Take out $$4$$ digits from $101112131415$ to make the rest of the digits into a new $8-$digit number without changing the order of digits. The least possible value of the new number is~\uline{~~~~~~~~~~}~.

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

To make it the least, you should make the digit in the first place as small as possible. Thus the $$2$$, $$3$$, $4$ and $$5$$ should be taken out.


How many two-digit numbers have digits whose sum is a perfect square? (2006 AMC 8 Problem, Question \#11)

Overseas Competition->Knowledge Point->Counting Modules

There is 1 integer whose digits sum to $1: 10$. There are 4 integers whose digits sum to $4: 13,22,31$, and 40 . There are 9 integers whose digits sum to $9: 18,27,36,45,54,63,72,81$, and 90 . There are 3 integers whose digits sum to $16: 79,88$, and 97 . Two digits cannot sum to 25 or any greater square since the greatest sum of digits of a twodigit number is $9+9=18$ Thus, the answer is $1+4+9+3=$ (C)17.


Three students go apple-picking. Eddie picks $30$ apples, Avril picks $27$ apples, and Mike picks $33$ apples. How many apples does each of them pick on average?

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

They pick $30+27+33=90$ apples in total. Therefore, on average, each of them picks $90\div3=30$ apples.


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

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


SASMO 2014 P2 Q1  What is 2014 + 2 x 0 x 1 x 4 equal to?

None of the above

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


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

red

green

Uncertain

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

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


Each of the 20 balls is tossed independently and at random into one of the 5 bins. Let $p$ be the probability that some bin ends up with 3 balls, another with 5 balls, and the other three with 4 balls each. Let $q$ be the probability that every bin ends up with 4 balls. What is $\frac{p}{q}$ ?

Solution 1: For simplicity purposes, we assume that the balls and the bins are both distinguishable. Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have $$ p=\frac{5 \cdot 4 \cdot\left(\begin{array}{c} 20 \textbackslash\textbackslash{} 3,5,4,4,4 \end{array}\right)}{5^{20}} \text { and } q=\frac{\left(\begin{array}{c} 20 \textbackslash\textbackslash{} 4,4,4,4,4 \end{array}\right)}{5^{20}} \text {. } $$ Therefore, the answer is $$ \frac{p}{q}=\frac{5 \cdot 4 \cdot\left(\begin{array}{c} 20 \textbackslash\textbackslash{} 3,5,4,4,4 \end{array}\right)}{\left(\begin{array}{c} 20 \textbackslash\textbackslash{} 4,4,4,4,4 \end{array}\right)}=\frac{5 \cdot 4 \cdot \frac{20 !}{3 ! \cdot 5 ! \cdot 4 ! \cdot 4 ! \cdot 4 !}}{\frac{20 !}{4 ! \cdot 4 ! \cdot 4 ! \cdot 4 ! \cdot 4 !}}=\frac{5 \cdot 4 \cdot(4 ! \cdot 4 ! \cdot 4 ! \cdot 4 ! \cdot 4 !)}{3 ! \cdot 5 ! \cdot 4 ! \cdot 4 ! \cdot 4 !}=\frac{5 \cdot 4 \cdot 4}{5}=(\mathbf{E}) 16 . $$  Solution 2:  For simplicity purposes, we assume that the balls and the bins are both distinguishable. Let $q=\frac{x}{a}$, where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with 4 balls.  We can take 1 ball from one bin and place it in another bin so that some bin ends up with 3 balls, another with 5 balls, and the other three with 4 balls each. Note that one configuration of $4$-$4$-$4$-$4$-$4$ corresponds to $5 \cdot 4 \cdot 4=80$ configurations of $3$-$5$-$4$-$4$-$4$. On the other hand, one configuration of $3$-$5$-$4$-$4$-$4$ corresponds to 5 configurations of $4$-$4$-$4$-$4$-$4$. Therefore, we have $$ p=\frac{80}{5} \cdot \frac{x}{a}=16 \cdot \frac{x}{a}, $$ from which $\frac{p}{q}=$ (E) 16 .


Yuki and Claire are looking for a place to eat dinner. They know there are four Chinese restaurants, three French restaurants, and two Peruvian restaurants nearby. There are~\uline{~~~~~~~~~~}~different choices in total for them to eat one meal in one place.

They can only choose one place, so it can only be either Chinese, French, or Peruvian restaurants. Therefore, we can add each one up to get $$4+3+2 = 9$$.~


There are $12$ black balls, $27$ red balls, and $11$ blue balls in an opaque box. The balls are the same except for their colors. Bob adds several red balls in the box and mixes the balls. Now, if the probability of taking out a black ball is $\frac15$, how many red balls does Bob add?

$\frac15=\frac{12}{60}$, which means there are $60$ balls in total. Thus, Bob adds $60-12-27-11=10$ red balls.


There are $$10$$ players in a chess tournament. If each game is played by $$2$$ players, and each player plays every other player exactly once, what is the total number of games played in the tournament?

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

There are $10$ players and each player plays $9$ games, so there are $10 \times 9 \div 2 = 45$ games in total.


Among $$30$$ children, there are at least~\uline{~~~~~~~~~~}~children born in the same month.

There are $$12$$ months. Thus, $$30\div 12=2R6$$ and hence $$2+1=3$$.


If the sum of $$9$$ numbers is $$1998$$, then their average is ．

$$9\times 1998$$

$$1998\div 9$$

$$9\div 1998$$

The average of any $$9$$ numbers is their sum divided by $$9$$.


In the Coin Game, you toss three coins at the same time. You win only if the 3 coins are all showing heads, or if the 3 coins are all showing tails. If you play the game once only, what is the probability of winning?

Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Typical Probability Problems->Tossing Coins

When tossing a single coin, there are two possible outcomes, a head (H) or a tail (T).  When tossing 2 coins, there are $$2 \times 2 = 4 $$possible outcomes.  These are HH, HT, TH, and TT.  When tossing 3 coins, there are $$2 \times 2 \times 2 = 8 $$possible outcomes. These are HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. Of these 8 possible outcomes, there are 2 winning outcomes, HHH and TTT.  Thus, the probability of winning the Coin Game is $\frac{1}{4}$ .  Answer: E


How many different four-digit numbers can be formed by rearranging the four digits in $2021$?~(Adapted from $2004$ AMC $8$ Problem, Question \#2)

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

When the thousand digit is $2$, there are $\_3P\_3=6$ ways. When the thousand digit is $1$, there are $\_3C\_1=3$ ways. So the answer is $3+6=9$.


Paula shoots arrows at the following target. When she misses, she obtains zero points. Paula shoots two arrows and adds the number of points. Which of the following sums cannot be her score?

A is possible as 30+30 = 60  B is possible as 70+0+0 = 70  C is possible as 30 + 50 = 80  E is ossible as 30 + 70 = 100.


How many different natural numbers can be formed with the numbers $$1$$, $$2$$, $$3$$?（without using the same number two times like $$33$$）

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

Classify the number first by the digits and then enumerate．  One-digit number：$$1$$、$$2$$、$$3$$，  Two-digit number：$$12$$、$$13$$、$$21$$、$$23$$、$$31$$、$$32$$；  Three-digit number：$$123$$、$$132$$、$$213$$、$$231$$、$$312$$、$$321$$；  So totally $$3+6+6=15$$ different natural numbers can be formed．


Three kids $$A$$、$$B$$、$$C$$ are playing the game "pass the ball". If it starts with $$A$$, he can pass the ball to~\uline{~~~~~~~~~~}~.

$$A$$ or $$B$$ or $$C$$

$$B$$ or $$C$$

$$A$$ or $$C$$

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


An acronym is a word formed from the first one or more letters of each word in a group of words. If "UFO"is an acronym for "unidentified flying object," then for how many of the following word groups could MATH be an acronym?  $$\text{I}$$. Multiply All Those Hundreds  $$\text{II}$$. MArtians Take Hostages  $$\text{III}$$. MATthew Hides  $$\text{IV}$$. Minutes After The Hour

one

two

three

four

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

For $$\text{I}$$ and $$\text{IV}$$, use first letter of each word; for $$\text{II}$$, use first two letters of first word and first letter of other words; and for $$\text{III}$$, use first three letters of first word and first letter of second word. MATH could be an acronym for all four word groups.  $$\text{I}$$. Multiply All Those Hundreds  $$\text{II}$$. MArtians Take Hostages  $$\text{III}$$. MATthew Hides  $$\text{IV}$$. Minutes After The Hour


Gregor forms two numbers with digits $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, and $$6$$. 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?

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 $$6$$ and $$5$$. The sum of the hundreds digits is $$11$$. For the same reason, the sum of the tens digits should be $$3+4=7$$ and the sum of the ones digits should be $$2 + 1 = 3$$. Therefore, $$\rm D$$ is the correct answer.


A fair $6$-sided die is rolled once. What is the probability that the number on the top is an odd number? (adapted from 2011 AMC 8 Problem, Question \#18)

There are $3$ odd numbers out of $6$, so the probability is $\frac36=\frac12$.


How many two-digit numbers are there where the ones digit is greater than the tens digit? (2008 Math Kangaroo Problem, Level 3-4, Question \#21)


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

Bill, Amy, Celine

Amy, Bill, Celine

Celine, Amy, Bill

Celine, Bill, Amy

Amy, Celine, Bill

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


$$30$$ children going to Adventure Park took part in at least one of two events. $$15$$ of them took part in the "moving bridge" contest, and $$20$$ of them went down the zip-wire. How many children from Adventure Park took part in both events? ($$2013$$ Math Kangaroo Problems, Level $$3-4$$, Question \#$$14$$)

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


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

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

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


How many different four-digit odd numbers can be made by using digits $$1$$, $$2$$, $$3$$, $$4$$, and $$5$$ (without digits be repeated)?

Rule of product: $$3\times 4\times 3\times 2=72$$; therefore we can choose D.


How many ways are there of making a total of 10 using three different positive numbers?

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


A conductor wanted to make a trio consisting of a fiddler, a pianist, and a drummer. He had to choose one of two fiddlers, one of two pianists, and one of two drummers. He decided to try each of the possible trios. How many attempts did he have to make?

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


Eve brings $$12$$ pieces of candy, Alice brings $$9$$ pieces of candy and Irene doesn\textquotesingle t bring any candy. They put all the pieces of candy together on a table and divide them equally among themselves. How many pieces of candy does each of the girls get? (2012 Math Kangaroo Problem, Levels 1-2, Question \#15）

They have $12+9=21$ pieces of candy in total. $21\div3=7$


Given that only one of the following statement is corect, which one is correct?  ($$1$$) All of the statements below are correct.  ($$2$$) None of the statement below is corect.  ($$3$$) One of the statements above is corect.  ($$4$$) All the statements above are correct.  ($$5$$) None of the statement above is corect.

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

Suppose ($$1$$) is correct, then ($$2$$) must be wrong which contradicts that only one statement is correct.  Suppose ($$2$$) is correct, then ($$5$$) is correct which contradicts that only one statement is correct.  Suppose ($$3$$) is correct, it also contradicts that only one statement is correct.  Suppose ($$4$$) is correct, it also contradicts that only one statement is correct.  Hence ($$5$$) is the correct statement.


The farmer has a fish, a cat, and a dog. He wants to take them cross the river by boat, and he can only take one animal each time. When the farmer is away, the cat cannot be put with the fish or the dog, or they will fight. The boat makes one trip from one side of the river to the other side, so it takes~\uline{~~~~~~~~~~}~trips in total to carry all the animals and farmers cross the river.

Overseas Competition->Knowledge Point->Counting Modules

We can use $A$ to represent the side of the river where the farmer is at the beginning, and use $B$ to represent the other side.  The first time, take the cat from $A$ to $B$.  The second time, the farmer go back to $A$.  The third time, take the fish from $A$ to $B$.  The fourth time, take the cat from $B$ to $A$.  The fifth time, take the dog from $A$ to $B$.  The sixth time, the farmer go back to $A$.  The seventh time, take the cat from $A$ to $B$.


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 \#$$ 18$$)

$$5$$ 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 $$\left( \frac{5}{6} \right)\times\left( \frac{5}{6} \right)=\frac{25}{36}$$. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of $$5$$, is $$1-\frac{25}{36}=\frac{11}{36}$$.


Mom has $190$ coins in total and wants to give them to Sana as the pocket money. She gives Sana $30$ coins for the first month, and $40$ coins for each of the following months. How many months in total can Sana get the pocket money?


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)

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

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


My train leaves Southampton and arrives in Birmingham at $$08:48$$ that morning, the duration of the whole journey is $$52$$ minutes, what time does the train leave Southampton?

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

The departure time is $$52$$ minutes forward from $$8:48$$. You can push it forward $$48$$ minutes, which is $$8:00$$, and then push it forward $$4$$ minutes, which is $$7:56$$.


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

Toss two identical coins and both land on heads.

Throw a fair die and number of the dot shown is $$3$$.

Sun sets in the west.

It must rain in cloudy days.

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

$$\text{A}$$. Throw a coin randomly. It has a $$50\textbackslash\%$$ chance of landing head up and a $$50\textbackslash\%$$ chance of landing tail up.  $$\text{B}$$. Throw a die, and number of the dot shown may be any number from $$1$$ to $$6$$.  $$\text{C}$$. Sun sets in the west.  $$\text{D}$$. It may not rain in cloudy days.  So $$\text{C}$$ is the answer.


The correct understanding of the statement "there is a $$75\textbackslash\%$$ chance that it will rain tomorrow in New York City" is~\uline{~~~~~~~~~~}~.

It will rain $$75\textbackslash\% $$ of the time tomorrow in New York City.

It will rain in $$75\textbackslash\%$$ of the regions in New York City tomorrow.

It will definitely rain tomorrow in New York City.

The probability of raining in New York City tomorrow is high.

The statement that "there is a $$75\textbackslash\% $$ chance that it will rain tomorrow in New York City"~shows that it is more likely to rain tomorrow in New York city.


The digits $1$, $2$, and $3$ can make~\uline{~~~~~~~~~~}~three-digit numbers without repeating digits.


If a turkey facing north turns $$225^{}\circ $$ clockwise, it will then face．

southwest

southeast

northwest

northeast

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

Since $$225^{}\circ =180^{}\circ +45^{}\circ $$, the bird turns $$45^{}\circ $$ past south. That\textquotesingle s southwest.


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 $$18$$?

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

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.


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


The number of days in July plus the number in August is twice the number of days in．

March

April

June

November

Together, July \& August have $$62$$ days. That\textquotesingle s twice the $$31$$ days in March.


Let $f$ be a linear function for which $f(3)-f(2)=5$. What is $f(8)-f(2)$? (  Adapted From 2003 AMC 12B Problems, Question \#9)

Overseas Competition->Knowledge Point->Calculation Modules

Let $f$ be a linear function with slope $m$.  $$ \begin{gathered} m=\frac{f(3)-f(2)}{\Delta x}=\frac{5}{3-2}=5\textbackslash\textbackslash{} f(8)-f(2)=m \Delta x=5(8-2)=30 \Rightarrow(D) \end{gathered}$$


Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences


What is the correct ordering of the three numbers $$\dfrac{5}{19}$$, $$\dfrac{7}{21}$$, and $$\dfrac{9}{23}$$, in increasing order? ($$2012$$ AMC $$8$$ Problem, Question \#$$4$$)

$$\dfrac{9}{23}\textless{} \dfrac{7}{21}\textless\dfrac{5}{19}$$

$$\dfrac{5}{19}\textless{} \dfrac{7}{21}\textless{} \dfrac{9}{23}$$

$$\dfrac{9}{23}\textless{} \dfrac{5}{19}\textless{} \dfrac{7}{21}$$

$$\dfrac{5}{19}\textless{} \dfrac{9}{23}\textless{} \dfrac{7}{21}$$

$$\dfrac{7}{21}\textless{} \dfrac{5}{19}\textless{} \dfrac{9}{23}$$

Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering

$$\rm Method$$ $$1$$:  The value of $$\dfrac{7}{21}$$ is $$\dfrac{1}{3}$$. Now we give all the fractions a common denominator.  $$\dfrac{5}{19} \Rightarrow \dfrac{345}{1311}$$,  $$\dfrac{1}{3} \Rightarrow \dfrac{437}{1311}$$,  $$\dfrac{9}{23} \Rightarrow \dfrac{513}{1311}$$.  Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, $$\frac{5}{19}\textless{} \frac{7}{21}\textless{} \frac{9}{23}$$.  $$\rm Method$$ $$2$$:  Instead of finding the LCD, we can subtract each fraction from $$1$$ to get a common numerator. Thus,  $$1- \dfrac{5}{19}= \dfrac{14}{19}$$,  $$1- \dfrac{7}{21}= \dfrac{14}{21}$$,  $$1- \dfrac{9}{23}= \dfrac{14}{23}$$.  All three fraction have common numerator $$14$$. Now the order of the fractions is obvious . $$\dfrac{14}{19}\textgreater\dfrac{14}{21}\textgreater\dfrac{14}{23}\Rightarrow\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$. Therefore, $$\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$.  $$\rm Method$$ $$3$$:  Change $$\frac7{21}$$ into $$\frac13$$, $$\dfrac{1}{3} \times \dfrac{5}{5}=\dfrac{5}{15}$$, $$\dfrac{5}{15}\textgreater\dfrac{5}{19}$$, $$\dfrac{7}{21}\textgreater\dfrac{5}{19}$$, and $$\dfrac{1}{3} \times \dfrac{9}{9}=\dfrac{9}{27}$$, $$\dfrac{9}{27}\textless\dfrac{9}{23}$$, $$\dfrac{7}{21} \textless{} \dfrac{9}{23}$$. Therefore, $$\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$.  $$\rm Method$$ $$4$$:  When $$\dfrac{x}{y}\textless1$$ and $$z\textgreater0$$, $$\dfrac{x+z}{y+z}\textgreater\dfrac{x}{y}$$. Hence, $$\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$.


Which of the following statements is true?

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction


Alysha and Julia have some biscuits. Altogether they have $28$ biscuits. Alysha has $4$ more biscuits than Julia. How many biscuits does Alysha have? (Adapted from 2021 Math Kangaroo Problem, Level 3-4, Question \#7)


What is the simplest form of $5$ minutes$: 30 $ seconds?

Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio

We need to make units same first. $5$ minutes equal to $300$ seconds. Now we could remove the same unit, second. We get $300:30$ and simplify it to $10:1$.


Let $f$ be a linear function for which $f(6)-f(2)=12$. What is $f(12)-f(2)$? (  2003 AMC 12B Problems, Question \#9)

Let $f$ be a linear function with slope $m$.  $$ \begin{gathered} m=\frac{f(6)-f(2)}{\Delta x}=\frac{12}{6-2}=3 \textbackslash\textbackslash{} f(12)-f(2)=m \Delta x=3(12-2)=30 \Rightarrow(D) \end{gathered}$$


The sum of five consecutive integers is $$2015$$. What is the smallest of these integers?

Let the five consecutive integers be $$n-2$$, $$n-1$$, $$n$$, $$n + 1$$ and $$n + 2$$. These have a sum of $$5n$$. Hence $$5n=2015$$ and therefore $$n=403$$. Therefore, the smallest integer is $$403-2=401$$.


\textbf{Megan wonders how the size of her beagle Herbie compares with other beagles. Herbie is 40.6cm tall. Megan learned on the internet that beagle heights are approximately normally distributed with a mean of 38.5 cm and a standard deviation of 1.25 cm. What is the percentile rank of Herbie's height?}


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations


Jack has eight chocolates. He gives Jimmy one, Luna two, and Tim three respectively. Then Jack\textquotesingle s mother gives Jack two more. How many chocolates are left for Jack?~(adapted from $$2005$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$6$$)

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition and Subtraction of Whole Numbers->Questions Involving Addition and Subtraction

The problem can be regarded as an equation $8$ is the number of chocolates Jack had, $1, 2, 3$ is the number of chocolates Jack gave out, and $2$ is the number of chocolates his mother gave Jack. So the remaining chocolate is $8-1-2-3+2=4$


One kind of computer can make $$4\times {{10}^{9}}$$ operations per second.~ How many operations can it make in $$5\times {{10}^{2}}$$ seconds?

$$4\times {{10}^{11}}$$

$$2\times {{10}^{11}}$$

$$2\times {{10}^{12}}$$

$$20\times {{10}^{18}}$$

$$2\times {{10}^{19}}$$

Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division


$$1\times 2\times 3\times 4$$


Lucas pays for $4$ pens and $3$ pencils. Bryan spends twice as much paying for $2$ pens and $16$ pencils. A pen is how many times as expensive as a pencil?


Abel and Bella are friends and they always want to sit next to each other. How many ways are there to arrange Abel, Bella and three of their acquaintances to sit in a row, while satisfying the condition?


When calculating $$16\times 29$$, which of the following choices is correct?

$$144$$ in the blue box represents $$9\times 16$$.

$$32$$ in the red box represents $$2\times 16$$.

The product is $$32+144 =176$$.

There is no regrouping in the calculation.

$$32$$ in the red box represents $$20 \times 16 = 320$$, and the product should be $$320 +144 = 464$$.


David measured the length of his garden.  It was $$15$$ metres to the nearest tenth of a metre.  Between what limits was the actual length?

$$14.995\leqslant $$ the actual length $$\leqslant 15.005$$

$$14.9\leqslant $$ the actual length $$\textless~15.1$$

$$14.95\leqslant $$ the actual length $$\textless{} 15.05$$

$$14.99\leqslant $$ the actual length $$\leqslant 15.01$$

$$14.5\textless$$ the actual length $$\textless15.5$$

Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals->Finding Approximate Values


Let $$a\square b=ab+a+b$$ for any integers $$a$$ and $$b$$.  The solution of the equation $$3\square 5=2\square x$$ is: $x = $~\uline{~~~~~~~~~~}~

Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Unknowns Using the Given Operations

$3\square 5=2\square x \implies 15+3+5=2x+2+x$, hence $x=7$.


If $A=1+3+5+7+\cdots +99$, and $B=2+4+6+\cdots +100$, what is the value of $B-A$?


A certain game uses a set of cards containing odd numbers. There are one 1, three 3s, two 5s, one 7, and three 9s. The mean value of this deck is 5.4, with a standard deviation of 2.8. If someone purchased three more decks of these cards and combined~them into single large deck, what would the new mean and standard deviation be?

Mean = 5.4; standard deviation = 2.8

Mean = 5.4; standard deviation = 2.95

Mean = 16.2; standard deviation = 2.90

Mean = 16.2; standard deviation = 8.4

Mean = 21.6standard deviation = 11.2

The mean is a measure of center. While it can be affected by adding new values ,the purchase of second deck would merely add more cards with the same values .As a result ,the mean would not change .Eliminate (C), (D),and (E). The standard deviation is a measure of sprees. As above, unique values will cause the standard deviation to change. However, since the new decks are identic -call to the old, the standard deviation will not change. Eliminate (B).


Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions


Boyuan bought $$3$$ types of candies. He bought $$5$$ times as many toffee as lollipop. He bought $$5$$ more chocolate bar than lollipop. If he has $$7$$ lollipop, how many candies did he bought in total?

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers


Tom and Jerry found $\frac{1}{2}$ of a steak in the kitchen. Tom ate $\frac{1}{2}$ of the leftover steak and Jerry ate $\frac{1}{4}$ of it. They decided to give the rest to their friend Spike. What portion of the original steak would Spike get?

Overseas Competition->Knowledge Point->Calculation Modules->Fractions


Free-response questions on the AP Statistics Exam are graded 4, 3, 2, 1, or~ 0. Question 2 on the exam was of moderate difficulty. The average score on question 2 was 2.05 with a standard deviation of 1. To the nearest tenth, what score was achieved by a student who was at the 90th percentile of all students on the test? You may assume that the scores on the question were approximately normally distributed.


Josie and Holly share $$48$$ dollars between them.  Josie has three times as much as Holly.  How much does Holly have?

$12$ dollars

$24$ dollars

$$36$$ dolllars

$$40$$ dollars

$$45$$ dollars


In an arithmetic sequence with $$13$$ terms, if the seventh term is $$20$$, what is the sum of all the terms?

Sum $$=$$ Middle Term $$\times$$ Number of Terms, $$20 \times 13 = 260$$.


Find the value of $$123\times (-129)+123\times130$$.

$$\begin{eqnarray}\&\&123\times (-129+130)\textbackslash\textbackslash{} \&=\&123\times (130-129)\textbackslash\textbackslash{} \&=\&123\times1\textbackslash\textbackslash{} \&=\&123.\end{eqnarray}$$.


Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly


The ratio of the perimeter of a rectangle to the length of one of its sides is~$14:3$. If the area is 27 square inches, how many inches long is one of the longer sides?

$\dfrac{2\left( x+y\right)}{x}=\dfrac{14}{3}\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} \dfrac{x+y}{x}$~$=\dfrac{7}{3}\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} 1+\dfrac{y}{x}=1+\dfrac{4}{3}$  $\Rightarrow\textbackslash{} \textbackslash{} \dfrac{y}{x}=\dfrac{4}{3}\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} \dfrac{y\times y}{x\times y}=\dfrac{4}{3}$  The area is~$x\times y=27\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} \dfrac{y\times y}{27}=\dfrac{4}{3}\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} y\times y=36\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} y=6$


The $2021^{st}$ digit at the right of the decimal point in the decimal expression of $\dfrac{6}{7}$ is~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals

$$\frac{6}{7}=0.\overline{857142}$$, it is a decimal which repeats in cycles of $6$ digits. Every $6$$^{th}$ digit goes back to $2$. The $2022$$$^{nd}$$ digit is $2$, so the $2021$$^{st}$ digit is $4$.


Two non-zero real numbers, $x$ and $y$, satisfy $x = 6-y$. Which of the following is a possible value of $$ \frac{{{x}^{2}}}{x-y}+\frac{{{y}^{2}}}{y-x}$$? (Adapted From 2000 AMC 12 Problems, Question \#11)


If $ a◆b$ means $(a\times b)+b$, then $2◆3$ has the value of~\uline{~~~~~~~~~~}~.

$2◆3=(2\times3)+3=9$.  So the answer is $\rm B$.


Each of two boxes contains three chips numbered $$1$$, $$2$$, and $$3$$. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even? (2015 AMC 8 Problem, Question \#7)

We can instead calculate the probability that their product is odd, and subtract this from $$1$$. In order to get an odd product, we have to draw an odd number from each box. We have a $$\frac23$$ probability of drawing an odd mumber from one box, so there is a $$\left(\frac{2}{3}\right)^{2}= \frac{4}{9}$$ probability of having an odd product. Thus, there is a $$1- \frac{4}{9}= \frac{5}{9}$$ probability of having an even product.


If $$8$$ watermelons cost as much as $$12$$ pears, then $$24$$ waternelons cost as much as pears.

Overseas Competition->Knowledge Point->Heuristics Skills-> Equivalent Substitution

$$8$$ watermelons = $$12$$ pears  $$8\times3$$ watermelons = $24$ watermelons  $$12\times3$$ pears = $36$ pears


An object of mass m is traveling at constant speed v in acircular path of radius r. How much work is done by the centripetal force during one-half of a revolution?

$\pi mv^{2}$

$\pi mv^{2}r$

$2\pi mv^{2}r$

Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables

Since the centripetal force always points along a raduis toward the center of the od the circle, and the velocity of the object is always tangent to the circle (and thus perpendicular to the radius), the work done by the centripetal force is zero. Alternatively, since the object\textquotesingle s speed remains constant, the Work-Energy Theorem tells you that no work is being performed.


When Xiao Dong did the subtraction equation, he saw the minuend as $$90$$ instead of $$75$$, and the difference he got was $$55$$. The correct result should be~\uline{~~~~~~~~~~}~.

None of the above

Overseas Competition->Knowledge Point->Calculation Modules


The number of papers I have is $$4\times 21$$ more than $6$ dogs. I have~\uline{~ ~?~ ~ ~ ~}~dogs．


$$16\times 2$$

$$12\times 3$$

$$7\times 5$$

$$38\times 1$$


What is the remainder when $$987\textasciitilde654\textasciitilde321$$ is divided by $$100$$?

If a number is divided by $$100$$, the remainder is the number\textquotesingle s last $$2$$ digits.


The distance between A and B is $$350$$km. Kin and Mary drive away from A and B respectively at $8$ a.m. and go towards each other at same time. Kin drives $$40$$km/h, and Mary drives $$50$$km/h. Mary rested for $$2$$ hours on her way and then continues driving. They will meet at~\uline{~~~~~~~~~~}~.

$$10$$ a.m.

$$11$$ a.m.

$$12$$ p.m.

$$1$$ p.m.


A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list? (2018 AMC 10B Problems, Question \#14)

To minimize the number of distinct values, we want to maximize the number of times they appear. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223+1+1=$ (D) $225$.


Which of the following is a solution of $$\begin{cases}2x-4=0 \textbackslash\textbackslash{} 4x-y=7 \end{cases}$$?

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with Multiple Variables


A bakery sells cakes in three different sizes: small ($S$), medium ($M$), and large ($L$). The medium size costs $40\textbackslash\%$ more than the small size and contains $25\textbackslash\%$ less cake than the large size. The large size contains twice as much cake as the small size and costs $60\textbackslash\%$ more than the medium size. Rank the three sizes from best to worst buy in terms of cost per unit of cake. (Adapted From 2005 AMC 8 Problems, Question \#22)

$MSL$

$SML$

$LSM$

$SLM$

Suppose the small size costs $$$1$$ and the large size has $10$ grams of cake. The medium size then costs $$$1.40$$ and has $7.5$ grams of cake. The small size has $5$ grams of cake and the large size costs $$$2.24$$. The small, medium, and large size cost respectively, $0.200$, $0.187$, $0.224$ dollars per gram. The sizes from best to worst buy are $MSL$.


Find the value of the expression $100-98+96-94+92-90+\cdots +8-6+4-2$.

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Grouping in Fast Addition and Subtraction of Whole Numbers


If~$\dfrac{y}{x-z}=\dfrac{x+y}{z}=\dfrac{x}{y}$~for three positive numbers x, y and z, all different, then what is the value of~$\dfrac{x}{y}$.

$\dfrac{x}{y}=\dfrac{y}{x-z}=\dfrac{x+y}{z}\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} \textbackslash{} \dfrac{x}{y}=\dfrac{x+y+(x+y)}{y+\left( x-z\right)+z}=\dfrac{2x+2y}{x+y}=\dfrac{2\left( x+y\right)}{x+y}=2.$


$$280$$ adults and $$445$$ children visited Universal Studios last week. $330$ visitors did not take the rollercoasters. How many visitors took the roller coaster last week?

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction


What is the simplest form of $3\dfrac{8}{12}$?

Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Basic Understanding of Fractions->Using Common Factors to Simplify Fractions

Factors of $8$ are $1, 2, 4, 8$ and factors of $12$ are $1, 2, 3, 4, 6, 12$.  The highest commom factor of $8$ and $12$ is $4$.  Dividing $8$ and $12$ by $4$ respectively, we get  $\frac{8 \div 4}{12 \div 4}$=$\frac{2}{3}$;  Write the whole and the simplified fraction together,  $3\dfrac{2}{3}$.


Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Units of Length


An automobile manufacturer wished to know which of two new paint colors were preferred on its newest line of vehicles. A~large simple random sample was taken from people throughout the USA who had purchased one of their vehicles in the~previous five years. What is the safest generalization of this survey?

Only those people who took this particular survey

All future customers of this automobile manufacturer

All people who have purchased from that manufacturer in the previous five years

Only those people who have purchased a vehicle in the previous five years

All vehicle owners in the USA

A safe generalization will broaden the survey to a population that closely resembles the group of people surveyed. Since the sample was taken from purchasers of a single auto manufacturer over the past five years, the safest generalization is a population with these same qualities.


14. Consider the following system of equations: $$\begin{cases} 5 x+7 y=8 \textbackslash\textbackslash{} 2 x+q y=r \end{cases}$$, where $q$ and $r$ are constants. If $q$ and $r$ are chosen such that this system has infinitely many solutions $(x, y)$, find $q+r=$~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Indefinite Equations->System of Indefinite Equations

For the system to have infinitely many solutions, the second equation must be a constant multiple of the first. Comparing the coefficients of $x$, the second equation must be $\frac{2}{5}$ times the first equation. Therefore, $q+r=\left(\frac{2}{5}\right)\times (7)+\left(\frac{2}{5}\right)\times(8)=6$.


For $\triangle ABC$, all of its side lengths are integers. The primeter of $\triangle ABC$ with a side of length $12$ and a side length of $7$ is at least ．

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities

We know from the triangle inequality that the last side, $s$, fulfills $s+7\textgreater12$. $P=s+7+12\textgreater12+12$. Therefore, $P\textgreater24+1=25$


What is the sum of the numbers that range from $$1$$ to $$100$$ when divide by $$3$$ give you a remainder of $$1$$ ?

Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences


A cup of boiling water ($212$\textsuperscript{◦}F) is placed to cool in a room whose temper- ature remains constant at $68$\textsuperscript{◦}F. Suppose the difference between the water temperature and the room temperature is halved every $5$ minutes. What is the water temperature, in degrees Fahrenheit, after $15$ minutes?


What is the missing number in the box?  $2:3=\boxed{?}:9$

Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Proportions


(the number of $$\rm cm$$ in $$1\rm m$$):(the number of $$\rm m$$ in $$1\rm km$$)$$=$$．


The house number of a community consists of $5$ digits: $1, 2, 3, 6, 7$. Which number is not used if these $5$ digits are filled in the squares below to make the equation correct? Each number can only be filled in once.  $$\huge\square -\square =\square +\square $$

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form

Pay attention to look at both sides of the formula. One side is addition, and the other side is subtraction.


Kangaroo came up with a new operation * for positive integers. He gave a few examples as shown below:  $$2$$ * $$3$$$$=\left( 2+1 \right)\times 3=9$$;  $$4$$ * $$2$$$$=\left( 4+3+2+1 \right)\times 2=20$$;  $$3$$ * $$5$$$$=\left( 3+2+1 \right)\times 5=30$$.  What is the value of the expression $$6$$ * $$5$$ ?

Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly->Ordinary Type


\textbf{A copy machine dealer has data on the number x of copy machines at each of 89 customer locations and the number y of service calls in a month at each location. Summary calculations give $$\bar{}x$$ = 8.4, $$S\_x$$ = 2.1, ȳ = 14.2, $$S\_y$$ = 3.8, and r = 0.86. What is the slope of the least squares regression line of number of service calls on number of copiers?}

None of the above

\textbf{Can't tell from the information given}


A firm had sales revenue of 1 million dollars last year. It spent 600,000 dollars on labor, 150,000 dollars on capital, and 200,000 dollars on materials. What was the firm\textquotesingle s accounting profit?

0 dollar

50,000 dollars

400,000 dollars

650,000 dollars

1,000,000 dollars

Accounting profit = total revenue - explicit cost


Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas, would like to cover the package with a large number of stamps. Suppose he has a collection of $5$-cent, $10$-cent, and $25$-cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $\textbackslash$7.10$ in postage?


\textbf{In a statistics course, a linear regression equation was computed to predict the final exam score from the score on the first test. The equation was y = 10 + .9x where y is the final exam score and x is the score on the first test. Carla scored 95 on the first test. What is the predicted value of her score on the final exam?}

None of the above

\textbf{$$\hat{y}$$= 10 + .9*95=95.5. The predicted final exam score is 95.5.}


A rectangular water tank is filled to a depth of $$70$$cm. It contains $$1050$$ litres of water. Some water is taken out of the tank. The water level drops by $$25$$cm. How much water is left in the tank?

$$625\rm L$$

$$375\rm L$$

$$525\rm L$$

$$270\rm L$$

$$675\rm L$$

Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions


Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Multiplication and Division of Decimals


Pip and Bud buys some £$$10$$ books and £$$15$$ books together. If they spent £$$90$$ on the books, how many books have they bought at most in total?

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Indefinite Equations

Let $$x$$ be the number of £$$10$$ books and $$y$$ be the number of £$$15$$ books  $$10x+15y=90$$, which has two positive integer solutions only.  $$\begin{cases} x=6 \textbackslash\textbackslash{} y=2 \end{cases}$$, $$\begin{cases} x=3 \textbackslash\textbackslash{} y=4 \end{cases}$$  $$6+2=8$$


For how many values of $a$ is it true that the line $y=x+a^{2}-a-3$ passes through the vertex of the parabola $y=x^{2}-6x+a^{2}$? (Adapted From 2005 AMC 12B Problem, Question \#8)

infinitely many

We see that the vertex of the quadratic function $y=x^{2}-6x+a^{2}$ is $\left(3, a^{2}-9\right)$. If $\left(2, a^{2}-1\right)$ will be on the line $y=x+a^{2}-a-3$, $a^{2} -9=3+a^{2}-a-3$. Solve for $a$, there is one solution, $a=9$.


For how many values of $a$ is it true that the line $y=x+a$ passes through the vertex of the parabola $y=x^{2}+a^{2}$? (2005 AMC 12B Problem, Question \#8)

infinitely many

We see that the vertex of the quadratic function $y=x^{2}+a^{2}$ is $\left(0, a^{2}\right)$. The $y$-intercept of the line $y=x+a$ is $(0, a)$. We want to find the values (if any) such that $a=a^{2}$. Solving for $a$, the only values that satisfy this are 0 and 1 , so the answer is $(\text{C}) 2$


For the ratio of $$3:8$$, if the consequent increases by $$24$$, the antecedent should~\uline{~~~~~~~~~~}~.

$$\times 4$$

$$\times 24$$

Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio


Seagram knows that $$1111\times 2222=2468642$$  Which of the following answers should he decide is $$3333\times 4444$$?

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers

$$14811852$$  It should be noted that the units digit of the product of $$3333$$ and $$4444$$ will be the same as the units digit of the product of $$3$$ and $$4$$, namely $$2$$: this succinctly identifies the correct option as $$14811852$$. Alternatively, the calculation $$3333\times 4444$$ will have an answer that is $$3\times 2$$ times greater than $$1111\times 2222$$. Now we can work out $$2468642\times 6$$ exactly, which is indeed $$14811852$$.


For each patient visiting a doctor\textquotesingle s office, the nurse records the patient\textquotesingle s body temperature. The plot above shows the temperatures of 28 patients for one particular day. Which of the following statements is true about the distribution of body temperatures?

The distribution is skewed to the left.

The median temperature could be 98.2$^{\circ}F$

The median temperature could be 98.4$^{\circ}F$

The minimum temperature is exactly 97.0$^{\circ}F$

The distribution is normal.

Overseas Competition->Knowledge Point->Calculation Modules

The correct answer is (b).Since there are 28 patients, the median will be between the 14th and 15th data points, which puts it in the bar centered at 98.25°$F$


Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^{2}+x^{4}=2 x^{2} y+1$. What is $\textbar a-b\textbar$?

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation

Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$.  $$ \begin{aligned} \&y^{2}+\sqrt{\pi}^{4}=2 \sqrt{\pi}^{2} y+1 \textbackslash\textbackslash{} \&y^{2}+\pi^{2}=2 \pi y+1 \textbackslash\textbackslash{} \&y^{2}-2 \pi y+\pi^{2}=1 \textbackslash\textbackslash{} \&(y-\pi)^{2}=1 \textbackslash\textbackslash{} \&y-\pi=\pm 1 \textbackslash\textbackslash{} \&y=\pi+1 \textbackslash\textbackslash{} \&y=\pi-1 \end{aligned} $$  There are only two solutions to the equation $(y-\pi)^{2}=1$, so one of them is the value of $a$ and the other is $b$. The order does not matter because of the absolute value sign. $$ \textbar(\pi+1)-(\pi-1)\textbar=2 $$


Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating $3$ inches of pasta from the middle of one piece. In the end, he has $10$ pieces of pasta whose total length is $17$ inches. How long, in inches, was the piece of pasta he started with?


A new operation $$⊕$$ is defined as $$a⊕b=\frac{2}{a^{2}}+\frac{1}{b}$$:  Which of the equations below is/are correct?

Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations


How many even numbers are there?  2, 3, 5, 6, 7, 9, 10.

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers


\textbf{If I toss a fair coin 5000 times}

\textbf{and I get anything other than 2500 heads, then something is wrong with the way I flip coins.}

\textbf{the proportion of heads will be close to 0.5}

\textbf{a run of 10 heads in a row will increase the probability of getting a run of 10 tails in a row.}

\textbf{the proportion of heads in these tosses is a parameter}

\textbf{the proportion of heads will be close to 50.}

\textbf{Since this is a fair coin, the probability to get a head is always 0.5.}


True or False: It is impossible for the $25$-th percentile to be equal to the median.

True

False

For example, consider a dataset with the following values: 0, 2, 2, 3, 3. The median of this dataset is $2$, and the $25$-th percentile is also $2$.


Find the value of the expression $100-98+96-94+92-90+\cdots +8-6+4-2$.

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Grouping in Fast Addition and Subtraction of Whole Numbers


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction


Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are 6 cm in diameter and 12 cm high. Felicia buys cat food in cylindrical cans that are 12 cm in diameter and 6 cm high. What is the ratio of the volume of one of Alex\textquotesingle s cans to the volume of one of Felicia\textquotesingle s cans? (2019 AMC 8, 9)

Solution 1  Using the formula for the volume of a cylinder, we get Alex,~$\pi108$, and Felicia,~$\pi216$. We can quickly notice that~~cancels out on both sides, and that Alex\textquotesingle s volume is~$\dfrac{1}{2}$~~of Felicia\textquotesingle s leaving~$\dfrac{1}{2}=\boxed{1:2}$~as the answer.  Solution 2  Using the formula for the volume of a cylinder, we get that the volume of Alex\textquotesingle s can is~$3^{2}\cdot12\cdot\pi$, and that the volume of Felicia\textquotesingle s can is~$6^{2}\cdot6\cdot\pi$. Now, we divide the volume of Alex\textquotesingle s can by the volume of Felicia\textquotesingle s can, so we get~$\dfrac{1}{2}$, which is$\boxed{\left( B\right)\textbackslash{} 1:2}.$  Solution 3  The ratio of the numbers is~$\dfrac{1}{2}$. Looking closely at the formula~$r^{2}*h*\pi$, we see that the~$r*h*\pi$~will cancel, meaning that the ratio of them will be~$\dfrac{1\left( 2\right)}{2\left( 2\right)}=\boxed{\left( B\right)\textbackslash{} 1:2}$  Solution 4  The second can is 2 size in each of 2 dimensions, and~$\dfrac{1}{2}$~size in 1 dimension.~$\dfrac{2^{2}}{2}=\boxed{\left( B\right)\textbackslash{} 1:2}$


For certain real numbers $a, b$, and $c$, the polynomial $$ g(x)=x^{3}+a x^{2}+x+10 $$ has three distinct roots, and each root of $g(x)$ is also a root of the polynomial $$ f(x)=x^{4}+x^{3}+b x^{2}+100 x+c . $$ What is $f(1)$?

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Indefinite Equations->System of Indefinite Equations

$f(x)$ must have four roots, three of which are roots of $g(x)$. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that  $$ f(x)=g(x)(x-r) $$  where $r \in \mathbb{C}$ is the fourth root of $f(x)$. Substituting $g(x)$ and expanding, we find that  $$ \begin{aligned} f(x) \&=\left(x^{3}+a x^{2}+x+10\right)(x-r) \textbackslash\textbackslash{} \&=x^{4}+(a-r) x^{3}+(1-a r) x^{2}+(10-r) x-10 r \end{aligned} $$  Comparing coefficients with $f(x)$, we see that  $$ \begin{aligned} a-r \&=1 \textbackslash\textbackslash{} 1-a r \&=b \textbackslash\textbackslash{} 10-r \&=100 \textbackslash\textbackslash{} -10 r \&=c . \end{aligned} $$  Let\textquotesingle s solve for $a, b, c$, and $r$. Since $10-r=100, r=-90$, so $c=(-10)(-90)=900$. Since $a-r=1, a=-89$. Then, since $b=1-a r, b=-8009$. Thus, we know that $$ f(x)=x^{4}+x^{3}-8009 x^{2}+100 x+900 . $$ Taking $f(1)$, we find that $$ \begin{aligned} f(1) \&=1^{4}+1^{3}-8009(1)^{2}+100(1)+900 \textbackslash\textbackslash{} \&=1+1-8009+100+900 \textbackslash\textbackslash{} \&=(\mathbf{C})-7007 . \end{aligned} $$


Find the value of $$\left\textbackslash{ \frac{2018+1}{5} \right\textbackslash} + \left\textbackslash{ \frac{2018+2}{5} \right\textbackslash} + \cdots \cdots + \left\textbackslash{ \frac{2018+2017}{5} \right\textbackslash} + \left\textbackslash{ \frac{2018+2018}{5} \right\textbackslash}$$.

None of the above

Overseas Competition->Knowledge Point->Calculation Modules->Fractions


\textbf{According to data from the United States Elections Project, only 36 percent of eligible voters voted in the 2014 elections. For random samples of size 40, which of the following best describes the sampling distribution of pˆ, the sample proportion of people who voted in the 2014 elections?}

\textbf{The sampling distribution is skewed to the left, with mean 0.36 and standard deviation 0.076.}

\textbf{The sampling distribution is skewed to the right, with mean 0.64 and standard deviation 0.006.~}

\textbf{The sampling distribution is approximately normal, with mean 0.36 and standard deviation 0.076.~}

\textbf{The sampling distribution is approximately normal, with mean 0.36 and standard deviation 0.006.~}

\textbf{The sampling distribution is approximately normal,with mean 0.64 and standard deviation 0.076.~}

\textbf{For a large n, the sampling distribution of $\hat{p}$ is approximately normal distribution with $E(\hat{p})=p$}  \textbf{$\sigma\_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$.~}  \textbf{$E(\hat{p})=0.36$}  \textbf{$\sigma\_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.36*0.64}{40}}=0.07589$}


How many digits after the decimal does the product of $3.222$ and $4.22$ have?

Overseas Competition->Knowledge Point->Calculation Modules->Decimals

Count the number of digits in each decimals, the product of the two decimals will have that many digits after the decimal point.


In your family, there are $$15$$ chairs, $$5$$ tables, and $$20$$ cups. What is the ratio of chairs to cups?

There are $$15$$ chairs and $$20$$ cups. So the ratio of chairs to cups is $$15:20$$. The simplest form is $$3:4$$.


What is the sum of the remainders of $$1234 \div5$$ and $$6789 \div 10$$?

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division

The remainders are $$4$$ and $$9$$. Their sum is $$4 +9= 13$$.


If the 9-digit number $$2017122\square2$$ can be divisible by $$4$$, then the number in $$\square$$ can be ．

We check if it is divisible by $$4$$ by looking at the last two digits.  $$72$$ is divisible by $$4$$.


A box has fewer than $50$ cookies in it. The cookies can be divided evenly between $2, 3,$ or $4$ children. However, they cannot be divided evenly between $7$ children because $6$ more cookies would be needed. How many cookies are there in the box? (2021 Math Kangaroo Problem, Level 3-4, Question \#20)

The cookies can be divided evenly between $2, 3,$ or $4$ children, which means the number of cookies can be divisible by $2, 3,$ and $4$ at the same time. The cookies cannot be divided evenly between $7$ children because $6$ more cookies would be needed, which means the number of cookies divided by $7$ with $1$ remaining. Thus, the right answer is $D$.


The following are the weights (in pounds) of seven people: $100, 115, 135, 140, 180, 197, 230$. Find the $80$-th percentile.

$np=7(0.8)=5.6 \uparrow 6$ The $80$-th percentile is $197$.


Meena had $$96$$ coconuts. She sold $$\frac{1}{3}$$ of them yesterday and $$\frac{1}{2}$$ of them today. How many coconuts did she sell altogether?

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


Mariam had ＄$$4y$$. After buying some cloth at ＄$$7$$ per metre, she had ＄$$y$$ left. How many metres of cloth did she buy?

Amount of money she spent to buy the cloth  $$\rightarrow$$＄$$4y-$$＄$$y$$  $$=$$＄$$3y$$,  Length of cloth she bought  $$\rightarrow$$＄$$3y\div $$＄$$7/\text{m}$$  $$= \frac{3y}{7}\text{m}$$.


The mid points of the four sides of a rectangle are $(−3,0)$, $(2,0)$, $(5,4)$, and $(0, 4)$. What is the area of the rectangle?


Pick two consecutive positive integers whose sum is less than $$100$$. Square both of those integers and then find the difference of the square numbers. Which of the following could be the difference? ($2007$ AMC $8$ Problem, Question \#$19$)

Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas-> Difference of Two Squares Formula

Let\textquotesingle s say that $x$ is the smaller of the two numbers. So the question is $$(x+1)+x\textless100(x+1)^{2}-x^{2}=x^{2}+2x+1-x^{2}=2x+1$$. ~$$2x+1$$ is obviously odd, so the answer could be $$\text{C}$$ or $$\text{E}$$.  $$2x+1=131$$ doesn\textquotesingle t match with $$2x+1\textless100$$, so the answer is $$79$$.  Therefore, the answer is $$\text{C}$$.


Two integers are inserted into the list $3,3,8,11,28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?


The $1986^{}\text{th}$ digit at the right of the decimal point in the decimal expression of $\dfrac{1}{7}$ is~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Basic Permutation

$$\frac{1}{7}=0.\overline{142857}$$, it is a decimal which repeats in cycles of $6$. Every $6$\textsuperscript{th}~digit is $7$. The $1986$\textsuperscript{th} digit is $7$.


There are four more girls than boys in Mr. Tse\textquotesingle s class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions

We can set up an equation with $x$ being the number of girls in the class. The number of boys in the class is equal to $x-4$. Since the total number of students is equal to 28 , we get $x+x-4=28$. Solving this equation, we get $x=16$. There are $16-4=12$ boys in our class, and our answer is $16: 12=$ (B) $4: 3$.


Cassandra is helping her mother to pack $$75$$ cupcakes. The boxes that her mother prepares can only fit $$7$$ cupcakes. She must ensure the box is full before she can use the next box. How many boxes she can fill up?

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division->Division of Whole Numbers


If $*abcd*=a\times d+b\times c$, then $*2543*=$~\uline{~~~~~~~~~~}~. ($2004$ Math League.com contest problem, $8$\textsuperscript{th} Grade, Question \#$33$)


Louis had $8$ sticks. He broke three of them into two pieces. How many sticks does he have now?


Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers->Exponentiation of Powers

Start with $3^{}p+3^{4}=90$. Then, change $3^{4}$ to $81$. Subtract from $81$ from both sides to get $3^{}p=9$ and see that $p$ is $2$. Now, solve for $r$. Since $2^{}r+44=76$, $2^{}r$ must equal $32$, so $r=5$. Now, solve for $s$. $5^{3}+6^{}s=1421$ can be simplified to $125+6^{}s=1421$, which simplifies further to $6^{}s=1296$. Therefore, $s=4$. $prs$ equals $2\times5\times4$ which equals $40$. So, the answer is $40$.


Hannah bought $$5$$ whole pizzas: $$1$$ for herself and $$1$$ for each of her $$4$$ students.  Hannah sliced her pizza into $$5$$ equal parts and ate only $$3$$ slices.  Student $$A$$ sliced his pizza into $$4$$ equal parts, but ate only $$3$$ slices.  Student $$B$$ sliced his pizza into $$8$$ equal parts, but ate only $$7$$ slices.  Student $$C$$ sliced her pizza into $$3$$ equal parts, but ate only $$2$$ slices.  Student $$D$$ sliced her pizza into $$6$$ equal parts, but ate only $$3$$ slices.  Who ate less pizza than Hannah?

Student $$A$$

Student $$B$$

Student $$C$$

Student $$D$$

Nobody

All of them eat more than $$\frac{1}{2}$$, except student $$D$$.


When Koko the Koala is awake, he eats $84$ grams of leaves per hour. He was awake for $2$ hours yesterday and $10$ hours today. How many grams of leaves did he eat in total in the two days? (adapted from 2014 Math Kangaroo Problem, Level 3-4, Question \#4)


Which of the following is not an equivalent ratio of $$4:12$$?

Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Proportions


In a math competition, each participant has a unique $5$-digit registration number of the form $\overline{BBCAC}$, where $0 \leq A \textless{} B\textless{} C \leq 9$ and $B$ is the average of $A$ and $C$. What is the maximum number of participant that can join this competition?

Overseas Competition->Knowledge Point->Calculation Modules


For $\triangle ABC$, all its side lengths are integers. The primeter of $\triangle ABC$ with a side of length $14$ and a side length of $8$ is at least ．

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities

We know from the triangle inequality that the last side, $s$, fulfills $s+8\textgreater14$. Therefore, $P\textgreater14+14$. The least integer value of $P$ is $29$


There are two positive integers, $$x$$ and $$y$$. $$x$$ equals to $$3^{2}$$, and $$y$$ is the base of $$5^{3}$$. What is the product of $$x$$ and $$y$$?


SASMO 2016 P2 Q5  What number does () stands for?

Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Patterns->Encryption and Decryption


For $\triangle ABC$, all its side lengths are integers. The perimeter of $\triangle ABC$ with a side of length $14$ and a side length of $8$ is at least ．


$2018$ is an interesting number. This is because when we add the first digit and the last digit, we will get the reverse of the middle two digits. Which of the options below is also an interesting number?

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers


Given: $a-b\textgreater a$ and $a+b\textless b$. Which of the following must be true?~\uline{~~~~~~~~~~}~  $I$. $a b$ is negative  $II$. $a+b$ is negative  $III$. $a-b$ is negative

$I$ only

$I$ and $III$ only

$II$ only

$I$, $II$, $III$

$a-b\textgreater a$ then $b\textless0$  $a+b\textless b$ then $a\textless0$.  So both $a$ and $b$ are negative. $I$ is false, since the product will be positive. $II$ is true since the sum of two negative numbers is negative. $III$ is false because if $a=-1$ and $b=-2$ then $a-b$ is positive. Only $II$ is true.


What is the remainder when $2^{2023}+2023$ is divided by $2^{20}+1$? (Adapted From 2020 AMC 10B Problems, Question \#22)

Let $x=2^{20}$.  We are now looking for the remainder of $\frac{8x^{101}+2023}{x+1}$. By Polynomial Remainder Theorem, the remainder is $8\times (-1)^{101} + 2023 = 2015$.


The last four digits in Andy\textquotesingle s ID card are $2025$. What is the the difference between the largest and the smallest digit in $2025$?~(adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$7$$)

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form

The largest digit is $5$, and the smallest digit is $0$.  $5-0=5$


Let $a$ and $b$ be relatively prime positive integers with $a\textgreater b\textgreater0$ and $\frac{a^{3}-b^{3}}{(a-b)^{3}}=\frac{73}{3}$. What is $a-b$?

Slightly expanding, we have that $\frac{(a-b)\left(a^{2}+a b+b^{2}\right)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$. Canceling the $(a-b)$, cross multiplying, and simplifying, we obtain that $0=70 a^{2}-149 a b+70 b^{2}$. Dividing everything by $b^{2}$, we get that $$ 0=70\left(\frac{a}{b}\right)^{2}-149\left(\frac{a}{b}\right)+70 \text {. } $$ Applying the quadratic formula and following the restriction that $a\textgreater b\textgreater0$ $$ \frac{a}{b}=\frac{10}{7} \text {. } $$ Hence, $7 a=10 b$. Since they are relatively prime, $a=10, b=7$. $$ 10-7= 3 \text {. } $$


Which number has to be subtracted from $17$ in order to obtain $-33$? (Adapted from 2017 Math Kangaroo Problem, Level 7-8, Question \#3)

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations


Fill in the blank:~\uline{~~~~~~~~~~}~is $$3$$ tens $$7$$ ones less than $$4$$ tens $$6$$ ones.

None of the above


There were $$16$$ monkeys in total in the animal school. After the whistle, they arranged themselves into $$8$$ rows. How many monkeys were there in each row after the whistle? (Adapted from 2005 Math Kangaroo Problem, Level 3-4, Question \#4)

There were $$16 \div 8 = 2$$ monkeys in each row.


Find all values of $x$ such that $\textbar3 x+12\textbar\textless9$ and $\textbar x+2\textbar\textless\textbar-3 x-6\textbar$.~\uline{~~~~~~~~~~}~

$-7\textless x\textless-1$

$-7\textless x\textless-2$

$-7\textless x\textless-1 ; x \neq-2$

Based on the first inequality, we know that $-9\textless3 x+12\textless9$ because $3 x+2$ must be less than 9 units away from zero. We can subtract 12 from all three parts of the inequality to arrive at $-21\textless3 x\textless-3 \rightarrow-7\textless x\textless-1$. From the second inequality we can rewrite $\textbar-3 x-6\textbar$ as $\textbar3 x+6\textbar=3\textbar x+2\textbar$ because they must be equal. The second inequality must be true for all numbers except for when $\textbar x+2\textbar=0$, or when $x=-2$. Thus the answer includes all numbers from $-7$ to $-1$ with the exception of $-2$. The answer is $\mathbf{D}$.


For $\triangle ABC$, all its side lengths are integers. The perimeter of $\triangle ABC$ with a side of length $3$ and a side length of $4$ is at least ．

We know from the triangle inequality that the last side, $s$, fulfills $s+3\textgreater4$. Therefore, $P\textgreater4+4$. The least integer value of $P$ is $9$


Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{3 x+y}{x-3 y}=-2$. What is the value of $\frac{x+3 y}{3 x-y}$?

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Proportional Equations

Rearranging, we find $3 x+y=-2 x+6 y$, or $5 x=5 y \Longrightarrow x=y$. Substituting, we can convert the second equation into $\frac{x+3 x}{3 x-x}=\frac{4 x}{2 x}= 2$ More step-by-step explanation:  $$ \begin{aligned} \&\frac{3 x+y}{x-3 y}=-2 \textbackslash\textbackslash{} \&3 x+y=-2(x-3 y) \textbackslash\textbackslash{} \&3 x+y=-2 x+6 y \textbackslash\textbackslash{} \&5 x=5 y \textbackslash\textbackslash{} \&x=y \textbackslash\textbackslash{} \&\frac{x+3 y}{3 x-y}=\frac{1+3(1)}{3(1)-1}=\frac{4}{2}=2 \end{aligned} $$


$$\textgreater$$, $$\textgreater$$

$$\textgreater$$, $$\textless$$

$$\textless$$, $$\textgreater$$

$$\textless$$, $$\textless$$

Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering


$$\textgreater$$, $$\textgreater$$

$$\textgreater$$, $$\textless$$

$$\textless$$, $$\textgreater$$

$$\textless$$, $$\textless$$


If $ a◆b$ means $(a\times b)+b$ , then $（2◆3）◆4$ has the value．

Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly


(2) Eddie had 120 dollars as his pocket money, and spent $$ \frac{3}{4} $$of it. How much money is left?

Overseas Competition->Knowledge Point->Calculation Modules->Fractions

left！！！not spent! So you need to calculate the fraction of left money. Then use T*F=C


Divide $$2$$、$$3$$、$$24$$、$$33$$、$$55$$ and $$60$$ into two groups with 3 numbers in each group to make the product of numbers in each group the same, so the product is．

Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers->Power of Products

$$2=2$$，  $$3=3$$，  $$24=2\times 2\times 2\times 3$$，  $$33=11\times 3$$，  $$55=11\times 5$$，  $$60=2\times 2\times 3\times 5$$，  $$2\times 60\times 33=24\times 55\times 3$$，  $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde2\times 60\times 33$$  $$=(2\times 33)\times 60$$  $$=66\times 60$$  $$=3960$$．  $$\text{C}$$．


Which of the following integers cannot be written as the sum of four consecutive odd integers? (2015 AMC 8 Problems, Question \#14)

Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables

Let our $4$ numbers be $n, n+2, n+4, n+6$, where $n$ is odd. Then our sum is $4 n+12$. The only answer choice that cannot be written as $4 n+12$, where $n$ is odd, is (D) 100 .


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction

$$\begin{eqnarray}\&\&\left( 100-99 \right)+\left( 98-97 \right)\cdots +\left( 4-3 \right)+\left( 2-1 \right)\textbackslash\textbackslash{} \&=\&50\times 1\textbackslash\textbackslash{} \&=\&50．\end{eqnarray}$$


Of the following, which has a value different from the others?

$$40\times 50$$

$$4\times 5000$$

$$50\times 400$$

$$40\times 500$$

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division

To see which value is different, count the total number of $$0$$\textquotesingle s in each product.


$$12 \times \left( \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right)=$$．

Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions


For $\triangle ABC$, all its side lengths are integers. The primeter of $\triangle ABC$ with a side of length $3$ and a side length of $4$ is at least ．


The fraction $$\dfrac{214}{263}$$ keeps the same value when both its numerator and denominator are~\uline{~~~~~~~~~~}~.

multiplied by $$2~ $$

increased by $$2$$

decreased by $$2$$

$$ $$squared$$ $$

Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Basic Understanding of Fractions

The problem is about the basic property of fractions, namely, multiplying or dividing the numerator and denominator by the same number (except $$0$$), the value of the fraction is unchanged.


For $\triangle ABC$, all its side lengths are integer. The primeter of $\triangle ABC$ with a side of length $12$ and a side length of $7$ is at least ．

We know from the triangle inequality that the last side, $s$, fulfills $s+7\textgreater12$. $P=s+7+12\textgreater12+12$. Therefore, $P\textgreater24+1=25$


What is the correct ordering of the three answers from $\frac5{19}\div \frac{25}{38}$, $1\frac12 \div \frac{15}8$, and $\frac74 \div \frac{35}{12}$, in increasing order? (adapted from 2012 AMC 8 Problems, Question \#20)

$\frac5{19}\div \frac{25}{38}$ \textless~$1\frac12 \div \frac{15}8$ \textless{} $\frac74 \div \frac{35}{12}$

$\frac74 \div \frac{35}{12}$ \textless~$1\frac12 \div \frac{15}8$ \textless~$\frac5{19}\div \frac{25}{38}$

$1\frac12 \div \frac{15}8$ \textless{} $\frac74 \div \frac{35}{12}$ \textless~$\frac5{19}\div \frac{25}{38}$

$\frac5{19}\div \frac{25}{38}$ \textless{} $\frac74 \div \frac{35}{12}$ \textless~$1\frac12 \div \frac{15}8$

$\frac74 \div \frac{35}{12}$ \textless~$\frac5{19}\div \frac{25}{38}$ \textless~$1\frac12 \div \frac{15}8$

Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions->Division of Fractions


Choose the answer in the simplest form.  $$\frac{9}{4}\times \frac{8}{27}=$$．


In the arithmetic sequence $3$, $7$, $11$, $15$, $\cdots$ , the $26$\textsuperscript{th} number is．

Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences


If a whole number is multiplied by itself, the ones\textquotesingle{} digit of the product \emph{cannot} be．

Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers

If a whole number is multiplied by itself, the ones\textquotesingle{} digit of the product could be $$1 (1\times1)$$ or $$5 (5\times5)$$ or $$9 (3\times3)$$, but not $$7$$.


Bob: " Hi, Stanley. What is the coefficient of the variable in this algebraic expression $3x^{2}-4$?"  Stanley:" I can give you a hint. The value of the coefficient is $7$ more than the constant."  Stanley\textquotesingle s hint is~\uline{~~~~~~~~~~}~and the coefficient of the variable is~\uline{~~~~~~~~~~}~.

correct, $3$

correct, $-4$

incorrect, $3$

incorrect, $4$

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions

Constant: $-4$  Coefficient of the variable: $3$


Gilda has a bag of marbles. She gives $20 \textbackslash\%$ of them to her friend Pedro. Then Gilda gives $25 \textbackslash\%$ of what is left to another friend, Ebony. Finally, Gilda gives $5 \textbackslash\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? (adapted 2019 AMC 8, Question \#8)

Overseas Competition->Knowledge Point->Calculation Modules->Percentage Calculation

After Gilda gives $20 \textbackslash\%$ of the marbles to Pedro, she has $80 \textbackslash\%$ of the marbles left. If she then gives $25 \textbackslash\%$ of what\textquotesingle s left to Ebony, she has $(0.75 * 0.8)=60 \textbackslash\%$ of what she had at the beginning. Finally, she gives $5 \textbackslash\%$ of what\textquotesingle s left to her brother, so she has $(0.6 * 0.95)$ (D) $57\textbackslash\%$~ of what she had in the beginning left.


Given that $$x\otimes y=6\times x-5\times y$$, find $$7\otimes 6$$.


Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities


A slug called Glug eats $$2$$ tomatoes for every $$3$$ strawberries. Yesterday it had eaten $$35$$ tomatoes and strawberries altogether. How many tomatoes did it eat?

Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions


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

Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers->Exponentiation of Powers


Steven subtracts the units digit from the tens digit for each two-digit number. He then finds the sum of all his answers.  What is the value of Steven\textquotesingle s sum?

The sum of all the tens digits is $$\left(1+2+3+4+5 +6 +7+8+9\right)\times10$$.  The sum of all the units digits is $$\left(0 + 1+2+3 +4+5 +6 +7+8+9\right)\times9$$.  Therefore Steven\textquotesingle s sum is $$\left(1+2+3 +4+5+6+7+8+9\right)\times1=45$$.


The product of two different nonzero integers cannot be．

prime

zero

even

composite

The product of two different nonzero integers can never be $$0$$.


Bob bought three kinds of meat: pork, beef and chicken with the total cost of $152. The ratio of the weight of pork, beef and chicken is~$2:4:3$. The ratio of the price per pound of pork, beef and chicken is~$6:5:2$. What is the sum of the last digits of the cost of each kind in dollars?

Let~$x,\textbackslash{} y\textbackslash{} and\textbackslash{} z$~be the weight of pork, beef and chicken, respectively.  ~$\textbackslash{} x:y:z=2:4:3$~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (1)  The ratio of the price will be~$\left( 6x\right):\left( 5y\right):\left( 2z\right).$  The costs of pork, beef and chicken are~$A,B\textbackslash{} and\textbackslash{} C,$~respectively.  ~$A=\dfrac{6x}{6x+5y+2z}\times152$~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~(2)  From (1), we get:  ~$y=2x,\textbackslash{} and\textbackslash{} z=\dfrac{3}{2}x$~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~(3)  (2) becomes:~$A=\dfrac{6x}{6x+5\left( 2x\right)+2\times\dfrac{3}{2}x}\times152=\dfrac{6x}{19x}\times152=48$  Similarly,~$B=\dfrac{5y}{6x+5y+2z}\times152=\dfrac{5y}{19x}\times152=\dfrac{10x}{19x}\times152=80.$  And~$C=\dfrac{2z}{6x+5y+2z}\times152=\dfrac{2z}{19x}\times152=\dfrac{3x}{19x}\times152=24.$  Pork costs $48 per pound, beef costs $80, and chicken costs $24.  The sum of the last digits of the costs is 8+0+4=12.


$$1000\text{m}$$ per second $$=$$$$\text{km}$$ per hour.

Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Compound Units


Solve the equation: $$16\times 25-13\left( 3x+2 \right)=179$$．

$$39x=400-26-179$$  $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde x=5$$．


Find the missing term in the following sequence: $$1, 2, 4, 7,~\uline{~~~~~~~~~~}~, 16$$.

Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences

The number sequence are in the pattern of $$+1, +2, +3, +4, +5, \cdots $$


On Kangaroo planet each kangyear has 20 kangmonths and each kangmonth has 6 kangweeks, How many kangweeks are there in one quarter of a kangyear?

Overseas Competition->Knowledge Point->Calculation Modules->Fractions

$\dfrac{1}{4}$kangyear x~$\dfrac{20\textbackslash{} kangmonths}{1\textbackslash{} kangyear}$~x~$\dfrac{6\textbackslash{} kangweeks}{1\textbackslash{} kangmonth}$~= 30 kangweeks.


A vase was being sold at the price of $250$ dollars. The store decides to sell it with a $40\textbackslash\%$ discount. If you buy the vase now, how much will you save?


Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Multiplication and Division of Decimals


Mary wanted to buy $12$ lemons. The lemons were sold either for $90$ cents each or at $3$ dollars for a bag of $4$ lemons. How much money would Mary save if she buys three bags of lemons instead of buying $12$ lemons separately?

$1.2$ dollars

$1.5$ dollars

$1.6$ dollars

$1.8$ dollars

$2.1$ dollars

Overseas Competition->Knowledge Point->Calculation Modules


The following are the weights (in pounds) of ten people: $100,115, 135, 140, 180, 197, 203, 230, x, y$ (not necessarily in increasing order). It is also given that the average weight of these ten people is $157$ pounds, and there is a unique mode of $135$. Find the $56$-th percentile.

First of all, we find the values of $x$ and $y$.  Since the average is $157$, we have  $(100+115+135+140+ 180+197+203+230+x+y = 157 \times 10 = 1570$  $1300+x+y =1570$  $x+y = 270$.  There is a unique mode, $135$, then $135$ must appear at least twice. Therefore, one of $x, y$ is $135$. It is easy to deduce that both $x$ and $y$ are $135$.  List the weights in increasing order: $100,115, 135, 135, 135, 140, 180, 197, 203, 230$.  $np=10(0.56)=5.6 \uparrow 6$. The $56$-th percentile is $140$.


Which of the following equations are NOT equivalent to $x+5=13$?

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation

Equations are equivalent when you can obtain one by subtracting, adding, dividing, or multiplying the same number on the other.


What is $$9762 + 7 \times 8 \times 9 \times 4 \times 99 \times 0$$

None of the above


In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is $15$. What is the greatest possible perimeter of the triangle? （2006 AMC10B, Question 10)

Let $x$ be the length of the first side. The lengths of the sides are: $x, 3 x$, and 15 . By the Triangle Inequality, $$ \begin{aligned} \&3 x\textless x+15 \textbackslash\textbackslash{} \&2 x\textless15 \textbackslash\textbackslash{} \&x\textless\frac{15}{2} \end{aligned} $$ The greatest integer satisfying this inequality is 7 . So the greatest possible perimeter is $7+3 \cdot 7+15=$ (A) 43


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction


The $$2022$$th digit to the right of the decimal point in the decimal representation of $$\frac 1{54}$$ is．

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Sequence Operations

The decimal is $$0.0185185\cdots $$. An "$$8$$" appears in the $$3$$rd, $$6$$th, $$9$$th, $$\cdots $$, $$2022$$th decimal place. So a "$$8$$" is in the $$2022$$th place.


According to the regulation of the pyramid series, the formula $$1+2+3+4+5+6+5+4+3+2+1=$$．

Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables

``Pyramid series''$$1+2+3+4+\cdots +\left( n-1 \right)+n+\left( n-1 \right)+\cdots +3+2+1$$  $$={{n}^{2}}$$．  So the answer is $$\text{A}$$．


Which of the following has a result of $183$?


Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions


Simplify the following expression: $$a^{2}\times a+b^{2}\times b^{3}$$.


After the first three numbers, each number is the sum of the previous three numbers. So the next number is $$11 + 20 + 37 = 68$$.


\textbf{Sean and Evan are college roommates who have part-time jobs as servers in restaurants. The distribution of Sean's weekly income is approximately normal with mean $\textbackslash$225$ and standard deviation $\textbackslash$25$. The distribution of Evan's weekly income is approximately normal with mean $\textbackslash$240$ and standard deviation $15. Assuming their weekly incomes are independent of each other, which of the following is closest to the probability that Sean will have a greater income than Evan in a randomly selected week?}

\textbf{Sean \textasciitilde{} N(225, 25)}  \textbf{Evan \textasciitilde{} N(240, 15)}  \textbf{→}  \textbf{Sean - Evan \textasciitilde{} N(-15, 29.155)}  \textbf{P(Sean-Evan \textgreater{} 0) = 1-P(X\textless0) = 0.303}


An amusement park has a collection of scale models, with ratio $1: 40$, of buildings and other sights from around the country. The height of empire state building is 1250 feet. What is the height in feet of its replica to the nearest whole number? (adapted from 2018 AMC 8, Question 1)

You can see that since the ratio of real building\textquotesingle s heights to the model building\textquotesingle s height is $1: 40$. We also know that the Empire State Building is 1250 feet, so to find the height of the model, we divide by 40 . That gives us $31.25$ which rounds to 31 . Therefore, to the nearest whole number, the duplicate is (B) 31 feet.


A basketball is on sale with 35\% off, the discounted price is $$52$$ dollars, the original price of the basketball wasdollars．

$$x \times 65\textbackslash\% = 52$$  $$x = 80$$


$$0.02 $$

$$0.05 $$

$$0.2 $$

Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Four Operations of Decimals


The ratio of $w$ to $x$ is $4: 3$, the ratio of $y$ to $z$ is $3: 2$, and the ratio of $z$ to $x$ is $1: 6$. What is the ratio of $w$ to $y$? (2020 AMC 10B Problems, Question \#3)

$4: 3$

$3: 2$

$8: 3$

$4: 1$

$16: 3$

We need to somehow link all three of the ratios together.  We can start by connecting the last two ratios together by multiplying the last ratio by two. $z: x=1: 6=2: 12$, and since $y: z=3: 2$, we can link them together to get $y: z: x=3: 2: 12$.  Finally, since $x: w=3: 4=12: 16$, we can link this again to get: $y: z: x: w=3: 2: 12: 16$, so $w: y=(\mathbf{E}) 16: 3$.


The digit $9$ in what number represents a value of $0.09$?

Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals

If the digit $$9$$ represents a value of $$0.09$$, it means the digit $$9$$ is located on the hundredth place on that number, which is the second digit after the decimal point.  Check Lesson 4 Concept 1 on textbook


What is the smallest whole number larger than the perimeter of any triangle with a side of length $9$ and a side of length $1$? (adapted from 2015 AMC8, Question 8)

We know from the triangle inequality that the last side, $s$, fulfills $s\textless9+1=10$. Adding $9+1$ to both sides of the inequality, we get $s+9+1\textless20$, and because $s+9+1$ is the perimeter of our triangle, (C) 20 is our answer.


The sum of three numbers is $20$. The first is four times the sum of the other two. The second is seven times the third. What is the product of all three numbers?


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations


Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions

From $$3y -1=5$$, then we can get $$y=2$$.  Therefore from $$2x-3=2+5$$, we obtain $$x=5$$.


Write each of the numbers 0,1,2,3,4,5,6 in the sqaures to make the addition correct. What digit will be in the grey square?  pic

Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Patterns->Encryption and Decryption


If 5 plates weigh as much as 9 mugs, then 99 mugs weigh as much as~\uline{~~~~~~~~~~}~plates.

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution


On the first day of a $7-$day holiday, Judy reads $9$ pages of a book. On the second day, she reads $12$. Then each day later, she reads $3$ pages more than the day before. On the last day of the holiday, she reads the corresponding number of pages and exactly finishes reading the book. How many pages does the book have?

Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences

Use the formula of arithmetic sequence to solve this problem.  On the last day, she reads $9+6\times3=27$ pages.  The book has $(9+27)\times7 \div2=126$ pages.


Linda, Alice, Joe, Lily, and Rachel lost their balloons. The numbers on their balloons were all smaller than $8$. The difference between the largest number and the smallest number was $5$. What are their balloon\textquotesingle s numbers? (adapted from $$2021$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$8$$）

$7, 5, 4, 3,2$

$4, 1, 7,9,6$

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Subtraction of Whole Numbers->Subtraction in Horizontal Form

In~ B, D, and E, the largest number more than $8$. In the C, the difference between the largest number and the smallest number is $4$.


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers


None of the above

Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Fast Calculation in Fractions


What is the next number in the sequence below?  $$47, 44, 38, 29, 17, \cdots $$

None of the above

Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences


$10000$ years ago, Owen the fisher traded $3$ fishes for $1$ rabbit from Oscar the hunter. Then, Oscar traded $2$ rabbits for $3$ packs of wheat from Dennis the farmer. How many fishes should Owen give Dennis for a pack of wheat?

Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Proportions->Simplifying Continued Ratios

fish:rabbit$=3:1$  wheat:rabbit$=3:2$  fish: wheat$=2:1$


Harry and Terry are each told to calculate $8-(2+5)$.~Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$.~If Harry\textquotesingle s answer is $H$~and Terry\textquotesingle s answer is $T$, what is $H-T$? (2014 AMC $8$ Problem, Question \#1)


Two non-zero real numbers, $x$ and $y$, satisfy $(x+y)^{2}=3xy$. Which of the following is a possible value of $$\frac{x+y}{y}-\frac{x-y}{x}$$? (Adapted From 2000 AMC 12 Problems, Question \#11)

Overseas Competition->Knowledge Point->Calculation Modules


Eddie has a card. The number on the card is a neighbouring number of 8, but is not a neighbouring number of 6. What is the number on Eddie\textquotesingle s card?~\uline{~~~~~~~~~~}~

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers


Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Basic Understanding of Fractions


Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions


$$0.1\times 0.2\times 0.3=$$．($1979-1980$ Math League.com contest problem, $7$\textsuperscript{th~}Grade, Question \#$12$)

Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Multiplication and Division of Decimals


If $$5$$ plates weigh as much as $$9$$ mugs, then $$45$$ mugs weigh as much as~\uline{~~~~~~~~~~}~plates.

Since $$9$$ mugs $$=5$$ plates  $$45\div9=5$$ times of the above equation.  $$45$$ mugs $$=25$$ plates


Find the missing number: $$512\times2 = 32\times $$．


A container had $27$ ℓ of longan drink. The drink is made up of three $2$-ℓ\textbf{~}bottles of longan syrup and some water. What is the volume of water used to make the drink?

6ℓ

Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Units of Volume and Capacity


Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio


Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division


Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions


In the following Figures $\left (a\right )$ and $\left (b\right )$, each number inside a small triangle is the sum of the numbers inside in the neighbouring small circles. The number inside each circle is either $1$, $2$, $3$, $4$, $5$, $6$, $7$ or $8$.  The sum of whole numbers inside the circles in Figure $\left (a\right )$ is $1+8+8+2+3+5=27$. What is the largest possible sum of whole numbers inside the circles in Figure $\left (b\right )$?

None of the above


$A=B$

$A\textgreater B$

$A\textless B$

All of the above

None of the above


The fraction $$\frac23$$ keeps the same value when both its numerator and denominator are．

multiplied by $$2~ $$

increased by $$2$$

decreased by $$2$$

$$ $$squared$$ $$

The problem is about the basic property of fractions, namely, multiplying or dividing the numerator and denominator by the same number (except $$0$$), the value of the fraction is unchanged.


The original price of a product was $$80$$ dollars, and it\textquotesingle s on sale for 30\% off, this product isdollars cheaper than before．

the new price is $$70\textbackslash\%$$ of the original price，$$so$$ the new price is $$80\times 70\textbackslash\%=56$$；  $$80-56=24$$．  so choose $$\text{B}$$．


What is the product of five fractions, whose denominators are $5-1$ in descending order, and the numerators are $1-5$ in ascending order？

Overseas Competition->Knowledge Point->Calculation Modules->Fractions


Tony and Mary are waiting in line to go shopping. Tony is fifth in line, while Mary is 365th. How many people are standing between Tony and Mary in the line?

Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering

The question is asking for the number of people between Tony and Mary, which is $$365-5-1=359$$.


Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Complex Fractions


Find the value of $$ 6\times \frac{5}{3}$$．

Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions


Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences


Teacher Nicole bought some sweets and divided it equally among $9$ children. If everyone got $6$ sweets, there would still be some sweets remaining. What is the most number of sweets Teacher Nicole could have bought? What is the least number of sweets Teacher Nicole could have bought?

~\uline{~~~~~~~~~~}~$\div 9=6$ $\text{R}$~\uline{~~~~~~~~~~}~  Greatest possible remainder is $8$ while smallest possible remainder is $1$.  Greatest possible number of sweets is $$9\times 6+8=62$$, while the least possible number of sweets is $$9\times 6+1=55$$.


What is the correct ordering of the three numbers $$\dfrac{5}{19}$$, $$\dfrac{7}{21}$$, and $$\dfrac{9}{23}$$, in increasing order? ($$2012$$ AMC $$8$$ Problem, Question \#$$4$$)

$$\dfrac{9}{23}\textless{} \dfrac{7}{21}\textless\dfrac{5}{19}$$

$$\dfrac{5}{19}\textless{} \dfrac{7}{21}\textless{} \dfrac{9}{23}$$

$$\dfrac{9}{23}\textless{} \dfrac{5}{19}\textless{} \dfrac{7}{21}$$

$$\dfrac{5}{19}\textless{} \dfrac{9}{23}\textless{} \dfrac{7}{21}$$

$$\dfrac{7}{21}\textless{} \dfrac{5}{19}\textless{} \dfrac{9}{23}$$

Method $$1$$:  The value of $$\dfrac{7}{21}$$ is $$\dfrac{1}{3}$$. Now we give all the fractions a common denominator.  $$\dfrac{5}{19} \Rightarrow \dfrac{345}{1311}$$,  $$\dfrac{1}{3} \Rightarrow \dfrac{437}{1311}$$,  $$\dfrac{9}{23} \Rightarrow \dfrac{513}{1311}$$.  Ordering the fractions from least to greatest, we find that they are in the order  listed. Therefore, $$\frac{5}{19}\textless{} \frac{7}{21}\textless{} \frac{9}{23}$$.  Method $$2$$:  Instead of finding the LCD, we can subtract each fraction from $$1$$ to get a common numerator. Thus,  $$1- \dfrac{5}{19}= \dfrac{14}{19}$$,  $$1- \dfrac{7}{21}= \dfrac{14}{21}$$,  $$1- \dfrac{9}{23}= \dfrac{14}{23}$$.  All three fraction have the common numerator $$14$$. Now the order of the fractions is obvious. $$\dfrac{14}{19}\textgreater\dfrac{14}{21}\textgreater\dfrac{14}{23}\Rightarrow\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$. Therefore, $$\dfrac{5}{19}\textless\dfrac{7}{21}\textless\dfrac{9}{23}$$.


Which of the following is a linear equation written in function form?

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with one Variable

An equation is in function form when it is solved for $y$.  A linear equation is also an equation in which the highest power of the variable is always $1$


The school has $$11200$$ books. The school puts them into 70 equal piles. How many books are in each pile?

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division->Division of Whole Numbers->Division within the Multiplication Tables


\textbf{The height and age of each child in a random sample of children was recorded. The value of the correlation coefficient between height and age for the children in the sample was 0.8. Based on the least-squares regression line created from the data to predict the height of a child based on age, which of the following is a correct statement?}

\textbf{~On average, the height of a child is 80\% of the age of the child.}

\textbf{The least-squares regression line of height versus age will have a slope of 0.8.}

\textbf{The proportion of the variation in height that is explained by a regression on age is 0.64.~}

\textbf{The least-squares regression line will correctly predict height based on age 80\% of the time.}

\textbf{The least-squares regression line will correctly predict height based on age 64\% of the time.}


Karen works part-time at a local convenience store and earns 10 dollars per hour. She wants to spend next Saturday afternoon attending a music concert. The full price of a concert ticket is 75 dollars, but Karen was able to get a discounted price of 50 dollars from a friend who purchased the ticket but has become unable to attend. If Karen took 4 hours off from her job to attend the concert, what was her opportunity cost of attending the concert?

Economic costs are the sum of explicit and implicit costs.


The measures of angles $A, B$ and $C$ are $32^{\circ}, 68^{\circ},90^{\circ}$. The triangle formed by the three angles will be~\uline{~~~~~~~~~~}~.

right triangle

acute triangle

obtuse triangle

no triangle

Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns of Number Tables->Triangle Number Table

$32^{\circ} + 68^{\circ} + 90^{\circ}=190^{\circ}$  The sum of a triangle\textquotesingle s interior angles is $180^{\circ}$


The zeroes of the function $f(x)=x^{2}-a x+a$ are integers. What is the sum of the possible values of $a$? (Adapted From 2015 AMC 10A Problems, Question \#23)

Overseas Competition->Knowledge Point->Calculation Modules

By Vieta\textquotesingle s Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.  Because the zeros are integral, the discriminant of the function, $a^{2}-4a$, is a perfect square, say $k^{2}$. Then adding $4$ to both sides and completing the square yields $$ (a-2)^{2}=k^{2}+4$$.  Therefore, $(a-2)^{2}-k^{2}=4$ and $$((a-2)-k)((a-2)+k)=4$$.  Let $(a-2)-k=u$ and $(a-2)+k=v$; then, $a-2=\frac{u+v}{2}$ and so $a=\frac{u+v}{2}+2$.  Listing all possible $(u, v)$ pairs such that $a$ is integral: $(2,2), (-2,-2)$ (not counting transpositions because this does not affect $u+v$). Then, $a=4,0$.  These $a$ sum to $4$.


Consider these two operations:  $a ~♦ ~b = a^{2}~−~b^{2}$  $a✭b = (a~−~b)^{2}$  What is the value of $(5♦3)✭6$?


What is the least possible value of $(x y-1)^{2}+(x+y)^{2}$ for real numbers $x$ and $y$?

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Indefinite Equations

We expand the original expression, then factor the result by grouping: $$ \begin{aligned} (x y-1)^{2}+(x+y)^{2} \&=\left(x^{2} y^{2}-2 x y+1\right)+\left(x^{2}+2 x y+y^{2}\right) \textbackslash\textbackslash{} \&=x^{2} y^{2}+x^{2}+y^{2}+1 \textbackslash\textbackslash{} \&=x^{2}\left(y^{2}+1\right)+\left(y^{2}+1\right) \textbackslash\textbackslash{} \&=\left(x^{2}+1\right)\left(y^{2}+1\right) . \end{aligned} $$ Clearly, both factors are positive. By the Trivial Inequality, we have $$ \left(x^{2}+1\right)\left(y^{2}+1\right) \geq(0+1)(0+1)= 1 . $$ Note that the least possible value of $(x y-1)^{2}+(x+y)^{2}$ occurs at $x=y=0$.


How many even numbers are there?  1, 3, 4, 6, 7, 9, 5, 8.

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers


Find the value of $$26+(12-9\div3)\times4-2$$.

Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations


Avril has a card. The number on the card is a neighbouring number of 10, but is not a neighbouring number of 12. What is the number on Avril\textquotesingle s card?~\uline{~~~~~~~~~~}~


$$\textgreater$$

$$\textless$$

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation

If $$☆=10$$, then $$△=5$$，$$☆\textgreater△$$．  So, the answer is $$\text{A}$$．


\textbf{In a population of university students, 20\% of the students have experienced feelings of math anxiety. If we select 8 students, what is the probability that exactly three have experienced math anxiety?}

$${8 \choose 3} (0.20)^{3} (0.80)^{8}$$

$${8 \choose 3} (0.20)^{3} (0.80)^{5}$$

$${5 \choose 3} (0.20)^{3} (0.80)^{5}$$

$$(0.20)^{3} (0.80)^{5}$$

$$(0.20)^{5} (0.80)^{3}$$

\textbf{Binomial distribution with parameters n = 8 and p = 0.20.}  \textbf{$$P(X = 3) ={8 \choose 3}(0.20)^{3}(0.80)^{5}$$}