is older than the engineer, the doctor is Ms Raja. Also, since Ms Beatty is older than the engineer she cannot be the engineer. Hence the engineer is Ms Omar. Therefore the doctor and the engineer in order are Raja and Omar.


There were $20$ ducks, pigs, and sheep in total in Sam\textquotesingle s farm. After Sam bought some new sheep, the number of sheep has doubled. There are $27$ ducks, pigs, and sheep in total. Originally, how many ducks and pigs were there?


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?

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 $\dfrac{15+6}{36}=\dfrac{21}{36}=\boxed {\left (\text{D}\right )\dfrac{7}{12}}$.


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 $

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 $\rm D$.


How many whole numbers between $$5000$$ and $$6000$$ consist of four different digits that decrease from left to right?


Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

We will use complementary counting. First, the frog can go left with probability $\frac{1}{4}$. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since $4 \cdot \frac{1}{4}=1$, we will ignore the leading probability. From the left, she either goes left to another edge $\left(\frac{1}{4}\right)$ or back to the center $\left(\frac{1}{4}\right)$. Time for some casework. Case 1: She goes back to the center. Now, she can go in any 4 directions, and then has 2 options from that edge. This gives $\frac{1}{2}$. -End case 1 Case 2: She goes to another edge (rightmost). Subcase 1: She goes back to the left edge. She now has 2 places to go, giving $\frac{1}{2}$ Subcase 2: She goes to the center. Now any move works. $\frac{1}{4} \cdot \frac{1}{2}+\frac{1}{4} \cdot 1=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$ for this case. -End case 2 She goes back to the center in Case 1 with probability $\frac{1}{4}$, and to the right edge with probability $\frac{1}{4}$ So, our answer is $\frac{1}{4} \cdot \frac{1}{2}+\frac{1}{4} \cdot \frac{3}{8}=\frac{1}{4}\left(\frac{1}{2}+\frac{3}{8}\right)=\frac{1}{4} \cdot \frac{7}{8}=\frac{7}{32}$ But, don\textquotesingle t forget complementary counting. So, we get $1-\frac{7}{32}=\frac{25}{32} \Longrightarrow D$.


Sam\textquotesingle s exam was meant to begin at $$13:45$$ but started $$17$$ minutes late. At what time did it begin?

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

Seventeen minutes after $$13:45$$ is $$14:02$$.


SASMO 2014 P2 Q6  A shop sells sweets where every 3 sweet wrappers can be excanged for one more sweet. Ali has enough money to buy only 7 sweets. What is the biggest number of sweets that he can get from the shop?


One day, Pip asks his parents: "What day is it today?"  His mother says: "Today is Monday."  His father says: "Today is Tuesday."  Which of the following is true?

One of these two sentences is definitely wrong and the other one is correct.

It is possible that both of Pip\textquotesingle s parents are wrong.

It is possible that both of Pip\textquotesingle s parents are right.

If Pip\textquotesingle s mother is wrong, then his father must be right

"Today is Monday" is not the opposite of "Today is Tuesday".i.e. they can both be false.  "Today is Monday" is the direct opposite of "Today is not Monday". One must be true and the other must be false.


Five students are lining up to take a picture. If Jessica is standing in the middle, and Ian is standing on either end, how many ways can the students line up? ．


A sergeant stands too long in the sun and gets confused. His troops are lined up facing north. Then he gives the order to "Right Turn $90^{}\circ$" $$70$$ times, and his troops do so. In which direction are the troops facing at the end?

$$$$north

$$$$east

$$$$south

$$$$west

$$$$west-south-west

Overseas Competition->Knowledge Point->Combinatorics->Directions and Coordinates->Directions

After every $$4$$ "Right turns" the troops will be facing north again; so after $$68$$ turns they are facing north, after $$69$$ turns east, and after $$70$$ turns south.


Hansel wants to buy 2 dice of different colours. If the available colours in a store are red, blue, green, yellow, pink and white, how many different combinations of 2 dice are there in the store? (Example: 1 combination is yellow and white)

None of the above

RB, RG, RY, RP, RW  BG, BY, BP, BW  GY, GP, GW,  YP, YW,  PW


There is a cuboid with a dimension of $3\times4\times5$. Now paint all the surfaces red and cut it into many $1\times1\times1$ cubes. How many cubes that have two faces painted red are there?

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

Remove the two cubes in both ends, and then we can get: $[(3-2)+(4-2)+(5-2)]\times4=24$


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->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)

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 $$688\div8=86$$.


Nina and four of her friends are lining up to take a picture. If Nina has to be in the middle, how many different ways can the five friends line up?

Since Nina has to be in the middle, then there are four friends who can line up in different ways. We can write the equation as $$4\times3\times2\times1=24$$.


Each of Basil\textquotesingle s friends added the number of the day and the number of the month of their birthdays and obtained $$35$$. Their birthdays all fall on different days. What is the greatest possible number of friends that Basil has?


There are $20$ cards numbered from $1$ to $20$, respectively. What is the probability of taking out a card with a prime number?

There are $8$ prime numbers in total.  So, answer $= \frac{8}{20} = \frac{2}{5}$


SASMO 2015 P2 Q7  Study the figures made with matchsticks below. How many matchsticks are needed to make figure 5?

Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Classifying and Enumerating->Counting Regular Figures->Counting Triangles

Figure 1 = 1 triangle = 3 matchsticks  Figure 2 = (1+2) triangle = 3 x 3 = 9 matchsticss.  Figure 3 = (1+2+3) triangle = 6 x 3 = 18 matchsticks.  Figure 4 =  Figure 5 = (1+2+3+4+5) triangle = 15 x 3 = 45 matchsticks.


Among the natural numbers $1-75$: How many are divisible by $3$ or $5$?

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

All natural numbers divisible by $3$ in the range of $1-75$ are：$75\div3=25$．  All natural numbers divisible by $5$ in the range of $1-75$ are：$75\div5=15$．  The natural numbers divisible by 3 and 5 that are divisible by $15$：$75\div15=5$．  The natural numbers divisible by $3$ or $5$ are：$25+15-5=35$．


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 got?

Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Splitting Whole Numbers

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


The product of three consecutive numbers is $$15600$$. What is their sum?

Given that the product, $$15600$$, is a multiple of $$25$$, the three numbers between  them must contribute two factors of $$5$$; since they are consecutive and there are only three of them, this can happen only if one number is itself a multiple of $$25$$. It is now worth observing that the product of three consecutive numbers is roughly the same as the cube of the middle of the three. Since $$20^{3}\textless{} 15 600 \textless{} 30^{3}$$, the middle number must lie between $$20$$ and $$30$$, hence one of the numbers must be $$25$$. The numbers can therefore be $$\left\textbackslash{ {23, 24, 25} \right\textbackslash}$$,~ $$\left\textbackslash{ {24, 25, 26} \right\textbackslash}$$ or $$\left\textbackslash{ {25, 26, 27} \right\textbackslash}$$. The product $$15600 =13 \times1200$$, so it has factors both $$4$$ and $$13$$. The triple $$\left\textbackslash{ {25, 26, 27} \right\textbackslash}$$ has no factor of $$4$$, and $$\left\textbackslash{ {23, 24, 25} \right\textbackslash}$$ no factor of $$13$$. So, by elimination, the numbers are $$24$$, $$25$$ and $$26$$, and their total is $$75$$.


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 $8$ Problem, Question \#$18$)

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

There are $6\times6=36$ ways to roll a die twice, 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 time is greater than the second should be the same as the number of rolls where the second time 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 $\dfrac{15+6}{36}=\dfrac{21}{36}=\frac{7}{12}$.


Wendy wrote all the whole numbers from $$13$$ to $$78$$ on her paper. How many times did she write all the digit "$$3$$ on her paper?


A school has 100 students and 5 teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50,20,20,5$, and 5 . Let $l$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$ ?

The formula for expected values is $$ \text { Expected Value }=\sum(\text { Outcome } \cdot \text { Probability }) . $$ We have $$ \begin{aligned} t \& =50 \cdot \frac{1}{5}+20 \cdot \frac{1}{5}+20 \cdot \frac{1}{5}+5 \cdot \frac{1}{5}+5 \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =(50+20+20+5+5) \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =100 \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =20, \textbackslash\textbackslash{} s \& =50 \cdot \frac{50}{100}+20 \cdot \frac{20}{100}+20 \cdot \frac{20}{100}+5 \cdot \frac{5}{100}+5 \cdot \frac{5}{100} \textbackslash\textbackslash{} \& =25+4+4+0.25+0.25 \textbackslash\textbackslash{} \& =33.5 . \end{aligned} $$ Therefore, the answer is $t-s=(\mathbf{B})-13.5$.


Julia, Kasia, Susanna, and Helena have their birthdays on March $$1^{\rm st}$$, May $$17^{\rm th}$$, July $$20^{\rm th}$$, and March $$20^{\rm th}$$. Kasia and Susanna were born in the same month. Julia and Susanna were born on the same day of a month. Which of the girls was born on May $$17^{\rm th}$$? ．

Julia

Kasia

Susanna

Helena

Susanna: March 20\textsuperscript{th}  Kasia: March 1\textsuperscript{st}  Julia: July 20\textsuperscript{th}  Helena: May 17\textsuperscript{th}


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 $\dfrac{1}{2}\cdot \dfrac{1}{3}=\dfrac{1}{6}$.


Two different numbers are randomly selected from the set $-5, -3, -1, 3, 5$~and multiplied together. What is the probability that the product is a negative number? (adapted from $2016$ AMC $8$ Problem, Question \#$13$)


Overseas Competition->Knowledge Point->Counting Modules


The total score of $8$ students is an even number. Each of $3$ of them gets an odd number score, and each of $4$ of them gets an even number score. Which of the following would be the possible score that the last student gets?

The sum of $3$ odd numbers is an odd number.  The sum of $4$ even numbers is an even number.  The sum of an odd number and an even number is an odd number, so the total score of the $7$ students is an odd number.  The total score of $8$ students is an even number, and the sum of $2$ odd numbers is an even number.  Therefore, the score of the last one should be an odd number.


A deck of cards contains a white card and two red cards. Take one card out randomly, record its colour, and put it back in the deck. Then, take another card out randomly.  Compare the probability of the following situations.  I. Both cards are of the same colour.  II. Both cards are red.  III. Two cards are of different colours.  Which one has the highest probability?

$$I$$

$$II$$

$$III$$

I. Both cards are in the same colour: $$ \dfrac{5}{9}$$;  II. Both cards are red: $$\dfrac{4}{9}$$;  III. Two cards are in different colours: $$\dfrac{4}{9}$$.


Each of the 5 sides and the 5 diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?

Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear. After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.  Case 1: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection.  Case $2: 3$ are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.  Case $3: 2$ are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are $\left(\begin{array}{l}4 \textbackslash\textbackslash{} 2\end{array}\right)$ ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of $6 \cdot 2=12$ ways to color the pentagon so that no such triangle has the same color for all of its sides. These are all the cases, and there are a total of $2^{10}$ ways to color the pentagon. Therefore the answer is $1-\frac{12}{1024}=1-\frac{3}{256}=\frac{253}{256}=D$


Amos is taller than Eugene.  Leo is shorter than James but taller than Eugene.  James is shorter than Amos.  is the tallest andis the shortest.

Amos, James

James, Eugene

James, Leo

Amos, Eugene

From clue $$2$$, James is taller than Leo and Leo is taller than Eugene.  From clue $$3$$, Amos is taller than James.  Rank from tallest to shortest: \textbf{Amos}, James, Leo, \textbf{Eugene}.


How many different positive integers at most can add up to $$80$$?

$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10+11+12 = 78$$, $$80 -- 78 = 2$$. Since all the numbers should be different, the remaining \textquotesingle$$2$$\textquotesingle{} cannot make up a new positive integer. It can only be added to the number(s) before. Therefore, there are at most $$12$$ different positive integers that can add up to $$80$$.


Which of the following statements is not correct?

Indefinite events include impossible events.

The probability of an impossible event to happen is $$0$$.

The probability of an indefinite event to happen is between $$0$$ and $$1$$.

The probability of a certain event to happen is $$1$$.

Impossible events are definite events.  $$\text{B}$$, $$\text{C}$$, and $$\text{D}$$ are right.  Thus, the answer is $$\text{A}$$.


A bag contains four pieces of paper, each labeled with one of the digits $$1$$, $$2$$, $$3$$ or $$4$$, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $$3$$? ($$2007$$ AMC $$8$$ Problem, Question \#$$24$$)

The combinations of digits that give multiples of $$3$$ are $$(1$$, $$2$$, $$3)$$ and $$(2$$, $$3$$, $$4)$$. For each of them, there are $3\times2\times1=6$ possibilities. Thus, there are $6+6=12$ possibilities in total. The number of ways to choose three digits out of four is $$4\times3\times2=24$$. Therefore, the probability is $$\frac{12}{24}=\frac12$$.


How many two-digit numbers are there where the ones digit is greater than the tens digit?．

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


There is a ball in a box. Three kids are guessing the colour of the ball.  Val says: "The ball is white."  John says: "The ball is blue."  Elvis says: "I agree with Val."  Then, they open the box and find only one of them guessed right.  The ball is  .

white

Blue

Uncertain

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

Val\textquotesingle s point and Elvis\textquotesingle{} points are identical, so both of them guessed incorrectly.  Therefore, John guessed correctly.


A number is formed by writing $2022$ consecutively for $3$ times. Which of the following statements is correct?

The number is a multiple of $3$, $6$ and $9$.

The number is a multiple of $3$ and $6$ but not a multiple of $9$.

The number is a multiple of $3$ and $9$ but not a multiple of $6$.

The number is a multiple of $3$ but not a multiple of $6$.

None of the above

The sum of the digits of 202220222022 is 2 x 9 = 18. Hence this number is divisible by 3 and 9. As its last digit is 2, it is also divisible by 6. The answer is \textbf{Option A.}


The average height of June, Ali and Linda is $$160\text{cm}$$. Ali is $$166 \text{cm}$$ tall. June and Linda are as tall as each other. What is Linda\textquotesingle s height?

$$\rm~~154 cm$$

$$\rm~~157 cm$$

$$\rm~~162 cm$$

$$\rm~~164 cm$$

The total height is $160\times 3 = 480\text{cm}$ and hence the height of both June and Linda is $480-166=314\text{cm}$.  Therefore the height of Linda is $314 \div 2 = 157\text{cm}$.


SASMO 2014 P2 Q8  Find the number A such that the following statement is true:  7 x A = 3 x 8 + 4 x 8


What is the probability of choosing a composite number from $0\sim10$?

There are $11$ numbers in total. Among them, $4,6, 8, 9,$ and $10$ are composite numbers.


There are $$30$$ pupils in my class. $$20$$ pupils like Maths and $$18$$ pupils like English. Twice as many pupils like both subjects as like neither of them. How many pupils like only Maths?

Let the number of pupils who like neither subject be $$x$$. Hence the number who like both subjects is $$2x$$. Therefore the number of pupils who like only Maths is $$20−2x$$ and the number who like only English is $$18−2x$$. Since there are $$30$$ pupils in my class, we have $$\left( 20-2x \right)+2x+\left( 18-2x \right)+x=30$$ and hence $$38−x = 30$$. This has solution $$x = 8$$ and hence the number of pupils who like only Maths is $$20-2\times 8=4$$.


Amos, James

James, Eugene

James, Leo

Amos, Eugene

From tallest to shortest: \textbf{Amos}, James, Leo, \textbf{Eugene}.


The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number? (2019 AMC 8 Problems, Question \#18)

We have a $2$ dice with $2$ evens and $4$ odds on both dice. For the sum to be even, the $2$ rolls can be $2$ odds or $2$ evens.  Ways to roll $2$ odds: The total number of ways to obtain $2$ odds on $2$ rolls is $4 * 4=16$, as there are $4$ possible odds on the first roll and $4$ possible odds on the second roll.  Ways to roll $2$ evens: Similarly, we have $2 * 2=4$ ways to obtain $2$ evens.  The probability is $\frac{20}{36}=\frac{5}{9}$.


How many digits are there from $5$ to $118$?


None of the above


In the UK it is now $$11$$ am. The time in San Francisco is $$8$$ hours behind the UK. What time do the clocks now show in San Francisco?

$$3$$ am

$$4$$ am

$$5$$ am

$$11$$ am

$$7$$ pm

Eight hours behind $$11$$ am is $$3$$ am.


Nick wants to bake a cake that consists of only $$1$$ flavour and $$1$$ topping. He gets to choose from $$3$$ different flavours and $$3$$ different toppings. How many different kinds of cake can he make?

None of the above

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


Two different numbers are randomly selected from the set $-2, -1, 0, 3, 4, 5$~and multiplied together. What is the probability that the product is $0$? ($2016$ AMC $8$ Problem, Question \#$13$)

The product can only be $0$ if one of the numbers is $0$. Once we choose $0$, there are $5$ ways of choosing the second number, and there are 15 ways of choosing $2$ numbers randomly. Thus $\frac{5}{15} = \frac{1}{3}$. The answer is $D$.


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

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 $$688\div8=86$$.


How many distinguishable arrangements are there of $1$ blue tile, $2$ green tiles, and $3$ yellow tiles in row from left to right? (Tiles of the same color are indistinguishable)


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


Use a simple method to compute the following.  $$25\times38$$=．

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


Moon and Archie played chess competitively. Both of them had same levels of skill. They agreed to play seven games, and the one that win four games first would be the ultimate winner. Now, they have already played three games, and Moon won two games while Archie won one game. What is the probability that Mon be the ultimately winner?


One day, Pip asks his parents: "What day is it today?"  His mother says: "Today is Monday."  His father says: "Today is Tuesday."  From the options below, \textbf{Pip agrees with B}. Do you agree with Pip?

If Pip\textquotesingle s mother is wrong, then his father must be right.

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

"Today is Monday" is not the opposite of "Today is Tuesday".i.e. they can both be false.  "Today is Monday" is the direct opposite of "Today is not Monday". One must be true and the other must be false.


Abe holds $1$ green and $1$ red jelly bean in his hand. Bob holds $1$ green, $1$ yellow, and $2$ red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match? (2013 AMC 8 Problem, Question \#14)

The probability that both show a green bean is $\dfrac{1}{2}\times\dfrac{1}{4}=\dfrac{1}{8}$. The probability that both show a red bean is $\dfrac{1}{2}\times \dfrac{2}{4}=\dfrac{1}{4}$. Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$.


Two tiles numbered $1$ and $2$ are turned face down, respectively. One tile is turned up at random, and throw a die to get a number from $1$ to $6$. What is the probability that the product of the numbers on the tile and the die is greater than or equal to $12$?

$\frac {1}{2}$

$\frac {1}{6}$

$\frac {1}{3}$

$\frac {1}{4}$

$\frac {1}{12}$

There are $12$ different combinations. The product of two numbers is greater than $12$ will be $2\times6$. Thus, the probability is $\frac 1{12}$ .


What is the value of the following sum?  $$299 + 297 + 295 + 296 + 298$$

None of the above


There are $$29$$ students in a certain class. $$12$$ of the students have a sister and $$18$$ of the students have a brother. In this class, only Tania, Barbara, and Anna do not have any siblings. How many students from this class have both a brother and a sister?

$$$$None


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


Julia, Kasia, Zuzanna, and Helena have their birthdays on March $$1^{\rm st}$$, May $$17^{\rm th}$$, July $$20^{\rm th}$$, and March $$20^{\rm th}$$. Kasia and Zuzanna were born in the same month. Julia and Zuzanna were born on the same day of a month. Which of the girls was born on May $$17^{\rm th}$$?

Julia

Kasia

Zuzanna

Helena

Given that Kasia and Zuzanna were born in the same month, their birth month must be March. Given that Julia and Zuzanna were born on the same day of a month, they must be born on the $$20^{\rm th}$$. Hence, Zuzanna was born on March $$20$$; Kasia was born on March $$1$$; and Julia was born on July $$20$$. Helena is therefore the one born on May $$17$$.


A cat divides $$24$$ fish into $$4$$ groups, and each group has at least $$1$$ fish. There are at most~\uline{~~~~~~~~~~}~fish in the group that has most of the fish.


What is the smallest possible sum of two positive integers with a product of $$100$$?

Since the product of the two positive integers is $$100$$, the possible pairs of integers are $$\left( 1,100 \right)$$, $$\left( 2,50 \right)$$, $$\left( 4,25 \right)$$, $$\left( 5,20 \right)$$, $$\left( 10,10 \right)$$, the smaller the difference, the smaller the sum, so the smallest sum is $$10+10=20$$.


The diagram shows some cubes of the same size stacked at a corner of a room. How many cubes are there altogether? (Note: The floor is horizontal and the two walls are vertical. There are no gaps or holes behind the visible cubes).


In how many ways can the letters in $CPCCKBY$ be rearranged so that $C$ cannot be put in both ends and two or more $C$s do not appear together?

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

There are $3$ $C$s in total now with other $4$ letters remaining.  There are $\_4P\_4$ ways for us to arrange the $4$ letters\textquotesingle{} positions.  So the answer is $\_4P\_4=24$.


A drawer contains ten identical yellow socks, eight identical blue socks and one hundred identical pink socks. Amrita picks socks from the drawer without looking.  What is the smallest number of socks she must pick to be sure that she has at least two pairs of matching socks?

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

Consider the worst case scenario: first 3 are drawn, all of different colors, when the fourth is drawn, no matter which color is drawn, a pair of the same color will definitely be matched. Continue to draw the fifth one. The worst case is that the last color socks are drawn again, and at this point there are three socks of the same color and one sock of each of the other two colors. If you continue to draw the sixth one, no matter which color you draw, you will be able to form two pairs of socks of the same color. So the answer is to draw 6 times.


In a box, there are $10$ balls of which $4$ are red, $3$ are black and $3$ are white. What is the probability of picking up $3$ balls randomly such that there are $2$ red balls and $1$ balck ball?

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


How many triangles of all sizes can be seen in the picture below?  [insert pic]


Let $(a, b, c, d)$ be integers where they are not necessarily different. If each one of them is in the set ($0$, $1$, $2$, $3$), what is the probability that $a \cdot d-b \cdot c$ is odd? (For example, ($0$, $3$, $1$, $1$) meet the condition, because $0 \cdot 1-3 \cdot 1=-3$ is odd.)

In order for $a \cdot d-b \cdot c$ to be odd, we need to consider the parity. We must have (even)-(odd) or (odd)-(even). There are $2\times2+2\times4=12$ ways to pick numbers to obtain an even product. There are $2 \cdot 2=4$ ways to obtain an odd product. Therefore, the total amount of ways to make $a \cdot d-b \cdot c$ odd is $2 \cdot(12 \cdot 4)=96.$ Thus, the answer is $\frac{96}{256}=\frac38$


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 $m+n$?

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 $$ \frac{6 \cdot 3 \cdot 2}{11 \cdot 6 \cdot 3 \cdot 2 \cdot 2}=\frac{1}{22} \Longrightarrow \frac{m}{n}=\frac{1}{22} \Longrightarrow m+n=1+22=\text { (E) } 23 $$.


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 $$285\div3 = 95$$.


A pair of fair 6 -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals 7 at least once is greater than $\frac{1}{2}$ ?

Rolling a pair of fair 6 -sided dice, the probability of getting a sum of 7 is $\frac{1}{6}$ : Regardless what the first die shows, the second die has exactly one outcome to make the sum 7 . We consider the complement: The probability of not getting a sum of 7 is $1-\frac{1}{6}=\frac{5}{6}$. Rolling the pair of dice $n$ times, the probability of getting a sum of 7 at least once is $1-\left(\frac{5}{6}\right)^{}n$. Therefore, we have $1-\left(\frac{5}{6}\right)^{}n\textgreater\frac{1}{2}$, or $$ \left(\frac{5}{6}\right)^{}n\textless\frac{1}{2} $$ Since $\left(\frac{5}{6}\right)^{4}\textless\frac{1}{2}\textless\left(\frac{5}{6}\right)^{3}$, the least integer $n$ satisfying the inequality is (C) 4 .


There are $$20$$ red balls, $$2$$ black balls, and $$1$$ white ball in a bag. They are of identical shape, size and quality except for colour. Take out $$1$$ ball without looking. Among the following statements, which one is true?

The ball taken out must be a black ball.

It is impossible that a white ball will be taken out.

It is very likely that a red ball will be taken out.

It is certain that a red ball will be taken out.

The number of red balls in the bag is the most compared with other coloured balls. If we take out one ball at random, the probability of taking out a red ball is larger. So $$\text{C}$$ is the answer.


There are some pieces of candy on a table.You are challenged by your friend to play the  following game: you both have to altemate moves, and in each move, you can take away either $$1$$, $$2$$, $$3$$,~ $$4$$ or $$5$$ pieces from the table. The person who takes away the final piece from the table wins. If you go second, how many pieces of candy should be on the table before the game starts such that you can ensure victory?

Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Game Strategy

Only $$18$$ is one of the multiples of $$5+1$$, and the second player can ensure victory.


Elsa is looking for some two-digit numbers. The difference between the digits in the tens place and the ones place is 5. How many of such numbers are there?


The faces of two identical fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number?

There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order). We can solve for how many ways the $2$ numbers add up to an odd number and subtract the answer from $1$.  How to solve the problem: The probability of getting an odd number first is $\dfrac{4}{6}=\dfrac{2}{3}$. In order to make the sum odd, we must select an even number next. The probability of getting an even number is $\dfrac{2}{6}=\dfrac{1}{3}$. Now we multiply the two fractions: $\dfrac{2}{3}\times\dfrac{1}{3}=\dfrac{2}{9}$. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except switched, and by the Communitive Property of Multiplication, it does not matter if the equations are switched. Thus we do $\dfrac{2}{9}\times2=\dfrac{4}{9}$. This is the probability of getting an odd-number sum. In order to get the probability of getting an even number we do $1-\dfrac{4}{9}=\left (\text{C}\right )\dfrac{5}{9}$.