There are many red balls, yellow balls, and blue balls of identical shape in a bag. Many students are playing a game in which each of them can take $$2$$ balls from the bag without observing the color of the balls. (After each student takes the balls, they should put them back in the bag.) The teacher finds that no matter how the students take the balls, there are always at least $$2$$ students who take the same balls. (For example, student $$A$$ and student $$B$$ both get one red ball and one yellow ball.) There would be at least~\uline{~~~~~~~~~~}~students playing the game.

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

There are $$6$$ possible cases when taking $$2$$ balls from the box. In the worst case, each of the first $$6$$ students takes different balls to the others. When the $${{7}^{\text{th}}}$$ student takes the balls, he/she would definitely take the same balls with someone among the first $$6$$ students. $$6+1=7$$.


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

$$13$$ minutes

$$23$$ minutes

$$39$$ minutes

$$47$$ minutes

$$73$$ minutes

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

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


Let $(a, b, c, d)$ be an ordered quadruple of not necessarily distinct integers, 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.) (Adapted from 2020 AMC 10A Problem, Question \#18)

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 $\frac{96}{256}=\frac38$


$$124$$ children attended a fair. $$87$$ of them tried out sports events. $$65$$ of them tried out art activities. How many children tried out both sport events and art activities?

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


None of the above

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


Professor Chang has nine language books lined up on a bookshelf: two different Arabic books, three different German books, and four different Spanish books. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together? ($2018$ AMC $8$ Problem, Question \#$16$)  $\textasciitilde$  $\textasciitilde$  $\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$

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

Since the two Arabic books and four Spanish books have to be kept together, respectively, we can treat them both as just one book. That means we\textquotesingle re trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $\_5P\_5$. Now we multiply this product by $\_2P\_2\times \_4P\_4$~because there are $\_2P\_2$~ways to arrange just two Arabic books, and $\_4P\_4$~ways to arrange just four Spanish books. Multiplying all these together, we have the answer $C$.


Amy, Bill, and Celine are friends with different ages. Exactly 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. ($$2004$$ AMC $$8$$ Problem, Question \#$$13$$)

Bill, Amy, Celine

Amy, Bill, Celine

Celine, Amy, Bill

Celine, Bill, Amy

Amy, Celine, Bill

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

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 lastly Bill is the youngest. This order is $$\rm E$$: Amy, Celine, Bill.


There is a three-digit number \textquotesingle$$502$$\textquotesingle. John writes a digit before this number to make it a new four-digit number. If he writes a~\uline{~~~~~~~~~~}~, the new four-digit number would be as small as possible.

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

Note that the digit \textquotesingle$$0$$\textquotesingle{} could not be the highest digit of a four-digit number. Therefore, the smallest number is $$1502$$.


The average of the \emph{different~}prime factors of $$2009$$ is．


In the expression $$1□2□3□4$$ each $$□$$ is to be replaced by either $$+$$ or $$\times$$. What is the largest value of all the expressions that can be obtained in this way?

If $$m$$ and $$n$$ are positive integers, then $$mn{}\textgreater m+n$$ unless at least one of $$m$$ or $$n$$ is equal to $$1$$, or $$m=n=2$$. So, to maximise the expression, we need to place multiplication signs between $$2$$ and $$3$$ and between $$3$$ and $$4$$. However, we need to place an addition sign between $$1$$ and $$2$$ because $$1+2\times3\times4=25$$, whereas $$1\times2\times3\times4=24$$.


There are $$12$$ gold coins with exactly the same appearance, including $$11$$ real coins and $$1$$ fake coin. The weight of the fake coin is different from that of the real coin, and whether the fake coin is lighter or heavier than the real coin is unknown. How many times at least do you need to weigh the coins using a balance to ensure that you can find the fake coin?

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

In the first weighing with the balance, put coins $$1$$, $$2$$, $$3$$, and $$4$$ on one end of the balance and coins $$5$$, $$6$$, $$7$$, and $$8$$ on the other end of the balance.  The balance has two situations: balanced or not.  Analyze the situation of the balance: if balanced, the fake coin is among the remaining $$4$$ coins.  In the second weighing with the balance, randomly take $$3$$ coins from coin $$1$$ to coin $$8$$ and put them on the left end of the balance and randomly take $$3$$ coins from coin $$9$$ to coin $$12$$ and put them on the right end of the balance (such as $$9$$, $$10$$, $$11$$). The balance also has two situations: balanced or not.  If balanced, coin $$12$$ is the coin of different weight. In the third weighing with the balance, comparing No. $$12$$ coin with any other coin, whether the coin is lighter or heavier can be known.  If not, it can be known that the coin of different weight is among the three coins $$9$$, $$10$$, and $$11$$, and that whether it is lighter or heavier than other coins can be known. In the third weighing with the balance, randomly take two of the coins (such as $$9$$ and $$10$$) and put them on the both ends of the balance. If balanced, the remaining coin (coin $$11$$) is the one we are looking for; if not, based on the previous judgement that whether the coin is lighter or heavier, it can be determined that which one of the coins on the balance is what we are looking for.  Analyze the first imbalanced situation as follows:  There are two situations: the right end weighs more or the left end weighs more. Assume the left end weighs more (which is the same for the situation that the right end weighs more.)  In the second weighing with the balance, take off $$3$$ coins randomly from the left end (such as $$1$$, $$2$$, and $$3$$) and move $$3$$ coins from the right end to the left end (such as $$5$$, $$6$$, and $$7$$), then take $$3$$ coins randomly from the $$4$$ coins left in the first weighing (such as $$9$$, $$10$$, and $$11$$) to the right end, and there sees $$3$$ situations for the balance: ① the left end weighs more, ② the two ends strike a balance, ③ the right end weighs more. Analyze the situations one by one as follows:  ① If the left end weighs more, the coin we are looking for must be coin $$4$$ or coin $$8$$. In the third time weighing with the balance, take one of the coins (such as coin $$4$$) and put it on the left end of the balance. Randomly take one of the remaining $$10$$ coins and put it on the right end. There are also $$3$$ situations.  $$a$$: If balanced, coin $$8$$ is the one we are looking for. Based on the result of using the balance for the second time, it is known that the coin weighs less than other coins.  $$b$$: If the left end weighs more, coin $$4$$ is the one we are looking for and it weighs more than other coins.  $$c$$: If the right end weighs more, coin $$4$$ is the one we are looking for and it weighs less than other coins.  ② If the two ends strike a balance, the coin we are looking for is among the three coins ($$1$$, $$2$$, and $$3$$) taken from the left end. Since the left end weighs more in the first weighing, it is known that the coin weighs more than other coins. The following analysis is the same as previous one and will not be repeated.  ③ If the right end weighs more, the coin we are looking for is among the three coins ($$5$$, $$6$$, and $$7$$) moved from the right end to the left end. Based on the results of weighing in the first two times (the left end weighs more in the first weighing and the right end weighs more in the second weighing), it is known that the coin weighs less than other coins. The following analysis is the same.


Two tiles numbered $3$ and $4$ 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 $20$?

$\frac {1}{2}$

$\frac {1}{6}$

$\frac {1}{3}$

$\frac {1}{4}$

$\frac {1}{12}$

There are $12$ different combinations. The product of two numbers is greater than $20$ will be $4\times5$ and $4\times6$. Thus, the probability is $\frac 2{12}$ = $\frac 16$.


A cat divides $$24$$ fish into $$4$$ groups, and each group has at least $$1$$ fish. There are  fish in the group that has the largest number of fish.


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

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 $\_4P\_4\times \_5C\_3 =240$.


The digits $1$, $2$, and $3$ can make~\uline{~~~~~~~~~~}~three-digit numbers.


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?

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

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.


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?

Overseas Competition->Knowledge Point->Counting Modules


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 or equal to $8$?

$\frac {1}{2}$

$\frac {1}{3}$

$\frac {1}{4}$

$\frac {1}{5}$

$\frac {1}{6}$

There are $2\times6=12$ different combinations. The product of two numbers is smaller than $8$ will be $5\times1$ and $6\times1$. Thus, the probability is $\frac 2{12}$ = $\frac 16$.


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

There are $4$ $A$s in total, which have $3$ intervals leaving for the other $3$ letters. Thus, the answer is $\_3P\_3=6$.


If Keyue fished every day from July $$17$$ through July $$31$$ (within the same year), she fished fordays．

Total number of days $$=31-17 +1= 15$$ days.


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

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


Two distinct numbers from $$1$$ to $$100$$ inclusive will form a pair if the sum of these two is a multiple of $$5$$. How many different pairs are there?


Today is Amy\textquotesingle s birthday! She takes photos with her four close friends. All of them stand in a line and they make Amy stand in the middle. There are~\uline{~~~~~~~~~~}~different ways for them to form the line.

$$4 \times 3 \times 1 \times 2 \times 1 = 24$$.


There are $$24$$ four-digit numbers which is formed using 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 the eighth position?

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


The maximum number of intersection points of $$4$$ different circles is．

Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math->Dotted Line Arrangement

Each pair has $$2$$ intersection points. The $$6$$ pairs have $$12$$ such points.


In how many ways can the digits in $3433256337$ be rearranged so that two or more $3$s do not appear together?

There are $5$ $3$s in total now with other $5$ digits remaining.  There are $\_5P\_5$ ways for us to arrange the $5$ letters\textquotesingle{} positions. Then, we can put the $5$ $3$s in the $6$ intervals.  So the answer is $\_5P\_5 \times \_6C\_5=720$.


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

One £5 note and three~£1 coins

Eight £1 coins

One £5 note and four 50p coins


Elvis is packing shirts for a trip. He just randomly grabs $3$ shirts from his closet without observing them. The closet contains $10$ shirts: $5$ striped, $3$ plaid, and $2$ pure-colored ones. What is the probability that he chooses $2$ striped shirts and $1$ pure-colored shirt?

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


We can formdifferent two-digit numbers with the numbers $$1$$, $$3$$ and $$0$$.（without using the same number two times like $$33$$ ）

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

$$13$$、$$10$$、$$31$$、$$30$$， notice that $$0$$ cannot be in the first place, so only $$4$$ numbers can be formed．  So the answer is $$\text{B}$$．


How many four-digit numbers can be made using the digits $1-8$ without repeating digits?


Eight cards are numbered from $$1$$ to $$8$$. The cards are placed in two boxes $$P$$ and $$Q$$ so that the sum of the numbers on the three cards in box $$P$$ is equal to the sum of the numbers on the five cards in box $$Q$$. Which of the following statements must be true?

The card numbered $$1$$ is not in box $$Q$$

Four cards in box $$Q$$ have even numbers on

The card numbered $$5$$ is in box $$Q$$

The card numbered $$2$$ is in box $$Q$$

Exactly three cards in box $$Q$$ have odd numbers on

Note first that the sum of the numbers on the eight cards is $$36$$. Therefore the sum of the numbers on the cards in each of the boxes is $$18$$. There are only three cards in box $$P$$ and hence the possible combinations for the numbers on the cards in box $$P$$ are $$\left( 8,7,3 \right)$$, $$\left( 8,6,4 \right)$$ and $$\left( 7,6,5 \right)$$ with the corresponding combinations for box $$Q$$ being $$\left( 6,5,4,2,1 \right)$$, $$\left( 7,5,3,2,1 \right)$$ and $$\left( 8,4,3,2,1 \right)$$. The only statement which is true for all three possible combinations for box $$Q$$ is that the card numbered $$2$$ is in box $$Q$$. Hence the only statement which must be true is statement $$\rm D$$.


Amy has $$7$$ gray balls, $$4$$ white balls and $$3$$ black balls in a bag. What is the least number of balls she has to take out of the bag with her eyes closed to be sure that she takes at least two balls of each color?

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


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 $\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$. The probability that both show a red bean is $\frac{1}{2} \cdot \frac{2}{4}=\frac{1}{4}$. Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\left(\right.$ C) $\frac{3}{8}$


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

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


A dinner set in a restaurant is free to choose a main course, a salad and a drink. Today, the restaurant offers five main courses, three salads, and six drinks for customers to choose. How many different sets can be matched? ．


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

The sum of dots is $$12$$.

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

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

The sum of dots is $$13$$.

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


2020年希望杯二年级竞赛模拟第30题 其它

There are $13$ kids in line. They are numbered following the pattern: $1$, $2$, $3$, $4$, $5$, $1$, $2$, $3$, $4$, $5$, and so on. How many kids are numbered with odd numbers?

$13\div5=2R3$  There are $3$ odd numbers in each cycle.  $3\times2+2=8$


A palindromic number is a number that reads the same when the order of its digits is reversed. What is the difference between the largest and smallest five-digit palindromic numbers that are both multiples of $$45$$?

For a number to be a multiple of $$45$$ it must be a multiple of $$5$$ and also of $$9$$. In order to be a multiple of $$5$$, a number\textquotesingle s units digit must be $$0$$ or $$5$$. However, the units digit of a palindromic number cannot be $$0$$, so it may be deduced that any palindromic number which is a multiple of $$45$$ both starts and ends in the digit $$5$$. In order to make the desired number as large as possible, its second digit should be $$9$$ and for it to be as small as possible its second digit should be $$0$$. So, if possible, the numbers required are of the form \textquotesingle$$59x95$$\textquotesingle{} and \textquotesingle$$50y05$$\textquotesingle{} . In addition, both numbers are to be multiples of $$9$$ which means the sum of the digits of both must be a multiple of $$9$$. For this to be the case, $$x=8$$ and $$y=8$$ , giving digit sums of $$36$$ and $$18$$ respectively. So the two required palindromic numbers are $$59895$$ and $$50805$$. Their difference is $$9090$$.


There are 70 beads.~ $$A$$ and $$B$$ take turns taking beads from the pile ($$A$$ goes first). Each person can take between 1 and 4 beads per turn. The person who takes the last bead loses. If $$A$$ wants to win, how many beads should he take on the first turn?

$$A$$ can\textquotesingle t win

Overseas Competition->Knowledge Point->Combinatorics->Math Game->Sudoku

$$(70-1)\div (1+4)=17\cdots 1$$  $$A$$ takes one bead first, and then regardless of how many beads $$B$$ takes, as long as the sum of $$A$$ and $$B$$ is 5, $$A$$ will win.


What is $$6$$ ones and $$15$$ thousandths?

$$6$$ ones $=6$  $$15$$ thousandths $=$ $1$ hundredth and $5$ thousandths $=0.015$  $\textasciitilde$  Total $=6.015$


There are $$20$$ students in a class standing in a line for recess. Mike is the $$7$$\textsuperscript{th} counting from front to back. How many students are behind Mike? What is his position counting backwards?

$$12$$, $$13$$

$$13$$, $$14$$

$$13$$, $$12$$

$$14$$, $$13$$

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

The number of students behind Mike plus the position of Mike equals the total number of students in the class. So there are $$13$$ students behind Mike: $$20-7=13$$. However Mike is not included in these $$13$$ students, which means Mike is the $$14$$\textsuperscript{th} student counting from back to front.


Sophia's average score on six tests is $$82$$. Her average scores on the $$7^{}\text{th}$$ and $$8^{}\text{th}$$ tests is $$98$$. Sophia then took another test and the average score of all $9$ tests is $87$?

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

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


In an opaque bag, there are $5$ red balls, $5$ white balls, and $5$ yellow balls. The balls are the same except for their colours. Take out one ball from the bag. What is the probability that you take out a yellow ball?


Maria had $$28$$ dreams last month. If $$16$$ of them involved monkeys, $$15$$ involved squirrels, and $$4$$ involved no animals, then at least how many dreams involved both monkeys and squirrels?

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

Maria had $$28$$ dreams last month, $$24$$ of which involved animals. Since $$16+ 15 =31$$ involved moneys or squirrels, then at least $$31 - 24 = 7$$ dreams involved both monkeys and squirrels.


$\frac{7}{9}$, $\frac{5}{4}$  Subtract the sum from the difference of the above two fractions.

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


Two dice are thrown. What is the probability that the product of the two numbers is a multiple of $5 ?$ (2001 AMC 8 Problem, Question \#14)

This is equivalent to asking for the probability that at least one of the numbers is a multiple of 5 , since if one of the numbers is a multiple of 5 , then the product with it and another integer is also a multiple of 5 , and if a number is a multiple of 5 , then since 5 is prime, one of the factors must also have a factor of 5 , and 5 is the only multiple of 5 on a die, so one of the numbers rolled must be a 5 . To find the probability of rolling at least one 5 , we can find the probability of not rolling a 5 and subtract that from 1, since you either roll a 5 or not roll a 5 . The probability of not rolling a 5 on either dice is $\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)=\frac{25}{36}$. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of 5 , is $1-\frac{25}{36}=\frac{11}{36}, \text{D}$


Louis chooses a whole number from $$1$$ to $$30$$ at random. What is the probability that the number he chooses is a prime number?


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

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


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 for both time?


How many numbers are there between $$0 \sim 50$$ that do not consist of digit \textquotesingle$$6$$\textquotesingle?．


How many $$4$$-digit numbers greater than $$1000$$ are there that use the four digits of $$2012$$? ($$2012$$ AMC $$8$$ Problems, Question \#$10$)

There are $3$ numbers that can be formed between $1000$ and $1999$, and $3\times2\times1=6$ numbers that can be formed with $2$ in the first place.  So the answer is $3+6=9$.


Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 Driver seat, 1 front passenger seat, and 2 back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there? (2003 AMC 8 Problem, Question \#16)

There are only 2 people who can go in the driver\textquotesingle s seat-Bonnie and Carlo. Any of the 3 remaining people can go in the front passenger seat. There are 2 people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are $2 \cdot 3 \cdot 2$ or 12 ways. The answer is then (D) 12 .


The sum of the digits of $$1993$$ is $$1+9+9+3$$, or $$22$$. At some time in the future, the sum of the digits of a year will be $$33$$. This will \emph{first} occur in the century.

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

The sum of the digits of $$1993$$ is $$1+9+9+3$$, or $$22$$. At some time in the future, the sum of the digits of a year will be $$33$$. Since $$9+9+9 = 27$$, this will \emph{first} occur in $$6999$$, the $$70\text{th}$$ century.


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 $$156-24=132$$.


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

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


There are some two-digit numbers. The sum of the digits in the tens place and the ones place is 7. How many such numbers are there?


There is a two-digit number. The product of its digits is $$18$$. The sum of the digits is $$11$$. The difference of the digits of this number is~\uline{~~~~~~~~~~}~.

$3\times6=18$, $3+6=9$  $2\times9=18$, $9+2=11$  Thus, the two digits of the number is $9$ and $2$.  The difference is $9-2=7$.


The clock is showing $$11\text{am}$$ now.  When the hour hand has turned through $$165{}^{}\circ$$, what time will it be?

Each hour, the hour hand turns $$30{}^{}\circ$$, so it will take $$5\frac{1}{2}$$ hours to turn $$165{}^{}\circ$$; the time will therefore be $$4.30\text{pm}$$.


In a particular game, each of 4 players rolls a standard 6 -sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo\textquotesingle s first roll was a 5 , given that he won the game?

Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a 5 is just the probability that the highest roll in the first round is 5 . Let $P(x)$ indicate the probability that event $x$ occurs. We find that $P($ No one rolls a 6$)-P($ No one rolls a 5 or 6$)=P($ The highest roll is a 5$)$, so $$ \begin{gathered} P(\text { No one rolls a } 6)=\left(\frac{5}{6}\right)^{4}, \textbackslash\textbackslash{} P(\text { No one rolls a } 5 \text { or } 6)=\left(\frac{2}{3}\right)^{4}, \textbackslash\textbackslash{} P(\text { The highest roll is a } 5)=\left(\frac{5}{6}\right)^{4}-\left(\frac{4}{6}\right)^{4}=\frac{5^{4}-4^{4}}{6^{4}}=\frac{369}{1296}=(\text { C }) \frac{41}{144} \text {. } \end{gathered} $$


Miley had some keychains at first. She gave $$\frac{6}{7}$$ of them to her friends. Her mother then gave her another $$8$$ keychains. She had $$15$$ keychains in the end. How many keychains did Miley give to her friends in total?

Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Word Problems Involving Fractions and Percentages->Finding a Whole Given a Part and the Percentage

$\frac{1}{7}$ of her keychains $=15-8=7$  $\textasciitilde$  Given to friends  $=\frac{6}{7}$ of her keychains  $=7\times6$  $=42$


Martin uses the five digits $$0$$, $$1$$, $$3$$, $$7$$, $$9$$ to make some numbers (each digit can only be used once).  How many different three-digit numbers can be made?


Ann sleeps just $$8$$ hrs. each day, so in $$10$$ days, she\textquotesingle s \emph{awake~}~\uline{~~~~~~~~~~}~hrs.

$$10\times 16$$

$$8\times 10$$

$$8\times 24$$

$$16\times 24$$

Ann\textquotesingle s awake $$16$$ hours each day. In $$10$$ days, that\textquotesingle s $$(10\times 16)$$ hours.


A bag contains 4 red chips, 2 blue chips, and 3 white chips. If one chip is drawn at random, what is the probability that the chip will not be red?


In how many ways can the letters in $CPCCKBY$ be rearranged so that two or more $C$s are not adjacent to each other?

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 remaining $4$ letters\textquotesingle{} positions.  So the answer is $\_4P\_4\times \_5C\_3 =240$.


In a particular game, each of $4$ players rolls a standard $6$-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo\textquotesingle s first roll was a $5$, given that he won the game?

Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a $5$ is just the probability that the highest roll in the first round is $5$. Let $P(x)$ indicate the probability that event $x$ occurs. We find that $P$ (No one rolls a $6$)-$P$ (No one rolls a $5$ or $6$)$=P$ (The highest roll is a $5$), so  $$ \begin{gathered} P(\text {No one rolls a } 6)=\left(\frac{5}{6}\right)^{4}, \textbackslash\textbackslash{} P(\text {No one rolls a } 5 \text { or } 6)=\left(\frac{2}{3}\right)^{4}, \textbackslash\textbackslash{} P(\text {The highest roll is a } 5)=\left(\frac{5}{6}\right)^{4}-\left(\frac{4}{6}\right)^{4}=\frac{5^{4}-4^{4}}{6^{4}}=\frac{369}{1296}=(\text {C}) \frac{41}{144} \text {. } \end{gathered} $$


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


How many three-digit numbers are there that have the sum of their digits equal to $5$? (For example, $122$ is such a number, because $1+2+2=5$.)．

Method 1:  $104, 113, 122, 131, 140$;  $203, 212, 221, 230$;  $302, 311, 320$;  $401, 410$;  $500$  $5+4+3+2+1=15$  Method 2:$5=0+0+5=0+1+4=0+2+3=1+1+3=1+2+2$  When the three-digit number is formed by $0$, $0$, $5$, it can only be $500$, therefore $1$ number;  when the three-digit number is formed by $0$, $1$, $4$, it can be $104$, $140$, $401$ or $410$, therefore $4$ numbers;  when the three-digit number is formed by $0$, $2$, $3$, it can be $203$, $230$, $302$, or $320$, therefore $4$ numbers；  when the three-digit number is formed by $1$, $1$, $3$, it can be $113$, $131$, or $311$, therefore $3$ numbers;  when the three-digit number is formed by $1$, $2$, $2$, it can be $122$, $212$, or $221$, therefore $3$ numbers;  in total there are $1+4+4+3+3=15$ numbers that meet the requirement of the question.


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


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 an end, how many different ways can they line up for the picture?  ②If Mr. Carter insists on standing on an end, how many different ways can they line up for the picture?

$72$ , $24$

$96$ , $24$

$72$ , $48$

$96$ , $48$

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


Adele has $1$ blue and $1$ red marble in her hand. Bobby has $1$ blue, $1$ yellow, and $2$ red marbles in his hand. Each of them randomly picks a marble to show the other. What is the probability that the colors are the same?

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


Which of the following statements is wrong  ?

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

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

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

Indefinite events include impossible events.

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


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, each one of Joe's statements is false. Bret is actually sitting in seat \#$$3$$. Who is sitting in seat \#$$2$$? ．

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 \#$$2$$.


Bruno Mars is packing $3$ shirts for a trip. He just randomly grabs $3$ shirts from his closet without observing them. The closet contains $10$ shirts: $5$ striped, $3$ plaid, and $2$ solid-colored ones. What is the probability that he chooses $2$ striped shirts and $1$ solid-colored shirt?


Eddie is ordering lunch at a fast food restaurant with the following menu.  Mains: Sandwich, Burger, Pizza  Sides: Chicken Wings, Salad  Drinks: Coffee, Tea, Coke  If Eddie chooses to buy one main, one side, and one drink, how many different ways can he order lunch?

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


A conductor wanted to make a trio consisting of a violinist, a pianist, and a drummer. He had to choose one of two violinists, one of two pianists, and one of two drummers. How many possible combinations did he have?


The room numbers of a hotel are all three-digit numbers. The first digit represents the floor and the last two digits represent the room number. The hotel has rooms on five floors, numbered $$1$$ to $$5$$. It has $$35$$ rooms on each floor, numbered $$n01$$ to $$n35$$ where $$n$$ is the number of the floor. In numbering all the rooms, how many times will the digit $$2$$ be used?

Each floor has $$35$$ rooms. On every floor except floor $$2$$, the digit $$2$$ will be used for rooms \textquotesingle$$n02$$\textquotesingle, \textquotesingle$$n12$$\textquotesingle, \textquotesingle$$n20$$\textquotesingle{} to \textquotesingle$$n29$$\textquotesingle{} (including \textquotesingle$$n22$$\textquotesingle) and \textquotesingle$$n32$$\textquotesingle. Hence the digit $$2$$ will be used $$14$$ times on each floor except floor $$2$$. On floor $$2$$, the digit $$2$$ will be used an extra $$35$$ times as the first digit of the room number. Therefore the total number of times the digit $$2$$ will be used is $$5\times14+35 = 105$$.


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

$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 $\rm D$.


How many numbers are there between $$0 \sim 50$$ that do not consist of digit \textquotesingle$$6$$\textquotesingle?．

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?

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 red chip.  The answer is $$\dfrac{4}{10}=\dfrac{2}{5}$$, or $$\boxed { (\text{B})\dfrac{2}{5}}$$.


There is a box containing~$7$ 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 $\frac37$.


When three line segments with the lengths of $3$, $$4$$, and $$7$$ are used to form a triangle, it is called a/an~\uline{~~~~~~~~~~}~.

Random event

Certain event

Impossible event

None of the above

∵$$3+4=7$$,  ∴ Three line segments measuring $$3$$, $$4$$, and $$7$$ cannot form a triangle,  ∴ Using three line segments measuring $$3$$, $$4$$, and $$7$$ to form a triangle is an impossible event.  So $$\text{C}$$ is the answer.


Josh started playing soccer at $$14 : 35$$. He stopped playing at $$16 : 05$$. How much time did he spend playing soccer?~\uline{~~~~~~~~~~}~$$\text{h}$$~\uline{~~~~~~~~~~}~$$\min $$


How many three-digit whole numbers have a ones digit equal to the sum of the hundreds digit and the tens digit?

If the ones digit is $$2$$, the $$2$$ numbers are $$202$$ and $$112$$. For each ones digit, the number of possible numbers is the same as the ones digit. In all, there are $$1+2+3+\cdots+8+9 =45$$ numbers.


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,  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.


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

$$90 $$

$$ 100 $$

$$ 110 $$

$$120 $$

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

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 $$75 + 35 = 110$$.


The code is a two-digit number. The ones digit is 6 more than the tens digit. How many possible codes are there in total?

Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers->Complex Forming Numbers (with special requirements)


If the correct time now is $$1:15$$ P.M., and if my clock stopped running three and one-half hours ago, when did it stop running?

$$8:45$$ A.M.

$$9:45$$ A.M.

$$10:45$$ A.M.

$$4:45$$ P.M.

One hour ago the time was $$12:15$$ P.M.; two hours ago the time was $$11:15$$ A.M.; three hours ago the time was $$10:15$$ A.M. Three and one-half hours ago the time was $$9:45$$ A.M.


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.


A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly? (2014 AMC 8 Problems, Question \#12)

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

There are $3 \times 2 \times 1 = 6$ ways assign the pictures to each of the celebrities. There is one favorable outcome where all of them are matched correctly, so the answer is (B) $\frac{1}{6}$.


Nini has $$6$$ identical balls and she wants to place them into $$3$$ identical baskets. How many different ways can she do so? (The baskets cannot be empty.)

Since the balls and baskets are all identical, the order does not matter.  $$6=1+1+4$$  $$6=1+2+3$$  $$6=2+2+2$$  So, there are $3$ ways to place them.


The numbers from $1$ to $40$ are written on $40$ pieces of paper, respectively, and put into a box. A piece of paper is chosen at random. Find the probability of choosing a number greater than $20$.

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

Half the numbers from $1$ to $40$ are greater than $20$, Thus, the probability of choosing numbers greater than $20$ is $\dfrac{1}{2}$.


How many different cubes are there with three faces coloured red and three faces coloured blue?

Consider the case where two opposite faces are coloured red. Whichever of the four remaining faces is also coloured red, the resulting arrangement is equivalent under rotation to a cube with top, bottom and front faces coloured red. Hence, there is only one distinct colouring of a cube consisting of three red and three blue faces with two opposite faces coloured red. Now consider the case where no two opposite faces are coloured red. This is only possible when the three red faces share a common vertex and, however these faces are arranged, the resulting arrangement is equivalent under rotation to a cube with top, front and right-hand faces coloured red. Hence there is also only one distinct colouring of a cube consisting of three red and three blue faces in which no two opposite faces are coloured red. Therefore there are exactly two different colourings of the cube as described in the question.


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.


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

Since the two Arabic books and four Spanish books have to be kept together, respectively, we can treat them both as just one book. That means we\textquotesingle re trying to find the number of ways that you can arrange one Arabic book, one Spanish book, and three German books, which is just $\_5P\_5$. Now we multiply this product by $\_2P\_2\times \_4P\_4$~because there are $\_2P\_2$~ways to arrange just two Arabic books, and $\_4P\_4$~ways to arrange just four Spanish books. Multiplying all these together, we have the answer $C$.


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 $t-s=(\mathbf{B})-13.5$.


Several distinct positive integers no larger than $$10$$ are listed in a row. After inspecting the row of numbers, Mark concluded with astonishment that in each pair of adjacent numbers one is a divisor of the other. At most how many numbers are in the list? ．

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

$$6$$, $$3$$, $$9$$, $$1$$, $$4$$, $$8$$, $$2$$, $$10$$, $$5$$. Only $$7$$, which is a prime number and doesn\textquotesingle t have multiples except itself among $$1$$\textasciitilde$$10$$, doesn\textquotesingle t appear.


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? ．

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


In the expression $$1□2□3□4$$ each $$□$$ is to be replaced by either $$+$$ or $$\times$$. What is the largest value of all the expressions that can be obtained in this way?

If $$m$$ and $$n$$ are positive integers, then $$mn{}\textgreater m+n$$ unless at least one of $$m$$ or $$n$$ is equal to $$1$$, or $$m=n=2$$. So, to maximise the expression, we need to place multiplication signs between $$2$$ and $$3$$ and between $$3$$ and $$4$$. However, we need to place an addition sign between $$1$$ and $$2$$ because $$1+2\times3\times4=25$$, whereas $$1\times2\times3\times4=24$$.


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

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


How many different $3-$digit numbers are there with the sum of digits as $15$ ?

Overseas Competition->Knowledge Point->Counting Modules

To divide $15+1+1=17$ into three groups, there are $\_{16}C\_{2}=120$ ways in total.  But, since all digits cannot be larger than $9$, except the situations that don\textquotesingle t meet the requirements, there are $120-\_7C\_2-\_6C\_2\times 2=69$ numbers that match the condition.


Gregor, the mathematician, forms two numbers with the digits $$0$$, $$1$$, $$2$$, $$3$$, $$4$$, and $$5$$. Both numbers have three digits, and each digit is used only once. He adds these two numbers. What is the greatest sum Gregor can get? (adapted from 2012 Math Kangaroo Problem, Level 3 - 4, Question \#19)

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

If we want to get the sum as large as possible, the hundred digits for both numbers should be as large as possible. Therefore, they should be $$5$$ and $$4$$. The sum of the hundreds digits is $$9$$. For the same reason, the sum of the tens digits should be $$3+2=5$$ and the sum of the ones digits should be $$1 + 0 = 1$$. Therefore, $$\rm C$$ is the correct answer.


There are $$10$$ cards in a box, which numbered $$1$$, $$3$$, $$4$$, $$6$$, $$7$$, $$11$$, $$15$$, $$16$$, $$18$$, and $$20$$, respectively. Cathy picks a card from the box randomly. What is the probability the number on the card that Cathy picks is a multiple of $3$?

$\frac 1{10}$

$\frac 3{10}$

$\frac 2{5}$

$\frac 1{2}$

$\frac 3{5}$

$3, 6, 15,$ and $18$ are multiples of $3$. Thus, the probability is $\frac 2{5}$.


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? ($$2003$$ Math Kangaroo Problems, Level $$3-4$$, Question \#$$17$$)

$$$$None

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


In how many ways can the digits in $3433256337$ be rearranged so that two or more $3$s are not adjacent to each other?

There are $5$ $3$s in total.  There are $\_5P\_5$ ways for us to arrange the other $5$ letters\textquotesingle{} positions. Then, we can put the five $3$s in the $6$ intervals.  So the answer is $\_5P\_5 \times \_6C\_5=720$.


$$31$$ children stand in a row from the shortest to the tallest. The last child is $$131$$ cm height. Tom stands the $${{10}^{\text{th}}}$$ place counting from the first. No two children have the same height. What is the maximum height of Tom? (The height of all the children are whole numbers.)

$$100$$ cm

$$105$$ cm

$$110$$ cm

$$111$$ cm

$$125$$ cm

Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Extreme Value in Inclusion-Exclusion for Multi-sets


In Bill\textquotesingle s class, every student can play either piano, or violin, or both. $24$ of his classmates can play piano, and the number of students that can play violin is half of that of playing piano. If there are $4$ students who can play both, how many students are there in the class?

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


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


There are 70 beads.~ $$A$$ and $$B$$ take turns taking beads from the pile ($$A$$ goes first). Each person can take between 1 and 4 beads per turn. The person who takes the last bead loses. If $$A$$ wants to win, how many beads should he take on the first turn?

$$A$$ can\textquotesingle t win

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

$$(70-1)\div (1+4)=13\cdots 4$$  $$A$$ takes four bead first, and then regardless of how many beads $$B$$ takes, as long as the sum of $$A$$ and $$B$$ is 5, $$A$$ will win.


The password for a bankbook has six digits (digits can be $$0$$). Mr. Wang forgets the first digit. The probability that he can get the first digit right on his first try is ．

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

∵ There are $$10$$ possible situations. However, there\textquotesingle s only one time for him to get the first digit right at the first try.  ∴ The probability that he can get the first digit right at the first try is $$\frac{1}{10}$$.  So $$\text{D}$$ is the answer.


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


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


A film on TV lasts $$2$$ hours and $$28$$ minutes, and finishes at $$11.18 \text{p}.\text{m}.$$ At what time does it start?


In a regular hexagonal prism, how many pairs of parallel edges can you find?


What is the sum of all the even numbers between $30$ and $40$?


How many of the following statements are wrongt?  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 $$\text{D}$$.


A drawer contains ten identical yellow socks, eight identical blue socks and four 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

Two matching pairs of socks could be obtained by choosing four socks, but this is not certain. For instance, two of one colour, one of a second colour and one of a third colour could be drawn. Combinations of five chosen socks would give two matching pairs unless three socks of one colour, one of a second colour and one of a third colour were chosen. When this is the case, drawing a sixth sock would guarantee that there would be two matching pairs as there would now be either four socks of one colour plus two other socks or three socks of one colour, two of a second colour plus one of a third colour.


Peter bought a carpet 36dm wide and 60dm long. The figure shows part of this carpet. As seen, the carpet has a small squares contianing either a sun or a moon. You can count that along the width there are nine squares. When the carpet is fully unrolled, how many moons will be seen?  [Insert pic]

Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Classifying and Enumerating->Counting Figures Formed by Grid Points

Each small square is 36/9 = 4 dm on a side. there are 60/4 = 15 squares along the length.  Both dimensions, 15 and 9, are odd. Also the corners are suns, so there is one more sun than there are moons.  Thus, there are a total of (15 x 9 - 1)/2 = 67 moons in the carpet when fully rolled.


There are $$12$$ gold coins with exactly the same appearance, including $$11$$ real coins and $$1$$ fake coin. The weight of the fake coin is different from that of the real coin, and whether the fake coin is lighter or heavier than the real coin is unknown. How many times at least do you need to weigh the coins using a balance before finding the fake coin?.

In the first weighing with the balance, put coins $$1$$, $$2$$, $$3$$, and $$4$$ on one end of the balance and coins $$5$$, $$6$$, $$7$$, and $$8$$ on the other end of the balance.  The balance has two situations: balanced or not.  Analyze the situation of the balance: if balanced, the fake coin is among the remaining $$4$$ coins.  In the second weighing with the balance, randomly take $$3$$ coins from coin $$1$$ to coin $$8$$ and put them on the left end of the balance and randomly take $$3$$ coins from coin $$9$$ to coin $$12$$ and put them on the right end of the balance (such as $$9$$, $$10$$, $$11$$). The balance also has two situations: balanced or not.  If balanced, coin $$12$$ is the coin of different weight. In the third weighing with the balance, comparing No. $$12$$ coin with any other coin, whether the coin is lighter or heavier can be known.  If not, it can be known that the coin of different weight is among the three coins $$9$$, $$10$$, and $$11$$, and that whether it is lighter or heavier than other coins can be known. In the third weighing with the balance, randomly take two of the coins (such as $$9$$ and $$10$$) and put them on the both ends of the balance. If balanced, the remaining coin (coin $$11$$) is the one we are looking for; if not, based on the previous judgement that whether the coin is lighter or heavier, it can be determined that which one of the coins on the balance is what we are looking for.  Analyze the first imbalanced situation as follows:  There are two situations: the right end weighs more or the left end weighs more. Assume the left end weighs more (which is the same for the situation that the right end weighs more.)  In the second weighing with the balance, take off $$3$$ coins randomly from the left end (such as $$1$$, $$2$$, and $$3$$) and move $$3$$ coins from the right end to the left end (such as $$5$$, $$6$$, and $$7$$), then take $$3$$ coins randomly from the $$4$$ coins left in the first weighing (such as $$9$$, $$10$$, and $$11$$) to the right end, and there sees $$3$$ situations for the balance: ① the left end weighs more, ② the two ends strike a balance, ③ the right end weighs more. Analyze the situations one by one as follows:  ① If the left end weighs more, the coin we are looking for must be coin $$4$$ or coin $$8$$. In the third time weighing with the balance, take one of the coins (such as coin $$4$$) and put it on the left end of the balance. Randomly take one of the remaining $$10$$ coins and put it on the right end. There also sees $$3$$ situations.  $$a$$: If balanced, coin $$8$$ is the one we are looking for. Based on the result of using the balance for the second time, it is known that the coin weighs less than other coins.  $$b$$: If the left end weighs more, coin $$4$$ is the one we are looking for and it weighs more than other coins.  $$c$$: If the right end weighs more, coin $$4$$ is the one we are looking for and it weighs less than other coins.  ② If the two ends strike a balance, the coin we are looking for is among the three coins ($$1$$, $$2$$, and $$3$$) taken from the left end. Since the left end weighs more in the first weighing, it is known that the coin weighs more than other coins. The following analysis is the same as previous one and will not be repeated.  ③ If the right end weighs more, the coin we are looking for is among the three coins ($$5$$, $$6$$, and $$7$$) moved from the right end to the left end. Based on the results of weighing in the first two times (the left end weighs more in the first weighing and the right end weighs more in the second weighing), it is known that the coin weighs less than other coins. The following analysis is the same.


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? (2016 AMC 8 Problems, Question \#21)

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 $(\mathbf{B}) \frac{2}{5}$.


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

$$90 $$

$$ 100 $$

$$ 110 $$

$$120 $$

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

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 $$75 + 35 = 110$$.


A box contains blue marbles. Another two boxes contain only white marbles.  Label on Box $$\rm A$$: white marbles  Label on Box $$\rm B$$: blue marbles  Label on Box $$\rm C$$: Box $$\rm B$$ contains blue marbles  Which box contains blue marbles if two of the above labels are wrong?

Box A

Box B

Box C

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

If Box $$\rm A$$ has the wrong label, it contains blue marbles. This indicates that Box $$\rm B$$ has the wrong label too, because only one box has blue marbles.  It also means that Box $$\rm C$$ has the wrong label. We will then have a scenario of three wrong labels, so Box $$\rm A$$ cannot be the one having the wrong label.  If Box $$\rm A$$ has the right label, Box $$\rm B$$ and Box $$\rm C$$ will have the wrong labels.  Box $$\rm A\rightarrow$$ white marbles  Box $$\rm B\rightarrow$$ white marbles  Box $$\rm C\rightarrow$$ blue marbles  Box $$\rm C$$ contains blue marbles.


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 $1-\frac{12}{1024}=1-\frac{3}{256}=\frac{253}{256}=D$.


In my hometown, city streets are numbered with odd numbers in increasing order from south to north and with even numbers in increasing order from west to east. In what direction must I travel if I want to go directly from $$241\text{st}$$ Street to $$225\text{th}$$ Street?

$$$$north

$$$$south

$$$$east

$$$$west

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

In my hometown, city streets are numbered with odd numbers in increasing order from south to north. I must travel from north to south to go directly from $$241\text{st}$$ Street to $$225\text{th}$$ Street.~ When I travel from north to south, I travel south.


Anna went to travel by bike. She started at $8$ am. But she had to stop for $15$ minutes on the way and arrived at $12$ am. How long did Anna ride her bicycle? (adapted from 2011 Math kangaroo Problems, Level 3-4 , Question \#4)

$3$ hours $$45$$ min

$3$ hours $$35$$ min

$2$ hours $$45$$ min

$3$ hours $$25$$ min

$3$ hours $$45$$ min


Peter rolls two standard six-sided dice at the same time. What is the probability that the sum of both rolls is a prime number less than $6$?

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


Abe has $1$ green and $1$ red jelly bean 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 $\dfrac{1}{2}\cdot \dfrac{1}{3}=\dfrac{1}{6}$.


Mary, Nancy, and Peter are playing archery. They have $$7$$ identical arrows in total. Each of them must shoot at least once. How many different ways can they shoot the arrows?

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

There are 3 different people here, so order does matter. Each of them must shoot at least once:  $$7=1+1+5$$  $$7=1+2+4$$  $$7=1+3+3$$  $$7=1+4+2$$  $$7=1+5+1$$  $\textasciitilde$  $$7=2+1+4$$  $$7=2+2+3$$  $$7=2+3+2$$  $$7=2+4+1$$  $\cdots$  And so on. If we keep listing, total will be:  $$5+4+3+2+1=15$$ ways


There are $2$ black balls and $3$ white balls in an opaque bag. They are of identical shape, size and quality except for color. Under the condition that the balls cannot be seen, take out $3$ balls from the bag at a time randomly. Among the following events, the certain one is~\uline{~~~~~~~~~~}~.

Among the $3$ balls taken out, at least $1$ is white.

Among the $3$ balls taken out, at least $1$ is black.

Among the $3$ balls taken out, at least $2$ are black.

Among the $3$ balls taken out, at least $2$ are white.

$$\text{A}$$ is a certain event.  $$\text{B}$$ is a random event, so it\textquotesingle s wrong.  $$\text{C}$$ is a random event, so it\textquotesingle s wrong.  $$\text{D}$$ is a random event, so it\textquotesingle s wrong.


During Black Friday, a store earned a revenue of $80000$ dollars in one week. How much money can the store earn in a year, approximately, considering there are $52$ weeks in a year?

4.16 million dollars

4.17 million dollars

4.18 million dollars

unable to determine

The sample is biased becuase Black Friday is a promotion which allowed the store to earn way much more than they usually earn.


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,~\uline{~~~~~~~~~~}~broke the vase.

Edwin

Fred

Gary

Howard

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.


None of the above

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


$$2$$ beds and $$4$$ chairs have a total mass of $$240$$$$\text{kg}$$. $$1$$ bed weighs as much as $$6$$ chairs. Find the mass of $$1$$ chair.

$$12$$ $$\text{kg}$$

$$15$$ $$\text{kg}$$

$$24$$ $$\text{kg}$$

$$30$$ $$\text{kg}$$

Overseas In-curriculum->Knowledge Point->Algebra-> Numbers, Letters and Equations->Equivalent Substitution->Direct Substitution

Overseas Competition->Knowledge Point->Counting Modules


A box contains seven cards, each with a different integer from $$1$$ to $$7$$ written on it. Avani takes three cards from the box and then Niamh takes two cards, leaving two cards in the box. Avani looks at her cards and then tells Niamh "I know the sum of the numbers on your cards is even." What is the sum of the numbers on Avani\textquotesingle s cards?

To obtain an even number when adding two integers, both integers must be even or both integers must be odd. Therefore the four integers remaining once Avani has removed her three integers must all be odd or all be even or there would be a possibility that the sum of Niamh\textquotesingle s two integers could be odd. Since there were four odd integers and three even integers on the cards in the box initially, the integers on the cards remaining once Avani has removed her cards are all odd. Therefore the cards Avani removed had the three even integers $$2$$, $$4$$ and $$6$$ written on them which have sum $$12$$.


Maria had $$28$$ dreams last month. If $$16$$ of them involved monkeys, $$15$$ involved squirrels, and $$4$$ involved no animals, then at least how many dreams involved both monkeys and squirrels?

Maria had $$28$$ dreams last month, $$24$$ of which involved animals. Since $$16+ 15 =31$$ involved moneys or squirrels, then at least $$31 - 24 = 7$$ dreams involved both monkeys and squirrels.


Nine chairs are in a straight line and numbered $$1$$ to $$9$$ from left to right. Five girls and four boys sit in the chairs so that no girl is next to another girl. A boy could be sitting in the chair with which number?

The girls must be in the odd-numbered seats; otherwise, two girls would be seated next to each other. This leaves only the even-numbered seats for the boys, so a boy is seated in seat $$4$$.


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

None of the above


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{~~~~~~~~~~}~.


Alice\textquotesingle s average score on the first four assignments is $$89$$. She got $$94$$ on the fifth assignment. What is Alice\textquotesingle s average score on all five assignments?

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


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

$$$$Jack

Sarah

Jimmy

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


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


Joann and Sana have $20$ dresses in total. Joann gives half of her dresses to Claire, and then she and Sana have $14$ dresses in total. How many dresses does Sana have?

Joann gives $20-14=6$ dresses to Claire.  Joann has $6+6=12$ dresses originally.  Sana has $20-12=8$ dresses.


Vansen left for school at $6:30$ am. He took half an hour to walk from his house to school. Ron reached school $1$ hour later than Vansen. At what time did Ron reach school? (adapted from 2009 Math Kangaroo Problem, Level 3-4, Question \#8)

$7:00$ am.

$8:00$ am.

$8:45$ am.

$9:00$ am.

$10:00$ am.


In a certain country, a part of the residents can speak English only, a part can speak French only and the rest can speak both languages. It is known that $$85\textbackslash\%$$ residents can speak English and $$75\textbackslash\%$$ can speak French. What percent of the residents of this country can speak both English and French? ($$2002$$ Math kangaroo Problem, Level $$7-8$$, Question \#$$17$$)


Tim, an adventurer, found three treasure chests in the cave. Only one of the chests contains the treasure. The following clues are written on the three treasure chests respectively:  \textbf{Treasure chest A: "The treasure is not in Treasure chest C!"}  \textbf{Treasure chest B: "The treasure is not here!"}  \textbf{Treasure chest C: "The treasure is here!"}  Given that only one sentence of the above three is true, which chest should Tim open to get the treasure?

A

B

C

The clues on treasure chest A and C are conflicting, so either A or C is telling the truth. Thus, the clue on treasure chest B is a lie. Therefore, the treasure is in chest B.


Eddie, Avril and Pip want to share 10 cakes. Eddie only wants to get 3 cakes. Avril and Pip want to get at least 1 cake. How many ways are there to share the cakes?


The correct information James can get from the statement "there is a $11\textbackslash\%$ chance that it will rain tomorrow in Los Angeles"~is~\uline{~~~~~~~~~~}~.

It will rain $11\textbackslash\%$ of the time tomorrow in Los Angeles.

It will rain in $11\textbackslash\%$ of the regions in Los Angeles tomorrow.

It will definitely rain tomorrow in Los Angeles.

The probability of raining in Los Angeles is high.

The probability of raining in Los Angeles is low.

It is less likely to rain tomorrow in Los Angeles.


Mia has a date tomorrow. She has $2$ hats, $3$ dresses, and $3$ pairs of shoes. Now she wants to choose one dress and one pair of shoes. How many options does she have?


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

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


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


There are $$30$$ pupils in my class. $$20$$ pupils like Maths and $$18$$ pupils like English. Twice as many pupils like both subjects as those that 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 $$20-2\times 8=4$$.


Betty and Abby are playing a game. They take turns writing numbers from $$1$$ to $$52 $$ on a blackboard. Each person can only write $$1$$, $$2$$, $$3$$ or $$4$$ numbers at a time, and each number can only be written once. The person who has no more numbers to write loses. Should Betty go first or second in order to win?

First

Second

$$52$$ is not a multiple of $$4+1$$, so the first player will win the game.


Ashley and Elvis are playing a game that requires them to drink a total of $$12$$ cups of coffee. They take turns drinking and each can drink either $$1$$ or $$2$$ cups at a time. The person who finishes the last cup of coffee wins this game. Should Elvis go first or second to ensure victory?

Go first

Go second

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

$$12$$ is a multiple of $$2+1$$, so, the second player should make the total number for each round to be $$3$$ to ensure victory. Therefore, Elvis should go second to ensure his vitory.


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

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

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


SASMO 2014 P2 Q9  Two $1 coins and ten 50c coins are randomly distributed among 4 children such that each child receives the same number of coins. What is the difference between the biggest amount and the smallest amount a child can receive?

None of the above

There are a total of 2+10 = 12 coins. so each child receives 12/4 = 3 coins.  Largest possible is $1 + $1 +50c= $2.50.  smallest possible is 50c + 50c + 50c= $1.50  $2.50 - $1.50 = $1


Maria had $$28$$ dreams last month. If $$16$$ of them involved monkeys, $$15$$ involved squirrels, and $$4$$ involved no animals, then at least how many dreams involved both monkeys and squirrels?


Amy, Bill and Celine are friends with different ages. Exactly 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

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

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 lastly Bill is the youngest. This order is $$\rm E$$. Amy, Celine, Bill.


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.  ($$2005$$ Math Kangaroo Problem, Level $$9-10$$, Question $$ \textbackslash\# $$$$15$$


Pip multiplies all the digits in the number $$145$$ to get $$20$$. What's the biggest three digit number whose digits multiply together to give $$30$$?~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations->Basic Operations of Combinations


Vicky 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 Vicky chooses one food item and one drink item from the lunch menu, she has~\uline{~~~~~~~~~~}~different ways to order lunch.


In the following events, the certain event is~\uline{~~~~~~~~~~}~.

Ame will get $$100$$ points in the final math exam.

Claire will eat breakfast tomorrow.

The news is on when you open the TV.

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

$$\text{A}$$. "Ame will get $100$ points in the final math exam" is a random event.  $$\text{B}$$. "Claire will eat breakfast tomorrow." is a random event.  $$\text{C}$$. "The news is on when you open the TV" is a random event.  $$\text{D}$$. There are two red balls and one white ball in a bag. If two balls are taken out, one must be red. It is a certain event.  So $$\text{D}$$ is the answer.


In a two-digit number, the digit in the ones place is 3. The digit in the tens place is 3 more than the digit in the ones place. What is the number?


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 $10$?

$\frac {1}{2}$

$\frac {1}{6}$

$\frac {1}{3}$

$\frac {1}{4}$

$\frac {1}{12}$

There are $12$ different combinations. The product of two numbers is greater than or equal to $10$ will be $2\times5$ and $2\times6$. Thus, the probability is $\frac 2{12}$ = $\frac 16$.


The Dragonvale Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?

Boy, Girl, Girl, Girl, Boy.There are $$2\times 1=2$$ ways of arranging two boys; there are $$3\times 2\times 1=6\textasciitilde$$ways of arranging $3$ girls. Therefore, there are $$2\times 6=12$$ ways of arranging all students.


Martina and Linda had $$68$$ seashells.  After Martina gave $$8$$ seashells to Linda, they had an equal amount of seashells.  How many seashells did Martina have at first? ．


There are $$60$$ pupils in Arnold\textquotesingle s class. $$35$$ pupils like Maths and $$38$$ pupils like English. Twice as many pupils like both subjects as those that like neither of them. How many pupils like only Maths?

Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Word Problems for Linear Equations with One Variable

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 $$35−2x$$ and the number who like only English is $$38−2x$$.  Since there are $$60$$ pupils in Arnold\textquotesingle s class, we have:  $$\left( 35-2x \right)+2x+\left( 38-2x \right)+x=60$$ and hence $$x = 13$$ and hence the number of pupils who like only Maths is $$35-(2\times 13)=9$$.


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 box contains blue marbles. Another two boxes contain only white marbles.  Label on Box $$\rm A$$: white marbles  Label on Box $$\rm B$$: blue marbles  Label on Box $$\rm C$$: Box $$\rm B$$ contains blue marbles  Which box contains blue marbles if two of the above labels are wrong?．

Box A

Box B

Box C

If two of the labels are wrong, \textbf{one label is correct.}  If the label on Box A is correct, Box A contains white marbles.  The label on Box B is wrong, so Box B contains white marbles.  The label on Box C is wrong, so Box B contains white marbles.  Therefore, Box A and Box B contain white marbles, while \textbf{Box C} contains blue marbles.


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

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

There are six different ways for three kids to line up.


An elementary school is arranging students\textquotesingle{} schedules. For the first class in the morning, it has to be Chinese or Maths or English. Students won\textquotesingle t start the day with the same class for two days in a row. If Wednesday starts with Chinese, then Thursday can start with~\uline{~~~~~~~~~~}~.

Chinese or Maths

Maths or English

English or Chinese

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


Granny has $$10$$ grandchildren. Alice is the oldest. One day, Granny notices that her grandchildren all have different ages. If the sum of her grandchildren\textquotesingle s ages is $$180$$, what is the youngest age that Alice can be．

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

Try to consider the problem of the youngest Alice in such way: make each grandchild has a similar age as possible, and the age of each grandchild should be different, that is, $$1 + 2 + 3 +\cdots 9 + 10 = 55$$; then $$180 - 55 = 125$$, $$125 \div 10 = 12\cdots\cdots5$$, and $$5$$ is left. If every child\textquotesingle s age is added by $$12$$, then $$5$$ is left. Give the extra year to each of the $$5$$ children who are the oldest. In doing so, the minimum number of Alice\textquotesingle s age is $$10+12+1=23$$.


$$\dfrac{1}{2}$$ of the pens in a box are red, $$\dfrac{1}{4}$$ of the remainder are blue and the rest are green. What percentage of the pens in the box are green?

Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Word Problems Involving Fractions and Percentages->Finding the Percentage Given a Part and a Whole

Remainder $=1-\frac{1}{2}=\frac{1}{2}$  Green $=1-\frac{1}{4}=\frac{3}{4}$ of remainder  $=\frac{3}{4}\times\frac{1}{2}$ of total  $=\frac{3}{8}$ of total  $=\frac{3}{8}\times100\textbackslash\%=37.5\textbackslash\%$ of total


Using the digits $$1$$, $$3$$, and $$9$$, we can form~\uline{~~~~~~~~~~}~different 3-digit numbers without repeating digits.


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

$72$ , $24$

$96$ , $24$

$72$ , $48$

$96$ , $48$

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


When Pinocchio lies, his nose gets 6 cm longer. When he tells the truth, his nose gets 2 cm shorter. When his nose was 9cm long, he told three lies and made twotrue statments. How long was Pinnochio\textquotesingle s nose afterwards?


Four boys each prepared $1$ gift for a party. How many ways can these $4$ gifts be \textsubscript{} distributed among the $4$ kids so that no one receives his own gift?

None of the above

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

Let the boys be A, B,~C and D and their gifts be 1, 2, 3 and 4. List down all possible arrangements:  (A2, B1, C4, D3), (A2, B3, C4, D1), (A2, B4, C1, D3), (A3, B1, C4, D2), (A3, B4, C2, D1), (A3, B4, C1, D2), (A4, B1, C2, D3), (A4, B3, C2, D1), (A4, B3, C1, D2)  There are \textbf{9} possible arrangements.


How many three-digit whole numbers have a ones digit equal to the sum of the hundreds digit and the tens digit?

If the ones digit is $$2$$, the $$2$$ numbers are $$202$$ and $$112$$. For each ones digit, the number of possible numbers is the same as the ones digit. In all, there are $$1+2+3+\cdots+8+9 =45$$ numbers.


What is the sum of all the digits in the number $2023$?


The correct information James can get from the statement ``there is a $11\textbackslash\%$ chance that it will rain tomorrow in Los Angeles'' is~\uline{~~~~~~~~~~}~.

It is less likely to rain tomorrow in Los Angeles.


Ron\textquotesingle s work day is half over at $$1:15$$ P.M. if he starts work at $$9:30$$ A.M., his work day ends at．

$$4:30$$ P.M.

$$5:00$$ P.M.

$$5:30$$ P.M.

$$9:30$$ P.M.

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


What time is $$11999$$ hours after $$3$$ P.M.?

$$2$$ A.M.

$$2$$ P.M.

$$4$$ A.M.

$$4$$ P.M.


Moon and Archie played chess competitively. Both of them are of the same level in terms of skill. They agreed to play in a best-of-seven games, where the one who wins four games first would be the ultimate winner. They have already played three games, with Moon winning two games and Archie winning just one game. What is the probability that Moon will be the ultimate winner?


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?

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 $$156-24=132$$.


Sandy, Sam, and Steve are observing a cat in the distance. The cat is eating something but it\textquotesingle s hard to tell what it is.  Sandy says:"The cat is eating fish."  Sam says:"I agree with Sandy!"  Steve says: "The cat is eating meat."  Then they come closer and realize only one of them guessed right.  The cat is eating .

meat

fish

Not certain

We can directly find that Sandy\textquotesingle s point and Sam\textquotesingle s point are identical,so both of them tell lies.Therefore, Steve tells the truth.


Today is Tuesday, and Cindy eats cakes today. She will eat cakes again after $5$ days. What day will it be?

Sunday

Saturday

Friday

Thursday

Wednesday

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

Today is Tuesday, and $5$ days later will be Sunday.


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, which one do you agree with?

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

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.


How many different three-digit numbers can we make using the digits $$2$$, $$4$$, and $$6$$? (Each digit can be used only once.)


Overseas Competition->Knowledge Point->Counting Modules

$$542$$ hundredths $=5$ ones, $4$ tenths and $2$ hundredths $=5.42$


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 $\frac 1{12}$ .


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


How many different four-digit numbers can be formed by rearranging the four digits in $2004$?~

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

0 shouldn\textquotesingle t be put in the highest digit. There are two 0s  The solutions are: 2400, 2040, 2004, 4200, 4020, 4002.


When Coco the Caterpillar is not sleeping, he eats $$5$$ grams of leaves per hour. Yesterday he slept $$20$$ hours. How many grams of leaves did he eat yesterday?

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

Yesterday he did not sleep for $$24 - 20 = 4$$ hours, so he ate $$4 \times 5 = 20$$ grams of leaves.


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?

None of the above

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 $=56 \times2=112$.


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

There are $3$ $E$s in total now with other $6$ letters remaining. But pay attention to $B$: there are $3$ $B$s here.  There are $\_6P\_3$ ways for us to arrange the $6$ letters\textquotesingle{} positions. Then, we can put the $3$ $E$s in the $7$ intervals.  So the answer is $\_6P\_3 \times \_7C\_3=4200$.


Pip multiplies all the digits in the number $$145$$ to get $$20$$. How many three digit numbers are there whose digits multiply to give $$20$$?


Ranson holds $3$ white and $9$ purple jelly beans in his hand. Nancy holds $2$ white, $5$ yellow, and $2$ purple jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match? (adapted from $2013$ AMC $8$ Problem, Question \# $14$)

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

The probability that both show a white bean is $\dfrac{3}{12}\times \dfrac{2}{9}=\dfrac{1}{18}$. The probability that both show a purple bean is $\dfrac{9}{12}\times \dfrac{2}{9}=\dfrac{1}{6}$. Therefore, the probability is $\dfrac{1}{18}+\dfrac{1}{6}=\frac29$.


Linda picks $3$ different numbers from $$1-15$$. To make the sum of the three numbers divisible by $3$, how many different groups are there for Linda to pick?

Group $A$: $$1$$, $$4$$, $$7$$, $$10$$, $13$;  Group $B$: $$2$$, $$5$$, $$8$$, $$11$$, $$14$$;  Group $C$: $$3$$, $$6$$, $$9$$, $$12$$, $$15$$.  Linda can choose three numbers from the same group, or choose each number from a different group to get the sum she needs.  There are $3\times\_5C\_3+5\times5\times5=155$ groups.


There are $$5$$ dancers. Every $$2$$ dancers will have a dance. How many dances would they have in total?


If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$, then your score on the seventh test was ．

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

The total score of the first six tests was $84\times6=504$, and the total score of the first seven tests was $85\times7=595$. Therefore, the score of the seventh test equals to the difference: $595-504=91$.