math-eval/TAL-SCQ5K huggingface dataset:
Problem: 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 $a$ ? 
Options:
aoVal: A
content: $$249$$ 
aoVal: B
content: $$250$$ 
aoVal: C
content: $$251$$ 
aoVal: D
content: $$252$$ 
aoVal: E
content: $$253$$ 
Answer: E
Analysis: 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, $a=253$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Addition and Subtraction of Equations
Source: 其它
Problem: What is the ones digit of the result of $3^{50}$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$9$$ 
Answer: E
Analysis: 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 $9$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers->Exponentiation of Powers
Source: 其它
Problem: What is the result of $1\div 1\frac{2}{2001}\div 1\frac{2}{2003}\cdots \div 1\frac{2}{2023}$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $\frac{2025}{2001}$ 
aoVal: C
content: $\frac{2021}{2025}$ 
aoVal: D
content: $\frac{2001}{2025}$ 
aoVal: E
content: $\frac{1}{2001}$ 
Answer: D
Analysis: $1\div 1\frac{2}{2001}\div 1\frac{2}{2003}\cdots \div 1\frac{2}{2023}$  $=1\div \frac{2003}{2001}\div \frac{2005}{2003}\cdots \div \frac{2025}{2023}$  $=1\times \frac{2001}{2003}\times \frac{2003}{2005}\cdots \times \frac{2023}{2025}$  $=\frac{2001}{2025}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Which of the numbers below is greatest? ． 
Options:
aoVal: A
content: $$22222$$ 
aoVal: B
content: $2222^{2}$ 
aoVal: C
content: $222^{22}$ 
aoVal: D
content: $22^{222}$ 
aoVal: E
content: $2^{2222}$ 
Answer: E
Analysis: $2^{2222}\textgreater2^{2220}=\left (2^{10}\right )^{222}=1024^{222}\textgreater22^{222}$. According to this rule, $\text{E}\textgreater\text{D}\textgreater\text{C}\textgreater\text{B}\textgreater\text{A}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering
Problem: 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) 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$50$$ 
aoVal: E
content: $$60$$ 
Answer: C
Analysis: 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. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The next number in the sequence $$1$$, $$1$$, $$2$$, $$3$$, $$5$$, $$8$$, $$13$$, $$\cdots$$ is． 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$19$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: $$1+1= 2$$, $$1+2 = 3$$, $$\cdots$$, $$5+8 = 13$$, $$8+13 = 21$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Problem: 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？ 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: A
Analysis: After round one, she may have $0$ point, $1$ point, $2$points  After two rounds, she may have $3$, $4$, $6$, $7$, $8$ points 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the next number below?  3, 5, 6, 10, 9, 15, 12,20,~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$13$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$25$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: 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. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Source: 其它
Problem: 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) 
Options:
aoVal: A
content: $6a+4$ 
aoVal: B
content: $2a+7$ 
aoVal: C
content: $6a+12$ 
aoVal: D
content: $32a+34$ 
Answer: C
Analysis: $3\times (2a+4)=3(2a+4)=6a+12$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Using a Letter to Represent an Unknown Number
Source: 其它
Problem: What is the sum of the various numbers of Number $$1074$$? (adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$7$$) 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$14$$ 
Answer: C
Analysis: Pay attention to the review questions and sum the numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form
Problem: 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{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $-5$ 
aoVal: B
content: $\frac{11}{2}$ 
aoVal: C
content: $10.5$ 
aoVal: D
content: $$21$$ 
Answer: B
Analysis: $2 a+10-4 x=-4 x+21$  $2 a=11 $  $a=\frac{11}{2}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation
Source: 其它
Problem: 15-9=, 29+7=． 
Options:
aoVal: A
content: 6, 36 
aoVal: B
content: 4, 35 
aoVal: C
content: 5, 36 
aoVal: D
content: 8, 35 
aoVal: E
content: 7, 39 
Answer: A
Analysis: 15-9=6, 29+7=36 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction 
Source: 其它
Problem: Given that $x$ and $y$ are whole numbers such that $24x-25y =8$, find the smallest value of $x$. 
Options:
aoVal: A
content: $$42$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: None of the above 
Answer: E
Analysis: E 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Indefinite Equations
Source: 其它
Problem: What is the result of $$\frac{{{2}^{2}}}{{{2}^{2}}-1}\times \frac{{{3}^{2}}}{{{3}^{2}}-1}\times \cdots \times \frac{{{99}^{2}}}{{{99}^{2}}-1}$$? 
Options:
aoVal: A
content: $$\frac{99}{50}$$ 
aoVal: B
content: $$\frac{99}{100}$$ 
aoVal: C
content: $$\frac{1}{99}$$ 
aoVal: D
content: $$\frac{99}{200}$$ 
aoVal: E
content: $$\frac{50}{99}$$ 
Answer: A
Analysis: $$=\frac{2\times 2}{(2+1)\times (2-1)}\times \frac{3\times 3}{(3+1)\times (3-1)}\times \frac{4\times 4}{(4+1)\times (4-1)}\times \cdots \times \frac{98\times 98}{(98+1)\times (98-1)}\times \frac{99\times 99}{(99+1)\times (99-1)}$$  $$=\frac{2\times 2}{3\times 1}\times \frac{3\times 3}{4\times 2}\times \frac{4\times 4}{5\times 3}\times \frac{5\times 5}{6\times 4}\times \cdots \times \frac{98\times 98}{99\times 97}\times \frac{99\times 99}{100\times 98}$$  $$=\frac{2}{1}\times \frac{2}{3}\times \frac{3}{2}\times \frac{3}{4}\times \frac{4}{3}\times \frac{4}{5}\times \cdots \times \frac{98}{97}\times \frac{98}{99}\times \frac{99}{98}\times \frac{99}{100}$$  $$=\frac{2}{1}\times \frac{99}{100}$$  $$=\frac{99}{50}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas
Problem: 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) 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$25$$ 
Answer: B
Analysis: 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 $k=22 \rightarrow(\mathbf{B})22$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: 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 \#$$10$$) 
Options:
aoVal: A
content: $$707$$ 
aoVal: B
content: $$907$$ 
aoVal: C
content: $$916$$ 
aoVal: D
content: $$1000$$ 
aoVal: E
content: $$1001$$ 
Answer: B
Analysis: The smallest three-digit number is $107$ and the largest one is $800$. $107+800=907$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form
Problem: How many whole numbers less than $$1000$$ can be written as the product of $$3$$ consecutive whole numbers? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$21$$ 
Answer: A
Analysis: 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 $$10$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers
Problem: 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) 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$0$$ 
aoVal: E
content: $$-3$$ 
Answer: D
Analysis: $f(5)-f(2)=0 \Rightarrow f(5)= f(2)$; thus, $f(x)$ is a constant function. Then, $f(8) - f(2) = 0$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: 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?  
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: D
Analysis: $$36 \div 5 = 7R1$$, so the answer is $$5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division->Division of Whole Numbers->Division with Remainders
Source: 其它
Problem: Evaluate$$\left\textbar{} 3-9 \right\textbar+\left\textbar{} 7-2 \right\textbar$$. 
Options:
aoVal: A
content: $$-11$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$-2$$ 
Answer: B
Analysis: $$\left\textbar{} 3-9 \right\textbar+\left\textbar{} 7-2 \right\textbar=\left\textbar{} -6 \right\textbar+\left\textbar{} 5 \right\textbar=6+5=11$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Mathematical Thoughts->Absolute Value
Problem: 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) 
Options:
aoVal: A
content: William 
aoVal: B
content: Mark 
aoVal: C
content: Diana 
aoVal: D
content: Jimmy 
Answer: A
Analysis: 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. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning
Source: 其它
Problem: How many days are there in August?~(adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$5$$) 
Options:
aoVal: A
content: $$28$$ 
aoVal: B
content: $$29$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$31$$ 
aoVal: E
content: $$32$$ 
Answer: D
Analysis: August have $31$ days. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Reading the Clock
Problem: If the day before yesterday was Sunday. How many days are there from today until Sunday? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: D
Analysis: Before yesterday: Sunday  Yesterday: Monday  Today: Tuesday  From Tuesday to Sunday, we have $$5$$ days based on the information given in the question. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning
Source: 其它
Problem: 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? 
Options:
aoVal: A
content: Yes, he has the winning strategy. 
aoVal: B
content: No, he does not have the winning strategy. 
Answer: B
Analysis: $49\textbackslash{} \div(1+6)=7$~groups, so the second mover will have the winning strategy. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Game Strategy
Source: 其它
Problem: SASMO 2015 P2 Q1  How many months of the year have 28 days? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$12$$ 
Answer: E
Analysis: every month has 28 days. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem
Source: 其它
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? 
Options:
aoVal: A
content: The described situation is impossible 
aoVal: B
content: $$$$Juan 
aoVal: C
content: $$$$Pablo 
aoVal: D
content: $$$$Marco 
aoVal: E
content: It cannot be determined from the information given 
Answer: C
Analysis: Pablo is the name of his son. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Conditions->Complex Reasoning
Problem: Shawn is making a meal. It will cost $1$ minute for preparing the vegetables, $2$ minutes for washing the pan for oven, $6$ minutes for the oven roasting the meal, $2$ minutes for cleaning the table, and $1$ minute for putting the meal on the table. At least how long will it take for Shawn to finish the meal? 
Options:
aoVal: A
content: $$7$$ minutes 
aoVal: B
content: $$8$$ minutes 
aoVal: C
content: $$9$$ minutes 
aoVal: D
content: $$10$$ minutes 
aoVal: E
content: $$12$$ minutes 
Answer: D
Analysis: Shawn can clean the table while the oven is working, so it will take $6$ minutes in total. So, he can finish the meal in $1+2+6+1=10$ minutes. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem
Problem: \textbf{In which place did each of the following shops come in the competition? Write the correct letter on the Prize list. What is the ranking of the pizza shop?} 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
Answer: A
Analysis: nil 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning
Source: 其它
Problem: Eve arranged cards in a line as it is shown in the figure below. At each move Eve is allowed to interchange any two cards. What is the smallest number of moves Eve needs to get the word KANGAROO?  insert pic 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: B
Analysis: First, switch K and 1st O to make KANGONOA.  Second, switch N and R to make KANGOROA.  Third, switch the last A and the 2nd O to make KANGAROO. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning
Source: 其它
Problem: How many months of the year have exactly $$31$$ days? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$12$$ 
Answer: D
Analysis: March, May, July, August, October and December. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning
Source: 其它
Problem: A certain play has three $30$-minute parts and two intermission among them. The play started at $8:30$ and ended at $10:15$. How many minutes long were the intermissions in total? (Adapted from 2010 Math Kangaroo Problem, Level 1-2, Question \#19) 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$25$$ 
Answer: B
Analysis: From $8:30$ to $10:15$ = $1$ hr $45$ min, $1$ hr $45$ min = $105$ min, three $30$-minute parts = $90$ min, $105 - 90$ = $15$ min. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem
Source: 其它
Problem: There were $$8$$ cups on the table with the sides up,  If you can only turn $$6$$ at a time,  Can it be done to make all the cups face down at the end? 
Options:
aoVal: A
content: Yes, it can 
aoVal: B
content: No, it can\textquotesingle t 
Answer: A
Analysis: To make all the $$8$$ cups face down, it only needs to turn them over with odd times, and 6 of them can be turned at a time, so it can be done. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Operational Problem->Turning Mugs over
Problem: There are two groups of flowers. Group $A$ has $17$ flowers and Group $B$ has $13$ flowers. Cathy and Ivy want to play a game with these flowers. They will take turns to take flowers from the two groups. Each person can take any number of flowers from a group at a time. The person who takes the last flower in two groups will win the game. If Ivy starts the game first, who has the winning strategy? 
Options:
aoVal: A
content: Ivy has the winning strategy. 
aoVal: B
content: Cathy has the winning strategy. 
Answer: A
Analysis: Two groups of flowers are not equal. Thus, Ivy needs to take away 4 flowers from Group A and two groups will have the same amount of flowers. Then, Ivy will become the second mover and she has the winning strategy. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Game Strategy
Source: 其它
Problem: A certain play has three $30$-minute parts and two intermissions among them. The play started at $8:30$ AM and ended at $10:15$ AM. How many minutes long were the intermissions in total? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$25$$ 
Answer: B
Analysis: From $8:30$ to $10:15$ = $1$ hr $45$ min, $1$ hr $45$ min = $105$ min, three $30$-minute parts = $90$ min, $105 - 90$ = $15$ min. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem
Source: 其它
Problem: Find the sum of all possible values of $x$ such that $\textbar x-\textbar{} x-1\textbar\textbar-x=1$ .~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: A
Analysis: We must check cases where $\textbar x-\textbar{} x-1\textbar\textbar$ is positive and cases where it is negative. $x\textgreater1$ and $x\textless1$ will cause different signs for $x-\textbar x-1\textbar$. Hence we must check both intervals. However, when $x\textless1$, we see that when $x=\frac{1}{2}$, the expression is equal to 0 , so we must also check the intervals $x\textless\frac{1}{2}$ and $\frac{1}{2}\textless x\textless1$. Solving on the interval $x\textless\frac{1}{2}$, we get $\textbar x+1+x\textbar-x=1 \Longleftrightarrow x=0$. Solving on the interval $\frac{1}{2}\textless x\textless1$, we get $\textbar x+1+x\textbar-x=1 \Longleftrightarrow x=0$. Solving on the interval $x\textgreater1$, we get $\textbar x-x+1\textbar-x=1 \Longleftrightarrow x=0$. Checking for extraneous solutions, we find that $x=0$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Mathematical Thoughts->Absolute Value
Source: 其它
Problem: In the calculation shown below, different letters represent different digits.  $AA\times AB\times C=ADDA$  What is the sum of $A+B+C+D$? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$16$$ 
aoVal: E
content: $$18$$ 
Answer: B
Analysis: $11\times13\times7=1001$  $1+3+7+0=11$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles
Source: 其它
Problem: Find the next number in the sequence below.  $$1,4,10,22,46,94,\cdots $$ 
Options:
aoVal: A
content: $$190$$ 
aoVal: B
content: $$188$$ 
aoVal: C
content: $$186$$ 
aoVal: D
content: $$142$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: 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.} 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Patterns of Figures
Source: 其它