The sum of five consecutive whole numbers is $$280$$. What is the sum of the next five consecutive whole numbers?

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

Each whole number in the second sequence is $$5$$ more than the corresponding number in the first sequence. We have $$280 +5 \times5 = 305$$.


What is the range of the number in tens place?

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 $25$ and a side length of $18$ 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 $s+18\textgreater25$. Therefore, $P\textgreater25+25$. The least integer value of $P$ is $51$


$a$ and $$b$$ are reciprocals， $$\frac{2}{a}\div \frac{b}{12}=$$．

Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering->Comparing and Ordering Fraction by Comparing Its Reciprocal


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


Let $Z$ be a $6$-digit positive integer, such as $247247$, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$? (adapted from 2017 AMC 8 Problem, Question \#$7$)

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


Which of the following is neither a positive nor a negative number?

Overseas Competition->Knowledge Point->Calculation Modules->Knowing the Number Lines

$$\text{A}$$ is a positive number, $$\text{B}$$ and $$\text{D}$$ are two negative numbers, $$\text{C}$$ is zero, which is neither positive nor negative. We choose $$\text{C}$$.


None of the above

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


We can observe that when we remove the parentheses, all the $$x$$s are offset. $$5-4+3-2+1-0= 3$$.


Simplify the following expression: $$4a-3(a-b)$$．

$$a-b$$

$$a+b$$

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Using a Letter to Represent an Unknown Number


There were $$9$$ pieces of paper. Some of them were cut into three pieces. As a result, there are now $$15$$ pieces of paper now. How many pieces of paper were cut? ($$2005$$ Math Kangaroo Problem, Level $$3-4$$, Question \#$$17$$)

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

$15-9\times1=6$  $6\div(3-1)=3$ pieces of paper were cut.


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 $(\text{C}) 2$.


Values for $A, B, C$, and $D$ are to be selected from $\textbackslash{1,2,3,4,5,6\textbackslash}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=A x^{2}+B$ and $y=C x^{2}+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $ A=4, B=1, C=3, D=2 $.)

Setting $y=A x^{2}+B=C x^{2}+D$, we find that $A x^{2}-C x^{2}=x^{2}(A-C)=D-B$, so $x^{2}=\frac{D-B}{A-C} \geq 0$ by the trivial inequality. This implies that $D-B$ and $A-C$ must both be positive or negative. If two distinct values are chosen for $(A, C)$ and $(B, D)$ respectively, there are 2 ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by 2 at the end, however, since the 2 curves aren\textquotesingle t considered distinct. Calculating, we get $$ \frac{1}{2} \cdot\left(\begin{array}{l} 6 \textbackslash\textbackslash{} 2 \end{array}\right)\left(\begin{array}{l} 4 \textbackslash\textbackslash{} 2 \end{array}\right) \cdot 2=(\text { C) } 90 $$


For how many integers $x$ is the number $x^{4}-7x^{2}+12$ negative? (  2014 AMC 10B Problems, Question \#20)

Factor the polynomial as $\left(x^{2}-4\right)\left(x^{2}-3\right)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^{2}-4\textless0\textless x^{2}-3$. Solving this inequality, we find $3\textless x^{2}\textless4$. There is no integer $x$ that satisfies this inequality.


Joe writes an expression $\frac59\times\frac9{13}\times\frac{13}{17}\cdots $ Following the pattern, he writes middle fraction is \frac{45}{49}. What is the result of the expression?

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

The last fraction should be $\frac{85}{89}$, so the answer is $\frac5{89}.$


There were three color cups in the shop. At the beginning, there was only one cup of each kind. Later, there were five more blue cups, three more red cups, and one more yellow cup. How many cups were there in total at last?~(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

To calculate the total number of cups, do not need to consider the color of the cup. At the beginning, there were three cups. The number at last is $3+5+3+1=12$.


One apple, one banana and two peaches together weigh $12$ lbs. One apple and one peach together weigh $5$ lbs. One banana and $2$ peaches together weigh $5$ lbs more than one apple and one peach weigh together. Each peach weighs the same. How many pounds does one banana weigh?

$3$ lbs

$4$ lbs

$5$ lbs

$6$ lbs

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


Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{x+y}{x}=2$, what is the value of $\frac{x}{y}$?

Rearranging, we find $x+y=2x, -x=y, \frac xy = -1$.


Victor has $3$ siblings, and his mom is buying them chocolate cake for afternoon tea. His mom decides that all kids should receive the same amount of cake. What percent of the cake will Victor get?

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


Think Store is selling boba plushies at $100$ dollars each. The plushies are so in demand that Think Store decides to increase the price of the plushies by $30\textbackslash\%$, but if a students buy $2$ boba plushies at the same time, there is a $40\textbackslash\%$ discount for the second plushie. How much does it costs to buy $2$ plushies?


The sum of the repeating decimals~$0.\overline{163}$~and~$0.\overline{614}$~is ．

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


Four numbers are written in a row. The average of the first two is $21$ , the average of the middle two is $26$ , and the average of the last two is $30$ . What is the average of the first and last of the numbers? (2022 AMC 8 Problems, Question \#16)

Note that the sum of the first two numbers is $21 \cdot 2=42$, the sum of the middle two numbers is $26 \cdot 2=52$, and the sum of the last two numbers is $30 \cdot 2=60$.  It follows that the sum of the four numbers is $42+60=102$, so the sum of the first and last numbers is $102-52=50$. Therefore, the average of the first and last numbers is $50 \div 2=$ (B) $25$.


$(1+2+3+4)\times 6$

$1\times1+2\times2+3\times2+4\times3+6+8\times2+12\times2+16$

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


Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?  Meat: beef, chicken, pork  Vegetables: baked beans, corn, potatoes, tomatoes  Dessert: brownies, chocolate cake, chocolate pudding, ice cream~($2001$ AMC $8$ Problem, Question \#$14$)


If the value of $$x+2y$$ is $$3$$, then what is the result of $$\frac{1}{2}x+y-1$$?．


What is the sum of the digits in the number $$3567$$? (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


Which of the following is the largest fraction?

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

Same numerator, so smaller denominator means larger fraction.


If I multiply $$333333333333333$$ by $$777777777777777$$ and add the first and last digits of the product, the sum is．

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

Just as for $$3\times7$$, $$33\times77$$, and $$333\times777$$, the first digit is a $$2$$, the last digit is a $$1$$, and the sum is $$2+1=3$$.


If $$5$$ apples cost as much as $$16$$ pineapples, then $$15$$ apples cost as much as pineapples.

$$5$$ apples = $$16$$ pineapples  $$5\times3$$ apples = $15$ apples  $$16\times3$$ pineapples = $48$ pineapples


None of the above

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


The outside temperature in a town is $-18$ $^{}o$$C$. What change in the temperature, would bring the outside temperature to $0$ $^{}o$$C$?

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


Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{5 x+y}{x-4y}=-1$. What is the value of $\frac{x+3 y}{3 x-y}$? (Adapted From 2017 AMC 10B Problems, Question \#4)

Overseas Competition->Knowledge Point->Calculation Modules

Rearranging, we find $5 x+y=-x+4y$, or $6x=3 y \Longrightarrow y=2x$. Substituting, we can convert the second equation into $$\frac{x+6 x}{3 x-2x}=\frac{7 x}{x}=\text { (C) } 7$$.


Debbie spent $12$ dollars on a book and a volleyball. The volleyball costed $4$ dollars than the book. How many dollars did she spend on the volleyball?


Let $a, b$, and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$?

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

Expanding the equation and combining like terms results in $2 x^{2}-(a+2 b+c) x+(a b+b c)=0$. By Vieta\textquotesingle s formula the sum of the roots is $\frac{-[-(a+2 b+c)]}{2}=\frac{a+2 b+c}{2}$. To maximize this expression we want $b$ to be the largest, and from there we can assign the next highest values to $a$ and $c$. So let $b=9, a=8$, and $c=7$. Then the answer is $\frac{8+18+7}{2}= 16.5$.


For how many integers $x$ is the number $x^{4}-7x^{2}+10$ negative? (  2014 AMC 10B Problems, Question \#20)

Factor the polynomial as $\left(x^{2}-5\right)\left(x^{2}-2\right)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^{2}-5\textless0\textless x^{2}-2$. Solving this inequality, we find $2\textless x^{2}\textless5$. There are exactly $2$ integers $x$ that satisfy this inequality, $\pm 2$. Thus our answer is $(\mathbf{A}) 2$.


In how many different ways can you select two digits out of $$ 0$$, $$1$$, $$ 2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, and $$9$$ and put into the two boxes to make the below equality correct?  $$ 20- \square =22-\square $$


$$\left( {} \right.$$The number of seconds in a week$$\left. {} \right)$$$$\div$$$$\left( {} \right.$$the number of minutes in a week$$\left. {} \right)=$$．

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

There are $$60$$ seconds in each minute, so the quotient of the two quantities is $$60$$.


If I round $$1315$$ to the nearest ten, then multiply the result by $$3$$, then round to the nearest hundred, what is my final result?

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

If I round $$1315$$ to the nearest ten, I get $$1320$$. Then, I multiply by $$3$$ to get $$3960$$. To the nearest hundred, this is $$4000$$.


Denise fired a silver rocket and a gold rocket at the same time. The rockets exploded into $20$ stars in total. The gold rocket exploded into $6$ more stars than the silver one. How many stars did the gold rocket explode into? (2021 Math Kangaroo Problem, Level 3-4, Question \#7)

The silver rocket exploded $(20-6)\div2=7$ stars, and the gold rocket exploded $20-7=13$ stars.


What is the missing number in the box?  $2:3=\boxed{?}:12$

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


When the number $$789678567456$$ is added to the number $$987876765654$$, how many digits does the sum have?

The number $$789678567456$$ is added to the number $$987876765654$$.  Since we carry a $$1$$ when adding the left-most digits, the sum has $$12+1$$ digits.


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? (2023 AMC 8 Problems, Question \#20)

To double the range, we must find the current range, which is $28-3=25$, to then double to: $2(25)=50$. Since we do not want to change the median, we need to get a value less than $8$ (as $8$ would change the mode) for the smaller, making $53$ fixed for the larger. Remember, anything less than $3$ is not beneficial to the optimization. So, taking our optimal values of $7$ and $53$, we have an answer of $7+53=$ (D) $60$.


What is the $5^{th}$ number in the $8^{th}$ row?

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


In a chinese chess competition, children would be awarded $$2$$ marks for every game won, $$1$$ mark will be awarded for every game that ended in a draw and $$0$$ marks will be awarded for every game that ended in a loss. $6$ students participated in the round robin and $5$ of the students\textquotesingle{} scores are as follows: $$7$$, $$6$$, $$5$$, $$4$$, $$3$$. How many marks did the last place student score?

$6$ students will play a total of $5+4+3+2+1=15$ games.  Total score $=15\times2=30$  Last student\textquotesingle s score $=30-7-6-5-4-3=5$


Three (different) numbers are selected from $$0$$, $$1$$, $$3$$ and $$5$$ to form a three-digit number, find the difference between the smallest three-digit number and the largest three-digit number formed.


Which of the following numbers is the multiple of $5$?


Usually, people spend $\frac{1}{4}$ of their sleeping time dreaming. If Poole slept $8$ hours last night, how much time did he spend dreaming yesterday?

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


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


This regression uses the number of times a new worker has spent practicing task (measured in Number Practice~Rounds) to predict how much time is needed to complete one round of the task (measured in Minutes). What is the equation of the least-squares regression line?

$\overset{\frown}{Rounds}=-0.6442+22.94\textbackslash{} Minutes$

$\overset{\frown}{Rounds}=22.94+0.5466\textbackslash{} Minutes$

$\overset{\frown}{Minutes}=22.94+2.866\textbackslash{} Rounds$

$\overset{\frown}{Minutes}=22.94-0.6442\textbackslash{} Rounds$

$\overset{\frown}{Minutes}=-0.6442+0.5466\textbackslash{} Rounds$

From Chapter 6, the correct answer is (d). The slope of the regression line, -0.6442, can be found under "Coef" to the right of "Number of Rounds" .The intercept of the regression line, 22.94, can be found under "Coef" to the right of "Constant." Rounds is the explanatory variable (x) and Time is the response variable (y).


None of the above

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


\textbf{Research indicates that the standard deviation of typical human body temperature is 0.4 degree Celsius (C). which of the following represents the standard deviation of typical human body temperature in degrees Fahrenheit (F), where $$F=\frac{9}{5}C+32$$?}

\textbf{Standard deviation for Y=ax+b is SD(Y)=\textbar a\textbar SD(x).}


On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?


Which of the following fractions is the largest?

$\frac 79 \textgreater{} \frac 34 \textgreater{} \frac 8{11}\textgreater\frac 57\textgreater\frac 13$


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


Let $a, b$, and $c$ be positive integers with $a \geq b \geq c$ such that $a^{2}-b^{2}-c^{2}+a b=2011$ and $a^{2}+3 b^{2}+3 c^{2}-3 a b-2 a c-2 b c=-1997$. What is $a$ ?

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Addition and Subtraction of Equations

Add the two equations. $$ 2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 a c-2 b c=14 . $$ Now, this can be rearranged and factored. $$ \begin{aligned} \&\left(a^{2}-2 a b+b^{2}\right)+\left(a^{2}-2 a c+c^{2}\right)+\left(b^{2}-2 b c+c^{2}\right)=14 \textbackslash\textbackslash{} \&(a-b)^{2}+(a-c)^{2}+(b-c)^{2}=14 \end{aligned} $$ $a, b$, and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. We see that the only is possibility is $14=9+4+1$ $(a-c)^{2}=9 \Rightarrow a-c=3$, since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b=1$ or 2 , or whether $b-c=1$ or 2 . We want to solve for $a$, so take the two cases and solve them each for an expression in terms of $a$. Our two cases are $(a, b, c)=(a, a-1, a-3)$ or $(a, a-2, a-3)$. Plug these values into one of the original equations to see if we can get an integer for $a$. $a^{2}-(a-1)^{2}-(a-3)^{2}+a(a-1)=2011$, after some algebra, simplifies to $7 a=2021$. 2021 is not divisible by 7 , so $a$ is not an integer. The other case gives $a^{2}-(a-2)^{2}-(a-3)^{2}+a(a-2)=2011$, which simplifies to $8 a=2024$. Thus, $a=253$.


What is the ones digit of the result of $3^{50}$?

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

The ones digits of exponents based on $3$ follow the rule: $3, 9, 7, 1, 3, 9, 7, 1\cdots $  $50\div4=12R2$, which means the ones digit is $9$.


What is the result of $1\div 1\frac{2}{2001}\div 1\frac{2}{2003}\cdots \div 1\frac{2}{2023}$?


Which of the numbers below is greatest? ．

$2^{2222}\textgreater2^{2220}=\left (2^{10}\right )^{222}=1024^{222}\textgreater22^{222}$. According to this rule, $\text{E}\textgreater\text{D}\textgreater\text{C}\textgreater\text{B}\textgreater\text{A}$.


In an ice-cream shop there was some money in a drawer. After selling $6$ ice-cream cones, there are $70$ dollars in the drawer. After selling a total of $16$ ice-cream cones, there are $120$ dollars in the drawer. How many dollars were there in the drawer at the start? (2021 Math Kangaroo Problem, Level 3-4, Question \#13)

The price of one ice-cream cone is $(120-70)\div(16-6)=5$ dollars. The price of $6$ ice cream cones is $5\times6=30$ dollars. There were $70-30=40$ dollars at the start.


None of the above


Susan participates a Math competition. In round one, she needs to solve $2$ problems. If she works out one problem successfully, she will get $1$ point. Otherwise, she will not lose any points. In round two, she also needs to solve $2$ problems. If she works out one problem successfully, she will get $3$ points. Otherwise, she will lose $1$ point. Which number of points is it impossible for her to have？

After round one, she may have $0$ point, $1$ point, $2$points  After two rounds, she may have $3$, $4$, $6$, $7$, $8$ points


None of the above

The numbers in the odd positions (1st, 3rd, 5th ,7th) are 3, 6, 9, 12. Each of these numbers is 3 more than the number before it. The numbers in the even positions (2nd, 4th, 6th, 8th) are 5, 10, 15, 20. Each of these numbers is 5 more than the number before it. Since the missing number is the gth position (odd), then the missing number is 12 + 3 = 15.


If a number is written as $2a+4$, then $3$ times the number is~\uline{~~~~~~~~~~}~. (adapted from 1977 Math League.com contest problem, 8\textsuperscript{th}~Grade, Question \#31)

Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Using a Letter to Represent an Unknown Number


What is the sum of the various numbers of Number $$1074$$? (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

Pay attention to the review questions and sum the numbers.


What value of $a$ would make the solution to the equation, $$ 2 a+3-4 x+7=3\left(-\frac{4}{3} x+7\right) $$,~"infinitely many solutions"?~\uline{~~~~~~~~~~}~

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


6, 36

4, 35

5, 36

8, 35

7, 39


Given that $x$ and $y$ are whole numbers such that $24x-25y =8$, find the smallest value of $x$.

None of the above

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


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


Let $f(x)=a x^{2}+b x+c$, where $a$, $b$, and $c$ are integers. Suppose that $f(1)=0,20\textless f(3)\textless24,36\textless f(4)\textless40$, $10 k\textless f(10)\textless10(k+1)$ for some integer $k$. What is $k$? (Adapted From2011 AMC 12A Problems, Question 20)

Overseas Competition->Knowledge Point->Calculation Modules

From $f(1)=0$, we know that $a+b+c=0$.  From the first inequality, we get $20\textless9 a+3 b+c\textless24$.  Subtracting $a+b+c=0$ from this gives us $20\textless8 a+2 b\textless24$, and thus $10\textless4 a+b\textless12$. Since $4a+b$ must be an integer, it follows that $4 a+b=11$.  Similarly, from the second inequality, we get $36\textless16a+4 b+c\textless40$. Again subtracting $a+b+c=0$ from this gives us $36\textless15 a+3 b\textless40$, or $12\textless5 a+b\textless\frac{40}{3}$.  It follows from this that $5 a+b=13$. We now have a system of three equations: $a+b+c=0,4a+b=11$, and $5a+b=13$. Solving gives us $(a, b, c)=(2,3,-5)$ and from this we find that $f(10)=2(10)^{2}+3(10)-5=225$. We find that $k=22 \rightarrow(\mathbf{B})22$.


The sum of the smallest three-digit numbers whose digits add up to $$8$$ and the largest three-digit numbers whose digits add up to $$8$$ is equal to~\uline{~~~~~~~~~~}~. ($$2011$$ Math kangaroo Problems, Level $$7-8$$, Question \#$$10$$)

The smallest three-digit number is $107$ and the largest one is $800$. $107+800=907$


How many whole numbers less than $$1000$$ can be written as the product of $$3$$ consecutive whole numbers?

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

The whole numbers less than $$1000$$ that can be written as such a product are $$0\times1\times2$$, $$1\times2\times3$$, $$2\times3\times4$$, $$3\times4\times5$$, $$4\times5\times6$$, $$5\times6\times7$$, $$6\times7\times8$$, $$7\times8\times9$$, $$8\times9\times10$$, and $$9\times10\times11$$. In all, that\textquotesingle s $$10$$.


Let $f$ be a linear function for which $f(5)-f(2)=0$. What is $f(8)-f(2)$? (  Adapted From 2003 AMC 12B Problems, Question \#9)

$f(5)-f(2)=0 \Rightarrow f(5)= f(2)$; thus, $f(x)$ is a constant function. Then, $f(8) - f(2) = 0$.


Daniel had a package of $$36$$ pieces of candy. Without breaking any pieces of candy, he divided all the candy equally among his friends without remaining. Which of the following was definitely not the number of his friends?

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


Evaluate$$\left\textbar{} 3-9 \right\textbar+\left\textbar{} 7-2 \right\textbar$$.

Overseas Competition->Knowledge Point->Combinatorics->Mathematical Thoughts->Absolute Value

$$\left\textbar{} 3-9 \right\textbar+\left\textbar{} 7-2 \right\textbar=\left\textbar{} -6 \right\textbar+\left\textbar{} 5 \right\textbar=6+5=11$$.


When William, Mark, Diana, and Jimmy checked their book bags, they found that Mark\textquotesingle s books were fewer than Diana\textquotesingle s and Jimmy\textquotesingle s were more than William\textquotesingle s. Mark\textquotesingle s is not the least. Do you know who has the least books? (adapted from 2009 Math Kangaroo Problems, Level 1-2, Question \#21)

William

Mark

Diana

Jimmy

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

Diana\textgreater Mark, so it is not Diana.  William \textless{} Jimmy, so it is not Jimmy.  We already know that Mark did not have the smallest number of books.  So, it is William.


How many days are there in August?~(adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$5$$)

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


If the day before yesterday was Sunday. How many days are there from today until Sunday?

Before yesterday: Sunday  Yesterday: Monday  Today: Tuesday  From Tuesday to Sunday, we have $$5$$ days based on the information given in the question.


There are $49$ matchsticks on the table. Kevin and Michael will take turns to take away the matchsticks from the first matchstick in order. Each person can take $1$ to $6$ matchsticks at a time. The person who takes the last matchstick on the table will win the game. If Kevin plays the game first, does he have the winning strategy?

Yes, he has the winning strategy.

No, he does not have the winning strategy.

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

$49\textbackslash{} \div(1+6)=7$~groups, so the second mover will have the winning strategy.


SASMO 2015 P2 Q1  How many months of the year have 28 days?

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


While fishing, Pablo caught as many fish as his son Marco. Juan caught three times as much fish as his son. Altogether, they caught $$35$$ fish. What\textquotesingle s the name of Juan\textquotesingle s son?

The described situation is impossible

$$$$Juan

$$$$Pablo

$$$$Marco

It cannot be determined from the information given

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


Shawn is making a meal. It will cost $1$ minute for preparing the vegetables, $2$ minutes for washing the pan for oven, $6$ minutes for the oven roasting the meal, $2$ minutes for cleaning the table, and $1$ minute for putting the meal on the table. At least how long will it take for Shawn to finish the meal?

$$7$$ minutes

$$8$$ minutes

$$9$$ minutes

$$10$$ minutes

$$12$$ minutes

Shawn can clean the table while the oven is working, so it will take $6$ minutes in total. So, he can finish the meal in $1+2+6+1=10$ minutes.


\textbf{In which place did each of the following shops come in the competition? Write the correct letter on the Prize list. What is the ranking of the pizza shop?}


Eve arranged cards in a line as it is shown in the figure below. At each move Eve is allowed to interchange any two cards. What is the smallest number of moves Eve needs to get the word KANGAROO?  insert pic

First, switch K and 1st O to make KANGONOA.  Second, switch N and R to make KANGOROA.  Third, switch the last A and the 2nd O to make KANGAROO.


How many months of the year have exactly $$31$$ days?

March, May, July, August, October and December.


A certain play has three $30$-minute parts and two intermission among them. The play started at $8:30$ and ended at $10:15$. How many minutes long were the intermissions in total? (Adapted from 2010 Math Kangaroo Problem, Level 1-2, Question \#19)


There were $$8$$ cups on the table with the sides up,  If you can only turn $$6$$ at a time,  Can it be done to make all the cups face down at the end?

Yes, it can

No, it can\textquotesingle t

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

To make all the $$8$$ cups face down, it only needs to turn them over with odd times, and 6 of them can be turned at a time, so it can be done.


There are two groups of flowers. Group $A$ has $17$ flowers and Group $B$ has $13$ flowers. Cathy and Ivy want to play a game with these flowers. They will take turns to take flowers from the two groups. Each person can take any number of flowers from a group at a time. The person who takes the last flower in two groups will win the game. If Ivy starts the game first, who has the winning strategy?

Ivy has the winning strategy.

Cathy has the winning strategy.

Two groups of flowers are not equal. Thus, Ivy needs to take away 4 flowers from Group A and two groups will have the same amount of flowers. Then, Ivy will become the second mover and she has the winning strategy.


A certain play has three $30$-minute parts and two intermissions among them. The play started at $8:30$ AM and ended at $10:15$ AM. How many minutes long were the intermissions in total?


Find the sum of all possible values of $x$ such that $\textbar x-\textbar{} x-1\textbar\textbar-x=1$ .~\uline{~~~~~~~~~~}~

We must check cases where $\textbar x-\textbar{} x-1\textbar\textbar$ is positive and cases where it is negative. $x\textgreater1$ and $x\textless1$ will cause different signs for $x-\textbar x-1\textbar$. Hence we must check both intervals. However, when $x\textless1$, we see that when $x=\frac{1}{2}$, the expression is equal to 0 , so we must also check the intervals $x\textless\frac{1}{2}$ and $\frac{1}{2}\textless x\textless1$. Solving on the interval $x\textless\frac{1}{2}$, we get $\textbar x+1+x\textbar-x=1 \Longleftrightarrow x=0$. Solving on the interval $\frac{1}{2}\textless x\textless1$, we get $\textbar x+1+x\textbar-x=1 \Longleftrightarrow x=0$. Solving on the interval $x\textgreater1$, we get $\textbar x-x+1\textbar-x=1 \Longleftrightarrow x=0$. Checking for extraneous solutions, we find that $x=0$


In the calculation shown below, different letters represent different digits.  $AA\times AB\times C=ADDA$  What is the sum of $A+B+C+D$?

Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles


Find the next number in the sequence below.  $$1,4,10,22,46,94,\cdots $$

None of the above

Overseas Competition->Knowledge Point->Combinatorics->Patterns of Figures

The pattern is as follows:  $1$ $\xrightarrow{+3}$ $4$ $\xrightarrow{+6}$ $10$ $\xrightarrow{+12}$ $22$ $\xrightarrow{+24}$ $46$ $\xrightarrow{+48}$ $94$ $\xrightarrow{+96}$ $190$  The differences start with 3 and double each time afterwards. The next number in the sequence is \textbf{190.}


What is the missing number in the sequence below?  $1, 3, 7, 15, 31,$~\uline{~~~~~~~~~~}~

None of the above


Rose and Kylie are playing a game. Here are the rules：  1. There are $$16$$ marbles placed in a row.  2. The players take turns removing either $1$ or $2$ marbles each turn.  3. Whoever picks the last marble wins the game.  Rose starts first and is followed by Kylie. To ensure her victory, how many marbles must Rose take away in the first turn?

If there are n marbles in total, the first player has a winning strategy for all $$n$$ that is not a multiple of $$3$$. For $$n$$ being a multiple of $$3$$, the second player can always win, regardless of what strategy the first player plays.  $$16\div (1+2)=5 \textbackslash{} \text{R} 1$$ Rose should take away the remainder, i.e. one marble, to make herself the second player when there is a multiple of $3$ marbles left.


Chris, Molly, Sara, and Lynn each has some coins. Chris gives Molly $20$ coins, Molly gives Sara $25$ coins, Sara gives Lynn $30$ coins, and Lynn gives Chris $30$ coins. Now, which of the following is correct if they each wants to have the same number of coins as in the beginning?

Lynn gives Sara $5$ coins, and Molly gives Sara $5$ coins.

Chris gives Molly $5$ coins, and Chris gives Lynn $10$ coins.

Chris gives Molly $5$ coins, and Chris gives Sara $5$ coins.

Chris gives Lynn $10$ coins.

Chris gives Sara $10$ coins.

Chris gives Molly $20$ coins, and Lynn gives Chris $30$ coins. Now, Chris has $10$ more.  Chris gives Molly $20$ coins, and Molly gives Sara $25$ coins. Now, Molly has $5$ less.  Molly gives Sara $25$ coins, and Sara gives Lynn $30$ coins. Now, Sara has $5$ less.  Sara gives Lynn $30$ coins, and Lynn gives Chris $30$ coins. Now, Lynn has the same as beginning.  Thus, Chris can give Molly $5$ coins, and give Sara $5$ coins.


In the following column puzzle, different letters represent different one-digit numbers. Find the three-digit number represented by $$ABC$$.  $$\begin{matrix}\& \& 8 \& 8\& \boxed C \textbackslash\textbackslash{} \&\&5\& \boxed B \&4 \textbackslash\textbackslash{} + \&\&\boxed A\&1 \&7 \textbackslash\textbackslash{} \hline \&2~\&0 \&2 \&1\textbackslash\textbackslash{} \end{matrix}$$


Tom wrote various words in code in such a way that different digits represent different letters and the same digit represents the same letter. For example, the word $BALL$ was coded as $$3488$$. One of the words below was coded as $$6155491$$. Which one?

$$SURGEON$$

$$HARBORS$$

$SWEATER$

$MESSAGE$

According to the pattern: the same digit represents the same letter, the code $$6155491$$ should represent a word with two same letters at the third and fourth place.


The area of a square is $16$, its side length is~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Combinatorics->Magic Square


2 foxes and 2 rabbits want to cross a river. There is a boat that can hold 2 animals. When the number of fox is less than the number of rabbits, the rabbits fight with the foxes. For 2 foxes and 2 rabbits to cross the river safely, at least how many times does the boat need to cross the river? (count the round trip as twice and every trip need one animal row the boat)~\uline{~~~~~~~~~~}~．

Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math->Fun Math Problems

Assume the foxes as A, and the rabbits as B. The way they cross the river as following:  $$AA\xrightarrow{BB}$$  $$AA\xleftarrow{B}B$$  $$B\xrightarrow{AA}B$$  $$B\xleftarrow{B}AA$$  $$\xrightarrow{BB}AA$$


There are two groups of balls. Group $A$ has $585$ balls and Group $B$ has $590$ balls. DQ and Justin want to play a game with these balls. They will take turns to take balls from the two groups. Each person can take $1$ to $6$ balls from a group at a time. The person who takes the last ball in the two groups will win the game. If DQ starts the game first, does she have the winning strategy?

Yes, she can take $5$ balls from group $A$ at first.

Yes, she can take $3$ balls from group $B$ at first.

Yes, she can take $6$ balls from group $A$ at first.

Yes, she can take $5$ balls from group $B$ at first.

No, she doesn\textquotesingle t have winning strategy.

DQ needs to take the $590-585=5$ balls in Group $B$ which will make the two groups have the same amount of balls. Then, no matter how many balls Justin takes in a group, DQ will take as many balls as Justin took before in the other group.


Jack and Edward want to take turns to count off numbers from $1$ to $124$ in order. Each person can count $1$ to $5$ numbers at a time. The person who counts off number $124$ will win the game. Does Jack have the winning strategy if he counts off numbers first?

Yes, he has.

No, he doesn\textquotesingle t.

$124\div(1+5)=20R4$. So, Jack should count off $1$ to $4$ and there will be $120$ numbers left. Jack will become the second mover and he has the winning strategy.


The ancient Romans used Roman numerals. We still use them today. Here are some examples: $$\rm I=1$$, $$\rm II=2$$, $$\rm V=5$$, $$\rm IX=9$$, $$\rm X=10$$, $$\rm XI=11$$, $$\rm XX=20$$. This year($2022$) we celebrate Math Kangaroo number $$\rm XX$$. What year was Math Kangaroo number $$\rm XV$$?

According to the pattern, $$\rm XX$$ is $$20$$, and $$\rm XV$$ is $$15$$.  So, Math Kangaroo number $$15$$ was $$5$$ less than real year, which was $$2017$$.


If May $6$ falls on a Friday, what day of the week will it be in $25$ days? (Adapted from 2015 Math Kangaroo Problem, Level 3-4, Question \#8)

Tuesday

Wednesday

Thursday

Friday

Sunday


If January $2$ falls on a Saturday, what day of the week will it be in $28$ days?

Tuesday

Wednesday

Thursday

Saturday

Sunday


James starts classes at $9$ A.M. He has $2$ classes, each class $$45$$ minutes long. After each class, he will take a $$15$$-minute break. When will he finish all $$2$$ classes?

$10:20$ A.M.

$11:00$ A.M.

$11:10$ A.M.

$11:20$ A.M.

$11:30$ A.M.


There are $13$ ping-pong players who will be divided into three teams, and they are going to have a men\textquotesingle s singles. The rule is that the players in the same team will not play with each other, and each of them only plays one game with each player in other teams. What is the maximum number of games happened in total?

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

Dividing the players into three teams as equally as possible can make the maximum number of games. So there are $4$, $4$, and $5$ players in the three teams and they will have $4\times4+4\times5+4\times5=56$ games in total.


We left for a summer camp at $4:30$ PM and got to our destination at $6:40$ PM. How long did we travel?~

$1$ hour $40$ minutes

$2$ hours $10$ minutes

$2$ hours $20$ minutes

$1$ hour $50$ minutes

$50$ minutes


There are $5$ boxes on the table. From left to right, each of them has $8,$ $4,$ $2,$ $1,$ and $5$ balls of the same size, respectively. Every time, Judy can take one ball each from the other four boxes, and then put them into the box with the smallest number of balls. She follows the rules and operates $2023$ times. Now how many balls are there in the first box counting from the left?

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

Without operation: $$8$$, $$4$$, $$2$$, $$1$$, $$5$$  After the first operation: $$7$$, $$3$$, $$1$$, $$5$$, $$4$$  After the second operation: $$6$$, $$2$$, $$5$$, $$4$$, $$3$$  After the third operation: $$5$$, $$6$$, $$4$$, $$3$$, $$2$$  After the fourth operation: $$4$$, $$5$$, $$3$$, $$2$$, $$6$$  After the fifth operation: $$3$$, $$4$$, $$2$$, $$6$$, $$5$$  After the sixth operation: $$2$$, $$3$$, $$6$$, $$5$$, $$4$$  After the seventh operation: $$6$$, $$2$$, $$5$$, $$4$$, $$3$$  We can find that $6-2-5-4-3$ is repeating starting from the second operation.  $$（2023-1）\div5 R 2$$  Thus, after the $2023$\textsuperscript{rd}~operation, the result will be the same as the third one, which means there are $5$ balls in the first box.


Alvin always tells the truth on Sundays, Tuesdays and Thursdays. He lies on every other days.  One day he said, "I told the truth yesterday."  On which day of the week did he make this statement?

Sunday

Wednesday

Thursday

Friday

Saturday

If "I told the truth yesterday" is a true statement -\/-\textgreater{} mean today I am telling the truth. But, the day that Alvin tell the truth are not consecutive day.  So, if "I told the truth yesterday" is not a true statement -\/-\textgreater{} mean yesterday I lie and today I am also lying.  Alvin lies on Monday, Wednesday, Friday and Saturday. Friday and Saturday is consecutive day he lies, so it must be Saturday.


There are two groups of balls. Group $A$ has $50$ balls and Group $B$ has $40$ balls. Candy and Nini want to play a game with these balls. They will take turns to take balls from the two groups. Each person can take any number of balls from a group at a time, but they should take at least one ball at a time. The person who takes the last ball in the two groups will win the game. If Nini starts the game first, how many balls should she take to guarantee her success? ．

Nini needs to take $$50-40=10$$ balls in Group A which will make the two groups have the same amount of balls. Then, no matter how many balls Candy takes in a group, Nini will take as many balls as Candy took before in the other group.


There are $7$ water glasses, all facing up. You need to turn exactly $2$ glasses over in each time. Is it possible to turn all $7$ of them upside down after several moves?

Yes, it is possible.

No, it is impossible.

I have no idea.

You have to flip a cup an odd number of times to turn it over.  To make all the $7$ cups face down, you have to make an odd number of flips.  Each time you flip exactly $2$ cups, the total number of flips is an even number of times.  Hence it is impossible.


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 difference equal to $$3$$? (adapted from $$2005$$ Math Kangaroo Problem, Level $$9-10$$, Question \#$$15$$)

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


Peter\textquotesingle s father has $$4$$ sons and Peter has $$5$$ brother(s) and sister(s). How many daughter(s) does Peter\textquotesingle s father have?


John and Olivia exchanged sweets. First John gave Olivia as many sweets as Olivia had. Then Olivia gave John as many sweets as John had after the first exchange. After these two exchanges, each had $$4$$ sweets. How many sweets did John have at the beginning?

Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Planning->Simple Time Planning Problems->Working Simultaneously


SASMO 2016 P2 Q2  How many hours are there in two weeks?

7 x 12

7 x 2 x 12

2 x 7 x 2 x 12

(7+7) x 12

None of the above


Alice began reading a 400-page book at 8 am and had read 50 pages by 9:30. If she continues to read at the same rate, when can she finish reading this book?

6 pm

7 pm

8 pm

9 pm

10 pm


Lucy, Maria, and Anna have a meeting at $$12:30$$. Lucy\textquotesingle s walk takes $$10$$ minutes, Maria\textquotesingle s walk takes a quarter of an hour, and Anna\textquotesingle s walk takes $$40$$ minutes. At what time must the person who needs the longest time to get to the meeting leave her house? (2006 Math Kangaroo Problem, Level 1-2, Question \#9)

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


A certain play has three $30$-minute parts and two intermissions among them. The play started at $8:30$ AM and ended at $10:15$ AM. How long did the commercials last for in total?

$$10 min$$

$$15 min$$

$$20 min$$

$$25 min$$


There are two containers, one with a capacity of 6 liters and the other with a capacity of 5 liters, and using them to get 1 liter of water from a bucket requires at leasttimes of operations.

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

Fill the 6-liter container first,  and then pour the water from the 6-liter container into the 5-liter container.  After the 5-liter container is filled, there is 1 liter of water left in the 6-liter container.  It takes two operations.  So the answer is $$A$$


A store opens at 9 a.m. and closes at 5p.m. each day. It is closed from 11.55 a.m. to 12.55 p.m. for a lunch break. How many hours does the store stay open each day?

$8$ hours and $5$ minutes

$$7$$ hours and $30$ minutes

$8$ hours

$7$ hours

$6$ hours


There are two piles of matches. Each pile has $$6$$ matches. Cindy and Doris take turns to pick up matches from either pile. There is no limit to how many matches they can pick up, but they must pick up at least one match each turn. The person who picks up the last match will be the winner. If Cindy starts picking up matches first,~\uline{~~~~~~~~~~}~will definitely be the winner (has a winning strategy).

Cindy

Doris

Doris will be the winner, since she can simply mirror the number of matches Cindy picks up every turn.


Taylor has $2$ brothers and $3$ sisters. How many brothers and sisters does his sister Lucy have?

3 brothers and 4 sisters

2 brothers and 3 sisters

3 brothers and 3 sisters

2 brothers and 4 sisters

3 brothers and 2 sisters


Among $42$ people, at least how many people were born in the same month with the most births?

Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle


Teacher wrote $$10$$ non-zero natural numbers in order on the blackboard, where the $$1$$st number is $$16$$, and the sum of any $$3$$ adjacent numbers is $$100$$. The $$8$$th biggest number is~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math


There are $$25$$ matches on the table. John and James take turns to remove $$1$$ to $$3$$ matches each time. The person who removes the last match will be the winner. If both of them were to use the best method and John removes first, then~\uline{~~~~~~~~~~}~will win.~

John

James

$$25\div 4=6\ldots 1$$ John removes $$1$$ match and $$24$$ is a multiple of $$4$$. So, the first player will win the game.


Light Bulb A lights up every $$15$$ minutes while Light Bulb B lights up every $$20$$ minutes. Both light bulbs lit up at the same time at $$8.30$$ a.m.  By noon, how many more times would both light bulbs have lit up at the same time?

The LCM of $$15$$ and $$20$$ is $$60$$.  Both light bulbs will light up at the same time every $$60$$ minutes, at $$9.30$$ a.m., $$10.30$$ a.m. and $$11.30$$ a.m. (total of $$3$$ times).


Adam built fewer sandcastles than Martin but more than Susan. Lucy built more sandcastles than Adam and more than Martin. Dana built more sancastles than Martin but fewer than Lucy.  Who of them built the most sandcastles?

Martin

Adam

Susan

Dana

Lucy

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

L \textgreater{} D\textgreater{} M \textgreater{} A \textgreater{} S


The Gellers went on a trip. They left home at 9:15 in the morning and arrived at the hotel at 8:22 in the evening. How long did the journey take?

$$11$$ hours $$22$$ minutes

$$11$$ hours $$15$$ minutes

$$11$$ hours $$7$$ minutes

$$1$$ hours $$7$$ minutes

$$12$$ hours $$7$$ minutes


The first student wrote the number $1$ on the board, the second student wrote the number $2$, and the third one and each of the following students wrote a number that was the quotient of the number written just before the last number and the last number. What did the tenth student write?


In Think Academy, there are $147$ students. Among these $147$ students, we can guarantee that at least~\uline{~~~~~~~~~~}~students belong to the zodiac sign with the most students.


From $1$ to $23$, how many integers can be chosen at most to ensure that no two of chosen numbers differ by $4$?

Overseas Competition->Knowledge Point->Combinatorics->Constructing and Proving


Rose and Kylie are playing a game. Here are the rules：  1. There are $$14$$ marbles placed in a row.  2. The players take turns removing either $1$ or $2$ marbles each turn.  3. Whoever picks the last marble wins the game.  Rose starts first and is followed by Kylie. To ensure her victory, how many marbles must Rose take away in the first turn?

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

If there are n marbles in total, the first player has a winning strategy for all $$n$$ that is not a multiple of $$3$$. For $$n$$ being a multiple of $$3$$, the second player can always win, regardless of what strategy the first player plays.  $$14\div (1+2)=4 \textbackslash{} \text{R} 2$$ Rose should take away the remainder, i.e. two marbles, to make herself the second player when there is a multiple of $3$ marbles left.


The time shown on the electronic clock is $6:08$. How long before the numbers appear again? (adapted from 2007 Math Kangaroo Problem, Level 3 - 4, Question \#16)

$$2$$ h $$2$$ min

$$3$$ h

$$3$$ h $$2$$ min

$$2$$ h $$12$$ min

$$2$$ h


We are given a certain number. We double the number and then subtract $1$ from the result. We repeat this operation $4$ more times. If the end result is $2^{6}+1$, then what is the number that we started with? (Adapted from $2005$ Math Kangaroo Problems, Level $11-12$, Question \#$29$)

After repeating the operation $4+1=5$ times, the result is $2^{6}+1$.  After repeating the operation $4$ times, the result is $(2^{6}+1+1)\div2=2^{5}+1$.  After repeating the operation $3$ times, the result is $2^{4}+1$.  $$\cdots $$  After doing the operation for the first time, the result is $2^{2}+1$.  Before we do anything, the number is $2^{1}+1$, which is $3$.


When number $\overline{OK}$ is divided by $K$, the result will give the same quotient as the divisor and leave the remainder of $O$. The same letter represents the same digit, and different letters represent different digits. Then, what is the product of the two digits, $O$ and $K$?


$$7\times 2\times 12$$

$$7\times 12 \times 60$$

$$7\times 2 \times 12 \times 60$$

None of the above

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

One week $$7$$ days.  $$24$$ hours per day.  $$60$$ minutes per hour.


We left for a summer camp at $4:32$ PM and got to our destination at $6:11$ PM. How long did we travel? (Adapted from 2015 Math Kangaroo Problem, Level 1--2, Question \#24)

$1$ hour $39$ minutes

$2$ hours $39$ minutes

$2$ hours $21$ minutes

$1$ hour $21$ minutes

$39$ minutes


Apply this operation to $$(50, 20)$$. What are the last two numbers when the operation stops?

None of the above

Overseas Competition->Knowledge Point->Combinatorics->Patterns of Figures->Special Changes


Amy calculate $47\times86$ wrong. Her teacher says her result which has a difference of $172$ from the correct one. After checking, Amy finds that she writes one of the four digits two multipliers the result as $9$. Which digits does she write incorrectly?

It\textquotesingle s impossible to determine.

Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles

The difference between the wrong answer and the correct one is $172$. $172\div86=2$, $172\div47=3R31$.That means Amy writes neither $ 8$ nor $6$ wrong. Then, compare $49\times86$ and $97\times86$, so we can get the correct answer. $49\times86-47\times86=172.$


Lucy, Maria, and Anna have a meeting at $$12:30$$. Lucy\textquotesingle s walk takes $$10$$ minutes, Maria\textquotesingle s walk takes a quarter of an hour, and Anna\textquotesingle s walk takes $$30$$ minutes. At what time must the person who needs the longest time to get to the meeting leave her house? (Adapted from 2006 Math Kangaroo Problem, Level 1-2, Question \#9)


From Monday to Wednesday Mark always lies. For the rest of the week he tells the truth. One day, Mark said to Mary:  $$1$$) "Yesterday I lied." and  $$2$$) "Starting the day after tomorrow, I will be lying for two consecutive days."  On what day of the week did Mark talk to Mary?

Monday

Tuesday

Wednesday

Thurday

Friday

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


Which of the following month has $30$ days? (adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$5$$)

$February$

$$June$$

$$July$$

$$October$$

$$December$$

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


A certain play has two $$45$$-minute parts and an intermission between them. The play started at $$10:50$$ and ended at $$12:40$$. How many minutes long was the intermission?


The distance between the school and the park is $$2400$$ $\text{m}$. Lily and Candy depart from the school towards the park at the same time, and return immediately after arriving at the park. Lily travels $$47$$ $\text{m/min}$, and Candy travels $$53$$ $\text{m/min}$. They will meet each other for the first time after~\uline{~~~~~~~~~~}~minutes.


If $a,b,c,d,e,f,g,h,j$ represents different digits from $1$ to $9$ such that $a+\frac{b}{c}+\frac{d}{e}\times f-\left(g+\frac{h}{j} \right) = N$, where $N$ is a whole number. What is the value of $\frac{d}{e}\times f$ to maximize $N$?

Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Forming a Maximum/Minimum Multi-Digit Numbers with Fixed Sums


True or False:  A number can\textquotesingle t be equal to its opposite.

True

False

Overseas Competition->Knowledge Point->Combinatorics->Mathematical Thoughts->Absolute Value

$0$\textquotesingle s opposite number is $0$


There are two groups of flowers. Group $A$ has $15$ flowers and Group $B$ has $13$ flowers. Cathy and Ivy want to play a game with these flowers. They will take turns to take flowers from the two groups. Each person can take any number of flowers from a group at a time. The person who takes the last flower in two groups will win the game. If Ivy starts the game first, who has the winning strategy?

Ivy has the winning strategy.

Cathy has the winning strategy.

Two groups of flowers are not equal. Thus, Ivy needs to take away 2 flowers from Group A and two groups will have the same amount of flowers. Then, Ivy will become the second mover and she has the winning strategy.


Bella got up at 8:10. It took 10 minutes to wash and 30 minutes to eat breakfast. At this moment, she began to read. After reading, she went out. She went out at 9:45. How long did she read?

$$10$$ minutes

$$15$$ minutes

$$20$$ minutes

$$25$$ minutes

$$30$$ minutes


London time is 7 hours behind of Hong Kong time. Mike is planning to have a flight to Hong Kong from London at $$1$$am of $$12$$th June. If the flight takes 13 hours. When will Mike arrive Hong Kong?(In Hong Kong time)

$$1$$am $$12$$th June

$$9$$pm $$12$$th June

$$8$$am $$12$$th June

$$7$$am $$12$$th June

Because the flight takes 13 hours, Mike will arrive Hong Kong at $$2$$pm of $$12$$th June(London time). As london time is 7 hours behind Hong Kong time, it will be~$2+7=9$pm(Hong Kong time)


Grace was going to meet her friends at the bus station. Right now, it is $8:40$. Grace arrived at the bus station half an hour ago. The trip from her home to the station was $1$ hour long. What time did Grace depart toward the bus station?

Grace arrived at the bus station at $8:10$, and the trip was $1$ hour long, so she departed at $7:10$.


In the calculation shown below, different letters represent different digits.  $AA\times AB=5467$  What is the sum of $A+B$?


Which of the following number is the opposite number of $-2022$

Opposite number is the number with opposite sign


In a speed skating competition 10 racers reached in final. Tom overtook 3 racers more than those who overtook him. Which place did Tom end up in?


A certain play has three $30$-minute parts and two intermissions among them. The play started at $8:30$ AM and ended at $10:15$ AM. How long did the intermissions last for in total?

$$10 min$$

$$15 min$$

$$20 min$$

$$25 min$$


Calculate the following.  $(220+130-310) \div (99-89) \times (2 \times 3) =$


When calculating $54\times 96$, Judy fails to write the correct column multiplication. She writes one of the four digits as $7$ and gets a result which has a difference of $1920$ from the correct one. Which digit does she write incorrectly?

The difference between the wrong answer and the correct one is $1920$, whose ones digit is $0$. That means Judy writes neither $4$ nor $6$ wrong. Then, compare $20\times54$ and $20\times96$, so we can get the correct answer. $74\times96-54\times96=1920.$


When calculating $63\times72$ , Judy fails to write the correct column multiplication. She writes one of the four digits as 9 and gets a result which has a difference of 432 from the correct one. Which digit does she write incorrectly?

The difference between the wrong answer and the correct one is $432$, whose ones digit is $2$. That means Judy writes neither $2$ nor $3$ wrong.  Then, compare $7\times63$ and $6\times72$, so we can get the correct answer.  $69\times72-63\times72=432.$


Four friends are competing in a math competition. They are Andy, Bob, Cindy and Daisy.  The organiser of the competition told you that:  ($$1$$) Andy performed better than Bob.  ($$2$$) Cindy\textquotesingle s score is higher than Andy\textquotesingle s score.  ($$3$$) Daisy\textquotesingle s score is lower than two of her friends.  Arrange their scores in descending order.

Andy, Bob, Cindy, Daisy

Andy, Cindy, Daisy, Bob

Cindy, Andy, Daisy, Bob

Cindy, Andy, Bob, Daisy

None of the above

The score from highest to lower:  First -\/-\textgreater{} Cindy  Second -\/-\textgreater{} Andy  Thrid -\/-\textgreater{} Daisy  Fourth -\/-\textgreater{} Bob


There are $5$ boxes on the table. From left to right, each of them has $8,$ $4,$ $2,$ $1,$ and $5$ balls of the same size, respectively. Every time, Judy can take one ball each from the other four boxes, and then put them into the box with the smallest number of balls. She follows the rules and operates $2023$ times. Now how many balls are there in the first box counting from the left?

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

Without operation: $$8$$, $$4$$, $$2$$, $$1$$, $$5$$  After the first operation: $$7$$, $$3$$, $$1$$, $$5$$, $$4$$  After the second operation: $$6$$, $$2$$, $$5$$, $$4$$, $$3$$  After the third operation: $$5$$, $$6$$, $$4$$, $$3$$, $$2$$  After the fourth operation: $$4$$, $$5$$, $$3$$, $$2$$, $$6$$  After the fifth operation: $$3$$, $$4$$, $$2$$, $$6$$, $$5$$  After the sixth operation: $$2$$, $$3$$, $$6$$, $$5$$, $$4$$  After the seventh operation: $$6$$, $$2$$, $$5$$, $$4$$, $$3$$  We can find that $6-2-5-4-3$ is repeating starting from the second operation.  $$（2023-1）\div5$$R$$2$$  Thus, after the $2023$\textsuperscript{rd}~operation, the result will be the same as the third one, which means there are $5$ balls in the first box.


Lily walks to school every day. She leaves home at $7:15$ and arrives at the school at $7:50$. How many minutes does she spend?


Which of the following number is the opposite number of $12$?


There is a cube whose six faces are marked with $3$, $4$, $5$, $6$, $7$, and $8$. If the sum of every two numbers that are on the opposite faces are identical, the number on the opposite face of $4$ is~\uline{~~~~~~~~~~}~

Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math->Fun Math Problems->Dice


Math workshops start at $5:00$ PM. Today Allan was $15$ minutes late to the workshop. What time did Allan come?

$5:00$ PM

$5:05$ PM

$5:15$ PM

$5:20$ PM

$15$ minutes past $5:00$ PM is $5:15$ PM.


insert pic  Seven children are standing in a circle. No two boys are standing next to each other. No three girls are standing next to each other. Which of these is true for the number of girls standing in the circle?

only 3 is possible

3 and 4 are possible

only 4 is possible

4 and 5 are possible

only 5 is possible

"No two boys stand next to each other" meaning there are a maximum of 3 boys only. Hence, there are a minimum of 4 girls in the circle.


Thirty apples are distributed among 4 children and each of them can get at least one apple. For the kid who gets the most apples, what\textquotesingle s the least possible number of apples he or she can get?

Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Simple Pigeonhole Principle Problems

$$30\div 4=7 \textbackslash{} \text{R}2$$, thus for the kid who gets the most apples, he or she can get $$7+1=8$$ apples at least.


Ivy, Vivian, and Candy are playing the truth or lie game. The rule is: the person who picks the truth card can only tell the truth, and the person who picks the lie card can only tell a lie.  Ivy said: "Vivian and Candy lied."  Vivian said: "I didn\textquotesingle t lie."  Candy said: "Vivian lied."  How many of them told the truth? How many of them told a lie?~\hspace{0pt}~．

$2$; $1$

$0$; $3$

$1$; $2$

$3$; $0$

Vivian and Candy told contradictory information, so one of them told the truth and the other one told the lie. Therefore, Ivy has definitely told the lie. So two people told the lie, and one people told the truth.


There are six balls numbered with $$1-6$$. Amy distributes them equally among three people and each time she needs the three people give her a digit from the two balls they got, then she will use the three digits to form a three$-$digit number. She does it three times, and the numbers are $$145$$, $$263$$, and $$651$$. Which two numbers below must belong to one person?

$4$ and $5$

$2$ and $1$

$2$ and $5$

$$3$$ and $$4$$

$4$ and $6$

According to $145$ and $651$, we can find that only $4$ and $6$ have changed with each other. So, they belong to one person.


There are two piles of bananas, six in one pile and six in the other pile. Mia and Leo take turns to take the bananas from any one of the two piles. The number of bananas they can take is unlimited, but they have to take at least one each turn. Whoever picks the last banana is the winner. If Mia takes the first match,  is guarantee to win.

Mia

Leo

The number of matches in the two piles is the same. No matter how many matches Andy takes from one pile, Bob just needs to take the same amount of matches from the other pile. As long as there are matches for Andy to take, Bob can definitely take away the same amount from the other pile. Therefore, Bob will take away the last match and he is guarantee to win.


Ken, Tom and Keven were playing a game together. At $1$ P.M., Ken got a score of $-75$, Tom got a score of $-32$, and Keven got a score of $3$. Half an hour later, Ken\textquotesingle s score was $-1$, Tom\textquotesingle s score was $41$, and Keven\textquotesingle s score was $-62$. Who had the greatest change of score?

Ken

Tom

Keven

They had the same change.

Overseas Competition->Knowledge Point->Combinatorics->Mathematical Thoughts

The changes of Ken\textquotesingle s, Tom\textquotesingle s and Keven\textquotesingle s scores were $74$, $73$ and $65$, respectively. Ken had the greatest change.


A certain movie is $$90$$ minutes long. It started at $$5:10\textasciitilde\text{PM}$$. During the movie, there were two commercial breaks, one lasting $$8$$ minutes and one lasting $$5$$ minutes. At what time did the movie finish? (2009 Math Kangaroo Problem, Level 3-4, Question \#8)

At $$6:13\textasciitilde\text{PM}$$

At $$6:27\textasciitilde\text{PM}$$

At $$6:47\textasciitilde\text{PM}$$

At $$6:53\textasciitilde\text{PM}$$

At $$7:13\textasciitilde\text{PM}$$

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


From City A to City B, the train passes through three other stations besides the starting and ending stations.~~The train leaves at 8:10 and arrives at the third station at 13:25. It will run for 45 minutes. How long does it take to reach the terminal?

$$4$$hours $$15$$minutes

$$5$$hours $$15$$minutes

$$4$$hours $$45$$minutes

$$5$$hours $$45$$minutes


There are $$3$$ matches on the table. Andy and Bob take turns to pick up $$1$$ to $$2$$ matches each time. The person who picks up the last match will be the winner. If both Andy and Bob were to use the best method and Andy wants to win, he should pick up the matches~\uline{~~~~~~~~~~}~.

First

Second

Andy should go second. Since Bob must pick up either $1$ or $2$ matches when he begins, there will be either $2$ or $1$ match(es) left. Both results lead to Andy\textquotesingle s victory


How many real solutions does $\textbar x+\textbar{} x+2\textbar x\textbar\textbar\textbar=1$ have?

We have to go through slowly and case by case in order to get every solution, then check for extraneous ones. First, we take off the outermost set of absolute value bars, leaving the equations $$ \left\textbackslash{\begin{array}{c} x+\textbar x+2\textbar{} x\textbar\textbar=1 \textbackslash\textbackslash{} x+\textbar x+2\textbar{} x\textbar\textbar=-1 \end{array}\right. $$ Now, moving the $x$ over and taking off the outermost absolute value bars on both equations leaves us with the equations $$ \left\textbackslash{\begin{array}{c} x+2\textbar x\textbar=1-x \textbackslash\textbackslash{} x+2\textbar x\textbar=-1+x \textbackslash\textbackslash{} x+2\textbar x\textbar=-1-x \textbackslash\textbackslash{} x+2\textbar x\textbar=1+x \end{array}\right. $$ Moving the $x$ over in every equation gives $$ \left\textbackslash{\begin{array}{c} 2\textbar x\textbar=1-2 x \textbackslash\textbackslash{} 2\textbar x\textbar=-1 \textbackslash\textbackslash{} 2\textbar x\textbar=-1-2 x \textbackslash\textbackslash{} 2\textbar x\textbar=1 \end{array}\right. $$ Obviously, $2\textbar x\textbar=-1$ cannot happen, so we can throw out that equation. Now, removing the absolute value bars on the three remaining equations gives us $$ \left\textbackslash{\begin{array}{c} 2 x=1-2 x \textbackslash\textbackslash{} 2 x=-1+2 x \textbackslash\textbackslash{} 2 x=-1-2 x \textbackslash\textbackslash{} 2 x=1+2 x \textbackslash\textbackslash{} 2 x=1 \textbackslash\textbackslash{} 2 x=-1 \end{array}\right. $$ Obviously, $2 x=-1+2 x$ and $2 x=1+2 x$ are not real equations, so we can throw those out leaving us with four final equations. Solving each one, we get that $x=\frac{1}{2},-\frac{1}{2}, \frac{1}{4}$, and $-\frac{1}{4}$. Plugging in all those values to the original equation tells us that only $\frac{1}{4}$ is a real solution while the others are all extraneous, so there is only one real solution.


$$\times$$

$$\div$$

Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles->Number Puzzles (sign of operations)->Filling the Symbol in the Equations


Mr. John plays a game with his two smart students.  He says: "I have a $2-$digit prime number. But I tell Andy only the ones digit and tell Fiona only the tens digit."  A few moments later, Andy says: "I don\textquotesingle t know what the number is, neither do you."  Fiona says: "I didn\textquotesingle t know what the number was before, but now I know it."  What is the ones digit of the prime number?

Andy knows neither of them know the number, which means the ones digit could not be $7.$  Then using elimination, Fiona knows it, which means the tens digit could be $3$ or $6.$ But no matter what it is, the ones digit will always be $1.$


Teacher wrote $$10$$ non-zero natural numbers in sequence on the blackboard, where the $$1$$st number is $$16$$, and the sum of any $$3$$ adjacent numbers is $$100$$. The biggest possibility of the $$8$$th number is~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math


Bud played basketball for half an hour and stopped at 10:15. When did Bud start to play basketball？


There are two piles of matches, seven in one pile and seven in the other pile. JoJo and Kevin take turns to take the matches from any pile. The number of matches they can take is unlimited, but they have to take at least one each turn. Whoever picks the last match is the winner. If JoJo takes the first match,  is guarantee to win.

JoJo

Kevin

The number of matches in the two piles is the same. No matter how many matches JoJo takes from one pile, Kevin just needs to take the same amount of matches from the other pile. As long as there are matches for JoJo to take, Kevin can definitely take away the same amount from the other pile. Therefore, Kevin will take away the last match and he is guarantee to win.


It is Thursday today. Mary\textquotesingle s birthday was sixteen days ago. On what day of the week was Mary\textquotesingle s birthday?

Tuesday

Wednesday

Thursday

Friday

Saturday

Fourteen days ago, it was Thursday again. Then, two days before Thursday was Tuesday.


Which of the following number is the opposite number of $-5$?

Opposite number is the number with opposite sign


The time in Thornsburg is $$6$$ hours ahead of London. The current time in London is $$8:27$$pm of $$15$$th April. What is the time in Thornsburg currently?  Choose the answer.

$$8:27$$pm $$15$$th April

$$14:27$$ $$15$$th April

$$2:27$$pm $$15$$th April

$$2:27$$am $$16$$th April

$$14:27$$am $$15$$th April

$6$ hours ahead of $08:27$pm is $02:27$am.


Amy has $$18$$ boxes of clay.  Bala has $$9$$ boxes of clay.  Cindy has $$6$$ boxes of clay.  How many boxes of clay do they have altogether?


Four friends are competing in a math competition. They are Andy, Bob, Cindy and Daisy.  The organiser of the competition told you that:  ($$1$$) Andy performed better than Bob.  ($$2$$) Cindy\textquotesingle s score is higher than Andy\textquotesingle s score.  ($$3$$) Daisy\textquotesingle s score is lower than two of her friends.  Rank them from the highest to the lowest.

Andy, Bob, Cindy, Daisy

Andy, Cindy, Daisy, Bob

Cindy, Andy, Daisy, Bob

Cindy, Andy, Bob, Daisy

None of the above

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


$$100$$ squares are placed in a row, each filled up with a digit among $$0$$, $$1$$, $$2$$, $$\cdots $$, $$9$$. Now if a digit appears $$5$$ times or more, all the squares filled up with that digit will be painted red. Find the smallest possible number of red squares.

None of the above

Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle


Ivy, Vivian, and Candy are playing the truth or lie game. The rule is: the person who picks the truth card can only tell the truth, and the person who picks the lie card must tell a lie.  \textbf{Ivy said: "Vivian and Candy lied."}  \textbf{Vivian said: "I didn\textquotesingle t lie."}  \textbf{Candy said: "Vivian lied."}  How many of them told the truth? How many of them told a lie?~\hspace{0pt}

$2$; $1$

$0$; $3$

$1$; $2$

$3$; $0$


Students $$A$$, $$B$$, $$C$$, $$D$$, $$E$$ and $$F$$ are standing in a row. We know that:  $$\quad \quad \quad 1$$) $$D$$ is standing between $$E$$ and $$F$$;  $$\quad \quad \quad 2$$) $$C$$ is standing between $$D$$ and $$E$$;  $$\quad \quad \quad 3$$) $$B$$ is standing between $$C$$ and $$D$$; and  $$\quad \quad \quad 4$$) $$A$$ is standing between $$B$$ and $$C$$.  Which of the folowing statements is true?

$$A$$ is positioned either at the right end of the row or the left end of the row.

$$A$$ is second from one of the ends of the row.

$$A$$ is third from one of the ends of the row.

The situation described in the problem is impossible.

$$E$$ and $$F$$ are right next to $$A$$.

The only true statement here is that $A$ is the third from one of the ends of the row, which is $C$.


Two sixth-grade students play chess with at least $$10$$ fifth-grade students, and each of the two men plays exactly one game against each other. There are three results: $$2$$ points for winning, $$1$$ point for drawing, and $$0$$ point for losing. After the competition, it is known that the sum of the two sixth grade students is $$20$$ points, and each fifth grader has scored $$N$$ points. $$N$$ is~\uline{~~~~~~~~~~}~.

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


Lucy, Maria and Anna have a meeting at $$12:30$$. Lucy\textquotesingle s walk takes $$10$$ minutes, Maria\textquotesingle s a quarter of an hour, and Anna\textquotesingle s $$40$$ minutes. At what time must the person who needs the longest time to get to the meeting leave her house? (2006 Math Kangaroo Problem, Level 1-2, Question \#9)


If you need $$6$$ minutes to cook $$2$$ eggs, you will need  minutes to cook $$4$$ eggs in a pot. (The pot can hold $$4$$ eggs.)

Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Planning

It takes $$6$$ minutes to cook a $$2$$ raw egg, and of course a $$4$$ raw egg will be cooked in a pot, and it will still take $$6$$ minutes to cook. So choose $$\text{B}$$.


Andy, Bob, Cindy, Daisy

Andy, Cindy, Daisy, Bob

Cindy, Andy, Daisy, Bob

Cindy, Andy, Bob, Daisy

None of the above

The score from highest to lower:  First -\/-\textgreater{} Cindy  Second -\/-\textgreater{} Andy  Thrid -\/-\textgreater{} Daisy  Fourth -\/-\textgreater{} Bob


Two men and two boys are planning to cross a river using a small boat that can only hold either one man or two boys. What is the least possible number of times the boat needs to cross the river in order to bring all four of them to the other side?

Two boys cross the river and one of them goes back.  A man crosses the river and the other boy goes back.  Two boys cross the river and one of them goes back.  The other man crosses the river and the other boy goes back.  Two boys cross the river.


Math workshop starts at $5:00$ PM. Today Allan was $15$ minutes late to the workshop. What time did Allan come?~

$5:00$ PM

$5:05$ PM

$5:15$ PM

$5:20$ PM

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

$15$ minutes past $5:00$ PM is $5:15$ PM.


Andrea needs an hour to get to the shopping center. If she leaves at $4$, she gets there half an hour after the store closes. When does the shopping center close?

If Andrea leaves at $4$ PM, she arrives at the store when it is $5$ PM, and the store closes at $4:30$ PM.


Which of the following numbers is the opposite of $-2$?


Which of the following calculations shows how to work out the number of seconds in~~24 days?

24 \times 24~

24 \times 60

24 \times 24 \times 60

24 \times 60 \times 60

24 \times 24 \times 60 \times 60

1 day = 24 hours，$$1$$ hour = 60 minutes，$$1$$ minute = 60 seconds，$$24$$ days = (24 \times 24) hours = [(24\times 24) \times 60] minutes = 24\times 24\times 60\times 60 seconds


Jacob writes down a three-digit number on a piece of paper. Lisa, Mike, and Henry are guessing the number Jacob writes down.  Lisa: "The number is $473$."  Mike: "The number is $623$."  Henry: "The number is $428$."  Among the three digits, each person guesses two of them correctly for both digits and their positions.  What is the number Jacob writes down?

In the hundreds place, $4$ appears twice, so $4$ is in the hundreds place.  In the tens place, $2$ appears twice, so $2$ is in the tens place.  In the ones place, $3$ appears twice, so $3$ is in the ones place.  Thus, the number Jacob writes down is $423$.


The math workshop started at $5:00$. Today Leo was $15$ minutes late to the workshop. What time did Leo arrive?


Which of the following number is the opposite number of $12$


What is the missing number in the sequence below?  1, 3, 7, 15, 31,~\uline{~~~~~~~~~~}~

None of the above

Overseas Competition->Knowledge Point->Combinatorics->Patterns of Figures


Grace was going to meet her friends at the bus station. Right now, it is $8:20$. Grace arrived at the bus station half an hour ago. The trip from her home to the station was $1$ hour long. What time did Grace depart toward the bus station?

Grace arrived at the bus station at $7:50$, and the trip was $1$ hour long, so she departed at $6:50$.


There are $1000$ students in Grade $3$ in Think Academy. Among these $1000$ students, at least how many students were born in the month with the most births?


$$\times$$

$$\div$$


In the calculation shown below, different letters represent different digits.  $AA\times BC\times ABC=ABCABC$  What is the product of $A\times B\times C$?

Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles


About the number 325, five boys said:  Andrei: "This is a 3-digit number"  Boris: "All digits are distinct"  Vick: "The sum of the digits is 10"  Greg: "The unit digits is 5"  Danny: "All digits are odd"  Which of the boys was wrong?

Andrei

Boris

Vick

Greg

Danny


A certain play has two $30$-minute parts and one intermission among them. The play started at $8:00$ and ended at $9:15$. How many minutes long were the intermission? (Adapted from 2010 Math Kangaroo Problem, Level 1-2, Question \#19)


Math workshops start at $5:00$ PM. Today Allan was $15$ minutes late to the workshop. When did Allan come?

$5:00$ PM

$5:05$ PM

$5:15$ PM

$5:20$ PM


A school has two classes in the morning, each class continues $$45$$ minutes, and there is a break between classes. The first class starts at $$10:00$$, and the second class ends at $$12:00$$. How long does the courseware rest?


Joseph has $$100$$ mice. Each of them is either white or grey. At least one of the mice is grey, and out of any seven of Joseph\textquotesingle s mice at least four are white. How many grey mice does Joseph have at the most?

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

The maximum is $3$ since any group of $7$ at most $3$ are grey.


There are $$25$$ matches on the table. John and James take turns to remove $$1$$ to $$3$$ matches each time. The person who removes the last match will be the winner. If both of them were to use the best method and John removes first, then~\uline{~~~~~~~~~~}~will win.~

John

James

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

$$25\div 4=6\ldots 1$$ John removes $$1$$ match and $$24$$ is a multiple of $$4$$. So, the first player will win the game.


Which of the following number is the opposite number of $-5$

Overseas Competition->Knowledge Point->Combinatorics->Mathematical Thoughts->Absolute Value


24 x24~

24 x~60

24 x~24 x~60

24 x~60 x~60

24 x~24 x~60 x~60

1 day = 24 hours，$$1$$ hour = 60 minutes，$$1minute$$ = 60 seconds，$$24$$ days = (24 x 24) hours = [(24x 24) x 60] minutes = 24x 24x 60x 60 seconds


The numbers $$1$$, $$2$$, $$4$$, $$5$$, $$8$$, $$9$$, $$10$$, $$13$$ and $$16$$ are divided into groups of one or more numbers. The sum of the numbers in each group is the same. What is the largest possible number of groups?

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


When calculating $63\times72$ , Judy fails to write the correct column multiplication. She writes one of the four digits as $9$ and gets a result which has a difference of $432$ from the correct one. Which digit does she write incorrectly?

The difference between the wrong answer and the correct one is $432$, whose ones digit is $2$. That means Judy writes neither $2$ nor $3$ wrong.  Compare $7\times63$ and $6\times72$, and we can get the correct answer.  $69\times72-63\times72=432.$


Given that $$2a+3b=84$$, the largest possible value of $$a\times b$$ is~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Forming a Maximum/Minimum Multi-Digit Numbers with Fixed Sums

Given that an unchanged sum, the smaller the difference the larger the product. $84=42+42$$$\Rightarrow (2a) \times (3b)=6ab\leqslant {{42}^{2}}\Rightarrow ab\leqslant 294$$. Thus, the largest possible value of $ab$ is $294$.


There is a cube whose six faces are marked with $3$, $4$, $5$, $6$, $7$, and $8$. If the sum of every two numbers that are on the opposite faces are the same, the number on the opposite face of $4$ is~\uline{~~~~~~~~~~}~

Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math->Fun Math Problems->Dice


Bob constructs a rectangular building by many $1\times1\times1$ cubes. The volume of the rectangular building is $210$. What is the least sum of all edges of the rectangular building?

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

Given the product, the smaller the difference, the smaller the sum.  $210=2\times3\times5\times7$, so the length, the width, and the height are $5,$ $6,$ and $7$. $(5+6+7)\times4=72.$


Apply this operation to $$(7, 28)$$. What are the last two numbers when the operation stops?

None of the above

Overseas Competition->Knowledge Point->Combinatorics->Patterns of Figures->Special Changes


Lucy, Maria and Anna have a meeting at $$12:30$$. Lucy\textquotesingle s walk takes $$10$$ minutes, Maria\textquotesingle s walk takes a quarter of an hour, and Anna\textquotesingle s walk takes $$40$$ minutes. At what time must the person who needs the longest time to get to the meeting leave her house?

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


A talk show has two $$30$$-minute parts and has few commercials between them. The TV show started at $$1:50$$ and ended at $$3:00$$. How many minutes long were the commercials in total?

$$30$$ min

$$20$$ min

$$15$$ min

$$10$$ min


Vera leaves home at $$9.15$$ a.m. to walk to Derek's house, which takes her $$25$$ minutes. Carl leaves his house $$5$$ minutes after Vera but only takes $$6$$ minutes to get to Derek's house.  When Carl arrives, how long will he and Derek have to wait for Vera to arrive?

$$11$$ min

$$14$$ min

$$21$$ min

$$24$$ min

Vera will arrive at $$9.40$$ a.m.  Carl will leave at $$9.20$$ a.m. and arrive at $$9.26$$ a.m.  So, Carl and Derek will have to wait for $$14$$ min.


The numbers $$1$$, $$5$$, $$8$$, $$9$$, $$10$$, $$12$$, and $$15$$ are divided into groups of one or more numbers. The sum of the numbers in each group is the same. What is the largest possible number of groups? (2016 Math Kangaroo Problem, Level 1-2, Question \#24)

$1+5+8+9+10+12+15=60$  $2$ groups: sum of each group is $30$, $30=15+5+10=1+8+9+12$  $3$ groups: sum of each group is $20$, $20=15+5=12+8=10+9+1$  $4$ groups: sum of each group is $15$, $15-12=3$, but there is no $3$ in given numbers  Thus, the largest possible number of groups is $3$.


Andrea needs an hour to get to the shopping center. If she leaves at $4$ PM, she gets there half an hour after the store closes. If she leaves at $8$ AM, she gets there half an hour before the store opens. What hours is the shopping center open? (2007 Math Kangaroo Problem, Level 1-2, Question \#24)

$7:30$ to $4:30$

$8:30$ to $5:30$

$7:30$ to $5:30$

$8:30$ to $4:30$

$9:30$ to $4:30$

If Andrea leaves at $4$ PM, she arrives at the store when it is $5$ PM, and the store closes at $4:30$ PM. If Andrea leaves at $8$ AM, she arrives at the store when it is $9$ AM, and the store opens at $9:30$ AM.


There are two piles of matches, six in one pile and six in the other pile. Andy and Bob take turns to take the matches from any one of the two piles. The number of matches they can take is unlimited, but they have to take at least one each turn. Whoever picks the last match is the winner. If Andy takes the first match,  is guarantee to win.

Andy

Bob

The number of matches in the two piles is the same. No matter how many matches Andy takes from one pile, Bob just needs to take the same amount of matches from the other pile. As long as there are matches for Andy to take, Bob can definitely take away the same amount from the other pile. Therefore, Bob will take away the last match and he is guarantee to win.


All numbers that are divisible by neither $$5$$ nor $$11$$ were removed from a sequence of consecutive natural numbers from $$1$$ to $$5500$$. A new sequence was formed. How many terms are there in this new sequence? (Adapted from $$2004$$ Math Kangaroo Problems, Level $$9-10$$, Question \#$$30$$)

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

Method $1$:  $5500\div5=1100$; $5500\div11=500$; $5500\div(5\times11)=100$.  $1100+500-100=1500$.  Method $2$:  In each group formed by $55$ consecutive numbers, $55-(11+5-1)=40$ numbers will be removed.  $5500\div55\times(55-40)=1500$ numbers will remain in this new sequence.


A tortoise is running along a $$5$$ metre track. It can run either half a metre or a metre every second. How many different ways can the tortoise finish the track?

Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Inductive Recursion


A cat divides $$24$$ cans of tuna into $$4$$ groups. Each group has at least $$1$$ can of tuna, and the number of cans in each group \uline{cannot be the same}. There are at mostcans of tuna in the group that has the largest number of cans.

Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles->Number Puzzles (sign of operations)->Obtaining Maximum/Minimum Values


Basil has several domino tiles, as shown in the figure. He wants to arrange them in a line according to the well-known "domino rules": in any two tiles that are next to each other, the squares that touch must have the same number of points. What is the largest number of tiles he can arrange in this way?  [insert picccc]

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

One possible configuration for 5 dominoes is shown.  [pic]


A school has three classes in the morning, and there is a $$15$$-minutes break between each two classes. Each class is of the same length. The first class starts at $$9:45 $$, and the third class ends at $$12:00 $$. How long is each class?


The weights of four boys are 45kg, 48kg, 52kg and 53kg. Mason\textquotesingle s weight is an even number. Joshua\textquotesingle s weight is a multiple of 5. Christopher is not the heaviest and Mateo is not the lightest. Who is the heaviest among the four boys?

Mason

Joshua

Christopher

Mateo

Impossible to determine

Mason\textquotesingle s weight is an even number = 48kg / 52kg.  Joshua\textquotesingle s weight is a multiple of 5 = 45kg.  Christopher is not the heaviest = x 53kg.  Mateo is not the lightest = x 45kg  Jason - \textgreater{} either Mason or Christpoher is 28/52kg -\textgreater{} Mateo.


From $1$ to $45$, how many numbers must be chosen at least to ensure that there are two of chosen numbers sum to $30$?

Overseas Competition->Knowledge Point->Combinatorics->Constructing and Proving


Teacher James distributes $82$ apples to $12$ students. At least how many apples can the student with the most apples get?

Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle


$A$ and $B$ play a game as follows. First $A$ says $1, 2$ or $3$. Then $B$ can add $1, 2$ or $3$ to the number the first player said. The game continues with the players playing alternately, in each turn adding $1, 2$ or $3$ to the previous number. For example, A can say $2$, then B can say $5$, then A could say $6$, and so on. The player who says $100$ wins. Who has the winning strategy?

A

B

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

Observe that B can always say a multiple of 4 in his turn. For example, consider the following sequence of moves: A-1; B-4; A-6; B-8 and so on. Regardless of what A says, B can always say a multiple of 4. Hence B will say 100 and will win


SASMO 2016 P2 Q7  It takes 5 mintues to boil an egg. What is the least amount of time it takes to boil 3 eggs?

None of the above

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


There are $9$ water glasses, all facing up. You need to turn exactly $2$ glasses over in each time. Is it possible to turn all $9$ of them upside down after several moves?

Yes, it is possible.

No, it is impossible.

I have no idea.

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

You have to flip a cup an odd number of times to turn it over.  To make all the $9$ cups face down, you have to make an odd number of flips.  Each time you flip exactly $2$ cups, the total number of flips is an even number of times.  Hence it is impossible.


David went to watch $Spider Man$ at the cinema on June $25$\textsuperscript{th}. The movie started at $6 : 32$ P.M. and finished at $7 : 54$ P.M.. On the same day, Mary needed to go abroad by plane. The plane took off at $11 : 40$ P.M., and the time she spent on plane was $2$ hours and $11$ minutes longer than that David spent at the cinema. When would Mary get off the plane?

$3 : 23$ P.M. on June $25$\textsuperscript{th}

$3 : 13$ A.M. on June $25$\textsuperscript{th}

$3 : 13$ P.M. on June $26$\textsuperscript{th}

$3 : 13$ A.M. on June $26$\textsuperscript{th}

$3 : 23$ A.M. on June $26$\textsuperscript{th}

From $6 : 32$ P.M. to $7 : 54$ P.M., it lasted $1$ hour and $22$ minutes.  The time Mary spent on plane: $1$ hour and $22$ minutes $+$ $2$ hours and $11$ minutes $=$ $3$ hours and $33$ minutes.  Thus, Mary would get off the plane at $3 : 13$ A.M. on June 26\textsuperscript{th}.


Cindy and Chloe are siblings. Each of them have $$2$$ brothers and $$2$$ sister in. How many children are there in the family?

Cindy is sister to Chloe and Chloe is sister to Cindy. Thus, they have one more common sister. There are a total of $$3$$ girls in the family.  Their brother are the same, thus $$2$$ boys in the family.  In total $$5$$ children.


Jacky wants to insert the digit 3 somehwere in the number 2014. Where should she insert digit 3 if she wants her five digit number to be as small as possible?

in front of 2014

between the 2 and the 0

between the 0 and the 1

between the 1 and the 4

behind 2014.

3 is a digit between 1 and 4.  Try to put into and see which number is the smallest.


Jack, Rebecca, Dustin and Steve write down one number, respectively, which are $44$, $85$, $72$ and $39$.Both Jack\textquotesingle s and Dustin\textquotesingle s numbers are odd. Both Jack\textquotesingle s and Steve\textquotesingle s numbers are more than $60$. What number is written by Rebecca?

Impossible to determine

Eliminate odd numbers and numbers more than $60$, so the only number left is $44$.


In class, Anna, Bell, and Julie are doing math homework at school. Among them, Anna took $$18$$ minutes and $$25$$ seconds. Bell took $$19$$ minutes and $$36$$ seconds, and Julie took $$20$$ minutes and $$15$$ seconds. How long did the three take in total? (adapted from 2008 Math Kangaroo Problem, Level 3 - 4, Question \#15)

$$57$$ minutes $$16$$ seconds

$$58$$ minutes $$16$$ seconds

$$58$$ minutes $$26$$ seconds

$$58$$ minutes $$6$$ seconds

$$57$$ minutes $$26$$ seconds

$18$ minutes + $19$ minutes + $20$ minutes + $25$ seconds $+$ $36$ seconds + $15$ seconds = $57$ minutes $76$ seconds = $58$ minutes $16$ seconds


The time in Thornsburg is $$6$$ hours ahead of London. The time in London is $$8.27$$ pm. What is the time in Thornsburg?  Choose the answer.

$$02.27$$ pm

$$14.27$$ pm

$$02.27$$ pm

$$02.27$$ am

$$14.27$$ am

$6$ hours ahead of $08:27$ pm is $02:27$ am.


Suppose $a, b$, and $c$ are nonzero real numbers, and $a+b+c=0$. What are the possible value(s) for $\frac{a}{\textbar a\textbar}+\frac{b}{\textbar b\textbar}+\frac{c}{\textbar c\textbar}+\frac{a b c}{\textbar a b c\textbar} ?$ (2017 AMC 8 Problem, Question \#21)

$1$ and $-1$

$2$ and $-2$

$0,2$ and $-2$

$0,1$ and $-1$

There are 2 cases to consider:  \textbf{Case 1:} 2 of $a, b$, and $c$ are positive and the other is negative. WLOG, we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that $$ \frac{a}{\textbar a\textbar}+\frac{b}{\textbar b\textbar}+\frac{c}{\textbar c\textbar}+\frac{a b c}{\textbar a b c\textbar}=1+1-1-1=0 \text {. } $$  \textbf{Case 2}: 2 of $a, b$, and $c$ are negative and the other is positive. Without loss of generality, we can assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that $$ \frac{a}{\textbar a\textbar}+\frac{b}{\textbar b\textbar}+\frac{c}{\textbar c\textbar}+\frac{a b c}{\textbar a b c\textbar}=-1-1+1+1=0 \text {. } $$  Note these are the only valid cases, for neither 3 negatives nor 3 positives would work, as they cannot sum up to 0 . In both cases, we get that the given expression equals $(\mathbf{A}) 0$.


The numbers $$2$$, $$4$$, $$6$$, $$8$$, $$9$$ and $$11$$ are divided into groups of one or more numbers. The sum of the numbers in each group is the same. Which number must be in the group with the number $$11$$?


Joy writes a four-digit number on a piece of paper and let Sandy guess.  $\textasciitilde$  Sandy asks: "Is it $$6031$$?"  Joy says: "You guess $1$ digit correct and it\textquotesingle s in the correct place."  Sandy asks: "Is it $$5672$$?"  Joy says: "You guess $2$ digits correct and they\textquotesingle re in the wrong places."  Sandy asks: "Is it $$4796$$?"  Joy says: "You guess $4$ digits correct and they\textquotesingle re in the wrong places."  $\textasciitilde$  According to the conversation above, what number does Joy write?

The number is formed by $4,7,9,6$ according to last question. Thus, according to the first question, $6$ should be in the thousands place. From the second and last question, $7$ is in ones place, $4$ is in tens place, and $9$ is in hundreds place.


Thirty apples are distributed among 4 children and each of them can get at least one apple. For the kid who gets the most apples, what\textquotesingle s the least possible number of apples he or she can get? ．

Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Simple Pigeonhole Principle Problems

$$30\div 4=7 \textbackslash{} \text{R}2$$, thus for the kid who gets the most apples, he or she can get $$7+1=8$$ apples at least.


Let\textquotesingle s convert!  $$1$$ hour $$45$$ minutes $$=$$~\uline{~~~~~~~~~~}~minutes


Which of the following calculations shows how to work out the number of seconds in~~24 days?

24 x~24~

24 x~60

24 x 24 x 60

24 x 60 x 60

24 x~24 x~60 x~60

1 day = 24 hours，$$1$$ hour = 60 minutes，$$1$$ minute = 60 seconds，$$24$$ days = (24 \times 24) hours = [(24\times 24) \times 60] minutes = 24\times 24\times 60\times 60 seconds


For any two numbers $$x$$ and $$y$$, the special operation $$^{}\otimes$$ is defined as follow $$x^{}\otimes y=x+y-\frac {2012}{335}$$.  Calculate $$\frac {2^{}\otimes 4^{}\otimes 6^{}\otimes \cdots ^{}\otimes 2010^{}\otimes 2012}{2012}$$.

None of the above


A train started from the starting station at 7:30, it takes 8 hours and 37 minutes to arrive at the terminal station. When will it arrive?

$$3:30$$ A.M.

$$3:37$$ P.M.

$$4:37$$ P.M.

$$4:07$$ P.M.

$$4:37$$ A.M.


The weights of four boys are $45\rm{kg}$, $48\rm{kg}$, $52\rm{kg}$ and $53\rm{kg}$. Mason\textquotesingle s weight is an even number. Joshua\textquotesingle s weight is a multiple of $5$. Christopher is not the heaviest and Mateo is not the lightest. Who is the heaviest among the four boys?

Mason

Joshua

Christopher

Mateo

Impossible to determine


Math workshop starts at $5:00$ PM. Today Allan was $15$ minutes late to the workshop. What time did Allan come? (2010 Math Kangaroo Problem, Level 1--2, Question \#12)

$5:00$ PM

$5:05$ PM

$5:15$ PM

$5:20$ PM

$15$ minutes past $5:00$ PM is $5:15$ PM.


Peter can draw a cartoon in $30$ minutes. How many hours does it take to paint four cartoons?~(adapted from 2011 Math kangaroo Problems, Level 1-2, Question \#4)

There are $60$ minutes in an hour.  It will take $120$ minutes to draw $4$ cartoons.~  $120$ minutes=$2$ hours.


Bob and Linda went mountain climbing together. They climbed the mountain from $8$:$30$. Bob reached the top at $14$:$00$, and Linda was $1$ hour and $40$ minutes behind him. How long did it take Linda climb?~(adapted from 2003 Math Kangaroo Problem, Level 3 - 4, Question \#10)

$3$ hr $30$ min

$4$ hr $30$ min

$7$ hr $10$ min

$6$ hr $20$ min

$8$ hr $40$ min

It lasted $3$ hours and $30$ minutes from $8$:$30$ to $12$:$00$. It lasted $2$ hours from $12$:$00$ to $14$:$00$. So Bob took $5$ hours and $30$ minutes. Linda was $1$ hour and $40$ minutes behind Bob. Linda\textquotesingle s time was $7$ hours and $10$ minutes.  Bob: $3$h+$30$min+$2$h=$5$h$30$min  Linda:$3$h+$30$min+$2$h+$1$h+$40$min=$7$h$10$min


A certain play has three $30$-minute parts and two intermission among them. The play started at $8:30$ and ended at $10:15$. How many minutes long were the intermissions in total?


There are two piles of matches. Each pile has $$6$$ sticks. Cindy and Doris take turns to pick up matches from either pile. There is no limit to how many matches they can pick up, but they must pick up at least one match each turn. The person who picks up the last match will be the winner. If Cindy starts picking up matches first,~\uline{~~~~~~~~~~}~will be the winner (has a winning strategy).

Cindy

Doris

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

Doris will be the winner, since she can simply mirror the number of matches Cindy picks up every turn.


Two cups with capacities of 9 liters and 3 liters, respectively, have different diameters at the top and bottom and are unmarked. If water can be taken arbitrarily, how many operations are required at least to pour out 6 liters of water? (Filling the cup is counted as one operation, pouring the water out is counted as another operation.)

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

2 operations at least.  1. Fill the 9 liter cup to the top.  2. Pour water from the 9 liter cup into the 3 liter cup until the 3 liter cup is full, leaving 6 liters of water in the 9 liter cup.  The answer is B


From $1$ to $30$, how many integers can be chosen at most to ensure that no two of chosen numbers sum to $36$?


Out of 8 bottles of sugar, 1 bottle has a lighter mass than the other 7. How many times can the balance be used to ensure that the lighter one is found?

Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math


Ivy, Vivian, and Candy are playing the truth or lie game. The rule is: the person who picks the truth card can only tell the truth, and the person who picks the lie card must tell a lie.  \textbf{Ivy said: "Vivian and Candy lied."}  \textbf{Vivian said: "I did\textquotesingle t lie."}  \textbf{Candy said: "Vivian lied."}  How many of them told the truth? How many of them told a lie?~\hspace{0pt}

$2$; $1$

$0$; $3$

$1$; $2$

$3$; $0$

Vivian and Candy told contradictory information, so one of them told the truth and the other one told the lie. Therefore, Ivy has definitely told the lie. So two people told the lie, and one people told the truth.


If the exam lasts for two days and ends on Wednesday, what day does the exam start?~(adapted from $$2008$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$4$$)

Monday

Tuesday

Friday

Saturday

Thursday

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time->Period of Dates

Tuesday to Wednesday is exactly two days.