math-eval/TAL-SCQ5K huggingface dataset:
Problem: After the trainer\textquotesingle s first whistle, the monkeys at the circus formed $$4$$ rows. There were $$4$$ monkeys in each row. After the second whistle, they rearranged themselves into $$8$$ rows. How many monkeys were there in each row after the second whistle? (2005 Math Kangaroo Problem, Level 3-4, Question \#4) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: B
Analysis: There were $$4 \times 4 = 16$$ monkeys in total. After the second whistle, there were $$16 \div 8 = 2$$ monkeys in each row. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Source: 其它
Problem: Let $X$ be the smallest positive integer, consisting of only the digit $4$ and $9$ (at least one of each), that is divisible by both $4$ and $9$. What is the last four digits of $X$? 
Options:
aoVal: A
content: $$4444$$ 
aoVal: B
content: $$4494$$ 
aoVal: C
content: $$4944$$ 
aoVal: D
content: $$9444$$ 
aoVal: E
content: $$9944$$ 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: How many of the following pairs of terms are like terms?  ($1$) $2$ and $6$;~($2$) $-2a$ and $-2c$;~($3$) $78x$ and $-200x$;~($4$) $8$ and $y$;~($5$) $4ab$ and $4b$ 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
Answer: B
Analysis: Like terms are terms that have the same variables. For each variable, the number of times it is multiplied by itself is also the same. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Source: 其它
Problem: Calculate: $$\frac{1}{3}\times \frac{3}{5}\times \frac{5}{7}\times \cdots \times \frac{2019}{2021}=$$． 
Options:
aoVal: A
content: $$\frac{1}{2021}$$ 
aoVal: B
content: $$\frac{1}{2019}$$ 
aoVal: C
content: $$\frac{2019}{2021}$$ 
aoVal: D
content: $$\frac{2019}{3}$$ 
Answer: A
Analysis: $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde\frac{1}{3}\times \frac{3}{5}\times \frac{5}{7}\times \cdots \cdots \times \frac{2019}{2021}$$  $$=\frac{1}{2021}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Problem: Erik and Ivy each writes down a fraction. They are surprised to find that the fractions they write are very similar: the denominator of each fraciton is exactly the same as the other\textquotesingle s numerator. What is the product of the two fractions? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $\frac12$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: It cannot be determined. 
Answer: C
Analysis: Their fractions are just reciprocal of the other one. Thus, the product of the two fractions should be $1.$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: What is the area of the triangle formed by the lines $y=-2$, $y=3+x$, and $y=3-x$? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$25$$ 
aoVal: D
content: $$49$$ 
aoVal: E
content: $$64$$ 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: \textbf{A basketball player makes 160 out of 200 free throws. We would estimate the probability that the player makes his next free throw to be heads the fourth time.} 
Options:
aoVal: A
content: \textbf{0.15} 
aoVal: B
content: \textbf{0.50; either he makes it or he doesn't} 
aoVal: C
content: \textbf{0.80} 
aoVal: D
content: \textbf{1.2} 
aoVal: E
content: $$80$$ 
Answer: C
Analysis: \textbf{160/200=0.8} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the sum of the following numbers?  $$5+15+17+23+5+15+17+23+5+15+17+23=$$ 
Options:
aoVal: A
content: $$140$$ 
aoVal: B
content: $$160$$ 
aoVal: C
content: $$180$$ 
aoVal: D
content: $$240$$ 
aoVal: E
content: None of the answer above 
Answer: C
Analysis: Grouping: $$5+15+17+23=60$$  $$60\times3=180$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Given that $$1^{2}+2^{2}+3^{2}+\cdots +n^{2}=\frac{n(n+1)(2n+1)}{6}$$, then $$1^{2}+2^{2}+3^{2}+\cdots +18^{2}=$$． 
Options:
aoVal: A
content: $$2016$$ 
aoVal: B
content: $$2107$$ 
aoVal: C
content: $$2018$$ 
aoVal: D
content: $$2109$$ 
aoVal: E
content: $$2020$$ 
Answer: D
Analysis: Directly substitute $$n$$ with $$18$$ into the formula to get $$1^{2}+2^{2}+3^{2}+\cdots +18^{2}=\frac{18\times 19\times 37}{6}=2109$$. So the answer is $$\text{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas->1²+2²+3²+......+n²=1/6n(n+1)(n+2)
Problem: The pages of a book are numbered $1, 2, 3, 4, 5$, and so on. The digit $5$ appears exactly $16$ times. What is the maximum number of pages this book could have? (2019 Math Kangaroo Problem, Level 3-4, Question \#19) 
Options:
aoVal: A
content: $$49$$ 
aoVal: B
content: $$64$$ 
aoVal: C
content: $$66$$ 
aoVal: D
content: $$74$$ 
aoVal: E
content: $$80$$ 
Answer: B
Analysis: Before page $50$, the digit $5$ appears $5$ times.  From page $50$ to $59$, the digit $5$ appears $11$ times.  Then, the digit $5$ appears the $17$\textsuperscript{th}~time in $65$. Thus, the maximum page number is $64$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: \textbf{Refer to the previous problem. On the final exam Carla scored 98. What is the value of her residual?} 
Options:
aoVal: A
content: $$98$$ 
aoVal: B
content: $$2.5$$ 
aoVal: C
content: $$-2.5$$ 
aoVal: D
content: $$0$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: \textbf{residual =$$\hat{y}$$-y = 98-95.5 = 2.5} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The base of a triangle increases by $30\textbackslash\%$ and its height decreases $20\textbackslash\%$ at the same time. The area of the new triangle is~\uline{~~~~~~~~~~}~$\textbackslash\%$ of the original triangle. 
Options:
aoVal: A
content: $110$ 
aoVal: B
content: $$104$$ 
aoVal: C
content: $$84$$ 
aoVal: D
content: $$50$$ 
Answer: B
Analysis: $1.3\times 0.8=1.04$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Percentage Calculation
Source: 其它
Problem: $$(2+4+6 +8+10)-(1+3+5 +7+9)=$$． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$15$$ 
Answer: B
Analysis: $$(2-1)+(4 -3)+(6-5)+(8-7)+(10-9)=5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Calculate:  $$401 + 402 + 403 + 404 + 405 = $$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$2010$$ 
aoVal: B
content: $$2008$$ 
aoVal: C
content: $$1610$$ 
aoVal: D
content: $$2015$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: $$400 \times 5 + 1+2+3+4+5 = 2000 + 15 = 2015$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Among these numbers, $-4.3$, $$12\%$$, $0$, $$\frac{2}{5}$$, $-9.97$, $$-\frac{20}{21}$$, how many numbers are negative? 
Options:
aoVal: A
content: two 
aoVal: B
content: three 
aoVal: C
content: four 
aoVal: D
content: five 
Answer: B
Analysis: $-4.3$, $-9.97$, and $$-\frac{20}{21}$$ are negative numbers. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Negative Numbers
Source: 其它
Problem: How many different isosceles triangles have integer side lengths and perimeter $23$ ? (2005 AMC 8 Problem, Question \#15) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$11$$ 
Answer: C
Analysis: let $a$ be legs, $b$ be the base.  $a: 1,2,3,4,5,6,7,8,9,10,11$  $b:21,19,17,15,13,11,9,7,5,3,1$  Since $2a\textgreater b$, $a\textgreater5$, there are $6$ possible $a$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: Robin had a great meal in a local Thai restaurant and he spent $34$ dollars on the dishes. The tax is $10\textbackslash\%$ of the dishes and Robin also needed to pay for the tip after paying the food and tax. Robin brought $60$ dollars with him and he\textquotesingle s thinking about how many tips should he give.  Which of the following is an inequality that represents the situation? 
Options:
aoVal: A
content: $x=60-34\times 10\textbackslash\%-34$, let $x$ be the amount of tips 
aoVal: B
content: $x=60-34$, let $x$ be the amount of tips 
aoVal: C
content: $x \leq 60-34\times 10\textbackslash\%-34$, let $x$ be the amount of tips 
aoVal: D
content: $x\leq 60-34$, let $x$ be the amount of tips 
Answer: C
Analysis: Robin needs to pay the food and taxes, the rest he can decide how much he wants to pay for tips. Therefore, the tip should be represented using inequality. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: The $2021^{}\text{st}$ digit to the right of the decimal point in the decimal expansion of $\dfrac{5}{37}$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: B
Analysis: $$\frac{5}{37}=0.\overline{135}$$, it is a decimal which repeats in cycles of $3$ digits. $2021\div 3=673$$R2$, so the $2021$$^{st}$ digit is $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals
Problem: Martin and Dai divide some sweets between them. There are $65$ sweets, and Martin takes $7$ more than Dai. How many does Martin take? 
Options:
aoVal: A
content: $$29$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$35$$ 
aoVal: E
content: $$36$$ 
Answer: E
Analysis: $(65 + 7) \div 2 = 36$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
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: $${{a}\_{n}}=\frac{{{\left( n+1 \right)}^{2}}}{\left( n+1+1 \right)\left( n+1-1 \right)}=\frac{{{\left( n+1 \right)}^{2}}}{n\left( n+2 \right)}$$.  $$=\frac{2\times 2}{(2+1)\times (2-1)}\times \frac{3\times 3}{(3+1)\times (3-1)}\times \frac{4\times 4}{(4+1)\times (4-1)}\times \cdots \times \frac{98\times 98}{(98+1)\times (98-1)}\times \frac{99\times 99}{(99+1)\times (99-1)}$$  $$=\frac{2\times 2}{3\times 1}\times \frac{3\times 3}{4\times 2}\times \frac{4\times 4}{5\times 3}\times \frac{5\times 5}{6\times 4}\times \cdots \times \frac{98\times 98}{99\times 97}\times \frac{99\times 99}{100\times 98}$$  $$=\frac{2}{1}\times \frac{2}{3}\times \frac{3}{2}\times \frac{3}{4}\times \frac{4}{3}\times \frac{4}{5}\times \cdots \times \frac{98}{97}\times \frac{98}{99}\times \frac{99}{98}\times \frac{99}{100}$$  $$=\frac{2}{1}\times \frac{99}{100}$$  $$=\frac{99}{50}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas
Problem: Three positive integers are equally spaced on a number line. The middle number is $15$, and the largest number is $4$ times the smallest number. What is the smallest of these three numbers? (2022 AMC 8 Problems, Question \#6) 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: C
Analysis: Let the smallest number be $x$. It follows that the largest number is $4 x$.  Since $x, 15$, and $4 x$ are equally spaced on a number line, we have  $$ \begin{aligned} 4 x-15 \& =15-x \textbackslash\textbackslash{} 5 x \& =30 \textbackslash\textbackslash{} x \& =(C) 6. \end{aligned} $$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Calculate：$$100-99+98-97+96-95+\cdots +4-3+2-1=$$． 
Options:
aoVal: A
content: $$40$$ 
aoVal: B
content: $$45$$ 
aoVal: C
content: $$49$$ 
aoVal: D
content: $$50$$ 
Answer: D
Analysis: $$=\left( 100-99 \right)+\left( 98-97 \right)+\left( 96-95 \right)+\cdots +\left( 4-3 \right)+\left( 2-1 \right)$$  $$=1+1+1+\cdots +1+1=50$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Problem: $$12:3=20:$$． 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$8$$ 
Answer: B
Analysis: $$\frac{12}{3}=\frac{4}{1}=\frac{20}{5}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Proportions
Problem: What is $$40\textbackslash\%$$ of $$28.5$$ plus $$28.5\textbackslash\%$$ of $$60$$? ． 
Options:
aoVal: A
content: $$14.25 $$ 
aoVal: B
content: $$28.5 $$ 
aoVal: C
content: $$42.75 $$ 
aoVal: D
content: $$57 $$ 
aoVal: E
content: $$71.25$$ 
Answer: B
Analysis: It is a general rule that $$x\textbackslash\%$$ of $$y$$ equals $$y\textbackslash\%$$ of $$x$$. This is because $$x\textbackslash\%$$ of $$y= \frac{x}{100} \times y= \frac{xy}{100}$$ and $$y \textbackslash\% $$ of $$x= \frac{y}{100} \times x= \frac{yx}{100}$$.  So $$40\textbackslash\%$$ of $$28.5$$ plus $$28.5\textbackslash\%$$ of $$60 = 40\textbackslash\%$$ of $$28.5$$ plus $$60\textbackslash\%$$ of $$28.5 = 100\textbackslash\%$$ of $$28.5 = 28.5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Extracting Common Factors from Decimals
Problem: In an arithmetic sequence, the $1$\textsuperscript{st} number is $5$, the $2$\textsuperscript{nd} number is $8$, the $3$\textsuperscript{rd} number is $11$, and so on. What is the $25$\textsuperscript{th} term of this sequence? 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$77$$ 
Answer: D
Analysis: The $25$\textsuperscript{th} number: $5+(25-1)\times3=77$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: In the arithmetic sequence: $77$, $86$, $95$, $104$, $\cdots$, the $14$\textsuperscript{th}~term is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$203$$ 
aoVal: B
content: $$181$$ 
aoVal: C
content: $$194$$ 
aoVal: D
content: $$212$$ 
aoVal: E
content: $$185$$ 
Answer: C
Analysis: $77+9\times (14-1)=194$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Source: 其它
Problem: In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term in the sequence is $4000$. What is the first term? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$10$$ 
Answer: A
Analysis: D 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$94$$ is $$49$$ more than． 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$54$$ 
aoVal: D
content: $$55$$ 
Answer: A
Analysis: $$94 - 49 = 45$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: What is $$0.9949$$ when rounded to the nearest hundredth? 
Options:
aoVal: A
content: $$0.99$$ 
aoVal: B
content: $$0.994$$ 
aoVal: C
content: $$0.995$$ 
aoVal: D
content: $$1.00$$ 
Answer: A
Analysis: Since the thousandth\textquotesingle s digit is $$4$$, round $$0.9949$$ down to $$0.99$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals
Problem: There are four types of number machines: $$A$$, $$B$$, $$C$$, $$D$$.  Device A: Add $$3$$ to the input number;  Device B: Divide the input number by $$3$$;  Device C: Subtract $$3$$ from the input;  Device D: Multiply the input number by $$3$$.  These devices can be connected. If the device $$A$$ is followed by the device $$B$$, it is written as $$A-B$$. For example, input $$6$$, $$6 + 3 = 9, 9 \div3 = 3 $$. Thus the output would be $$3$$. If you input a number in the device $$B$$-$$A$$-$$D$$-$$C$$, and get $$60$$, the input number was ． 
Options:
aoVal: A
content: $$54$$ 
aoVal: B
content: $$55$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$32$$ 
Answer: A
Analysis: $$B$$-$$A$$-$$D$$-$$C$$ has the rule: divide the input number by $$3$$; add $$3$$ to the input number;~multiply the input number by $$3$$; subtract $$3$$ from the input. So the inverse operation gives $$((60+3)\div3-3)\times3=54$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition
Problem: $$1+3\frac{1}{6}+5\frac{1}{12}+7\frac{1}{20}+9\frac{1}{30}+11\frac{1}{42}=$$． 
Options:
aoVal: A
content: $$36\frac{5}{14}$$ 
aoVal: B
content: $$25\frac{5}{14}$$ 
aoVal: C
content: $$36\frac{1}{3}$$ 
aoVal: D
content: $$25\frac{1}{3}$$ 
Answer: A
Analysis: $$\begin{eqnarray}\&\&1+3\frac{1}{6}+5\frac{1}{12}+7\frac{1}{20}+9\frac{1}{30}+11\frac{1}{42}\textbackslash\textbackslash{} \&=\&1+3+5+7+9+11+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\textbackslash\textbackslash{} \&=\&\left[ (1+9)+(3+7)+(5+11) \right]+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\left( \frac{1}{4}-\frac{1}{5} \right)+\left( \frac{1}{5}-\frac{1}{6} \right)+\left( \frac{1}{6}-\frac{1}{7} \right)\textbackslash\textbackslash{} \&=\&36+\left( \frac{1}{2}-\frac{1}{7} \right)\textbackslash\textbackslash{} \&=\&36\frac{5}{14}\end{eqnarray}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Problem: There are some identical candies on a electronic scale. The little bear wants to eat some of them. After eating $3$ candies, the scale shows $105$ grams. After eating a total of $5$ candies, the scale shows $75$ grams. How many candies are on the scale at first? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$11$$ 
Answer: D
Analysis: The quality of each candy: $(105 - 75) \div 2 = 15$ g  The amount of the apples at first: $75~ \div 15 + 5 = 10$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas, would like to cover the package with a large number of stamps. Suppose he has a collection of $5$-cent, $10$-cent, and $25$-cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $\textbackslash$7.10$ in postage? 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$46$$ 
aoVal: C
content: $$51$$ 
aoVal: D
content: $$54$$ 
aoVal: E
content: $$55$$ 
Answer: E
Analysis: E 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the second to last digit when $$5^{7}$$ is calculated? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: The second to last digit is always $$2$$. For example $$5^{3}$$ is $$125$$. When multiplying this by $$5$$, we see that is it inevitable that the second to last digit remains $$2$$. Try repeated multiplication of $$5$$ on a calculator. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers
Problem: If $$x=10$$, what is the value of $$(8x+2)^{2}$$? 
Options:
aoVal: A
content: $$6724$$ 
aoVal: B
content: $$6402$$ 
aoVal: C
content: $$6416$$ 
aoVal: D
content: $$6714$$ 
Answer: A
Analysis: $$(8x)^{2}+2\cdot(8x)\cdot2+2^{2}=8^{2}x^{2}+32x+4=8^{2}10^{2}+32\times10+4=6724$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas->Perfect Square Factorization
Problem: What is the value of the letter $H$? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$0$$ 
Answer: A
Analysis: A 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Patterns->Encryption and Decryption
Source: 其它
Problem: Starting with some gold coins and some empty treasure chests, I tried to put $8$ gold coins in each treasure chest, but that makes $1$ gold short to fill all the chests. So instead I put $6$ gold coins in each treasure chest, but then I had $15$ gold coins left over. How many gold coins did I have? ( adapted from 2017 AMC8, Questions \#17) 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$49$$ 
aoVal: D
content: $$63$$ 
aoVal: E
content: $$81$$ 
Answer: D
Analysis: We can represent the amount of gold with $g$ and the amount of chests with $c$. We can use the problem to make the following equations: $$ \begin{gathered} 8 c-1=g \textbackslash\textbackslash{} 6 c+15=g \end{gathered} $$  Therefore, $8 c-1=6 c+15$. This implies that $c=8$. We therefore have $g=63$. So, our answer is (D) $63$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with Multiple Variables
Source: 其它
Problem: Which does NOT have $-1$ as a solution?~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $2 x-4\textless-5$ 
aoVal: B
content: $1-x \geq 2$ 
aoVal: C
content: $2+x=1$ 
aoVal: D
content: $\frac{2}{x}\textgreater x$ 
Answer: D
Analysis: $A$ gives $-6\textless-5$. True.  $B$ gives $2 \geq 2$. True.  $C$ gives 1=1. True.  $D$ gives $-2\textgreater-1$. False. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\frac{1}{3}$ of all the ninth graders are paired with $\frac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy? (2015 AMC 8 Problems, Question \#16) 
Options:
aoVal: A
content: $\frac{2}{15}$ 
aoVal: B
content: $\frac{4}{11}$ 
aoVal: C
content: $\frac{11}{30}$ 
aoVal: D
content: $\frac{3}{8}$ 
aoVal: E
content: $\frac{11}{15}$ 
Answer: B
Analysis: Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\frac{n}{3}=\frac{2 s}{5}$, which simplifies to $n=\frac{6 s}{5}$. Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2 s}{5}}{n+s}$, we can just substitute $\frac{6 s}{5}$ for $n$ into the equation. We then get a value of $\frac{\frac{6 s}{5}+\frac{2 s}{5}}{\frac{6 s}{5}+s}=\frac{\frac{6 s+6 s}{15}}{\frac{11 s}{5}}=\frac{\frac{4 s}{5}}{\frac{11 s}{5}}=\left(\right.$ B) $\frac{4}{11}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$10-9+8-7+6-5+ 4-3+2-1=$$． 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$1$$ 
Answer: B
Analysis: $$10-9+8-7+6-5+4-3+2-1=1+1+1+1+1=5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Twelve friends met for dinner at Oscar\textquotesingle s Overstuffed Oyster House, and each ordered one meal. The portions were so large, there was enough food for $18$ people. If they shared, how many meals should they have ordered to have just enough food for the $12$ of them? (2004 AMC 8 Problems, Question \#3) 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$18$$ 
Answer: A
Analysis: Set up the proportion $\frac{12 \text { meals }}{18 \text { people }}=\frac{x \text { meals }}{12 \text { people }}$. Solving for $x$ gives us $x=8$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它