math-eval/TAL-SCQ5K huggingface dataset:
Problem: Which numbers should be filled in the parentheses to make the equation correct?  $\frac14=\frac{(\textasciitilde\textasciitilde\textasciitilde\textasciitilde)}{16}=\frac8{(\textasciitilde\textasciitilde\textasciitilde\textasciitilde)}$ 
Options:
aoVal: A
content: $4; 20$ 
aoVal: B
content: $16; 32$ 
aoVal: C
content: $4; 32$ 
aoVal: D
content: $4; 40$ 
Answer: C
Analysis: When the denominator and numerator are multiplied by the same number, the value of the fraction remains equal. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^{\text {th }}$-graders to $6^{\text {th }}$-graders is $5: 3$, and the the ratio of $8^{\text {th }}$-graders to $7^{\text {th }}$ graders is $8: 5$. What is the smallest number of students that could be participating in the project? (2013 AMC 8, Question 16) 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$55$$ 
aoVal: D
content: $$79$$ 
aoVal: E
content: $$89$$ 
Answer: E
Analysis: We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8 th graders, in order that we can put the two ratios together: $$ \begin{aligned} \&5: 3=5(8): 3(8)=40: 24 \textbackslash\textbackslash{} \&8: 5=8(5): 5(5)=40: 25 \end{aligned} $$ Therefore, the ratio of 8th graders to 7th graders to 6th graders is $40: 25: 24$. Since the ratio is in lowest terms, the smallest number of students participating in the project is $$ 40+25+24=\text { (E) } 89 $$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: What is the largest number which is both less than $$2\times3\times5\times7$$ and also a divisor of $$2 \times 3 \times 5\times 7$$? 
Options:
aoVal: A
content: $$357$$ 
aoVal: B
content: $$2357$$ 
aoVal: C
content: $$2 \times5 \times7$$ 
aoVal: D
content: $$3 \times5 \times7$$ 
Answer: D
Analysis: The largest divisor of $$2\times3\times5\times7$$ that is less than~ is $$3\times5\times7$$, which we get by dropping the smallest factor. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: Which of the following is not an algebraic expression? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $\frac{1}{h}$ 
aoVal: C
content: $3x=5y$ 
aoVal: D
content: $xyzabc$ 
Answer: C
Analysis: equation is not algebraic expression 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Source: 其它
Problem: $$2^{5}$$ means $$2$$ multipled by itself $$5$$ times, i.e. $$2^{5}=2\times2\times2\times2\times2=32$$. What is $$3^{4}$$ equal to? 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$81$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: $$3\times3\times3\times3=81$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Given that $$\left[ x+0.19 \right]+\left[ x+0.20 \right]+\left[ x+0.21 \right]+\cdots \cdots +\left[ x+0.91 \right]=546$$. Find $$\left[ 100x \right]$$ 
Options:
aoVal: A
content: $$793$$ 
aoVal: B
content: $$737$$ 
aoVal: C
content: $$757$$ 
aoVal: D
content: $$743$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Problem: Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132$. (2016 AMC 8 Problems, Question \#11) 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$12$$ 
Answer: B
Analysis: We can write the two digit number in the form of $10 a+b$; reverse of $10 a+b$ is $10 b+a$. The sum of those numbers is:  $$ \begin{gathered} (10 a+b)+(10 b+a)=132 \textbackslash\textbackslash{} 11 a+11 b=132 \textbackslash\textbackslash{} a+b=12 \end{gathered} $$  We can use brute force to find order pairs $(a, b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$.  Thus our ordered pairs are $(3,9) ;(4,8) ;(5,7) ;(6,6) ;(7,5) ;(8,4) ;(9,3)$ or $(\mathbf{B}) 7$ ordered pairs. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers
Source: 其它
Problem: There is an equal number of motorcycles and cars in the parking lot. They have $$48$$ wheels in total. How many cars are there in the parking lot? 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$24$$ 
Answer: A
Analysis: Let $$x$$ be the number of cars.  $$2x+4x=48$$  $$6x=48$$  $$x=8$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with one Variable
Source: 其它
Problem: Judy~ is waiting in a line to buy some toys. There are $7$ people in front of her, and $3$ people behind her. In total, how many people are there in the line? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$13$$ 
Answer: B
Analysis: $3 + 7 + 1 = 11$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the value of $1+3+5+\ldots+2017+2019-2 -4-6-\ldots-2016-2018$? (Adapted from $2018$ AMC 8 Problem, Question \#5) 
Options:
aoVal: A
content: $$1007$$ 
aoVal: B
content: $$1008$$ 
aoVal: C
content: $$1009$$ 
aoVal: D
content: $$1010$$ 
aoVal: E
content: $$1011$$ 
Answer: D
Analysis: $1+3+5+\ldots+2017+2019-2 -4-6-\ldots-2016-2018$  $=1+(3-2)+(5-4)+\cdots +(2017-2016)+(2019-2018)$  $=1+1+1+\cdots +1+1$  $=1010$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Grouping in Fast Addition and Subtraction of Whole Numbers
Source: 其它
Problem: At an ice cream shop, a sundae is made by selecting two flavors of ice cream and topping them with fudge, whipped cream, nuts, and a cherry. The available flavors of ice cream are chocolate, vanilla, strawberry, rocky road, chocolate chip cookie dough, and mint chip. How many difference sundae combinations are possible? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$120$$ 
aoVal: E
content: $$720$$ 
Answer: B
Analysis: ${6\choose 2} = \frac{6!}{2!(6-2)!} =15$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Six pencils and two erasers cost $$3.20$$ dollars. One pencil costs $$40$$ cents. How much does one eraser cost? (Adapted from 2000 Math Kangaroo Problem, Level 3-4, Question \#5) 
Options:
aoVal: A
content: $$20$$ cents 
aoVal: B
content: $$30$$ cents 
aoVal: C
content: $$40$$ cents 
aoVal: D
content: $$50$$ cents 
aoVal: E
content: $$60$$ cents 
Answer: C
Analysis: $3.20$ dollars = $320$ cents  $(320-40\times6)\div2=40$ cents 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division->Division of Whole Numbers->Division out of the Multiplication Table
Problem: Kitty writes down a sequence of five integers. The rule she uses is, "after the first two terms, each term is the sum of the two previous terms." She sequence is~\uline{~~~~~~~~~~}~,~\uline{~~~~~~~~~~}~,~\uline{~~~~~~~~~~}~, $$18$$, $$29$$.  What is her first term? ． 
Options:
aoVal: A
content: $$0 $$ 
aoVal: B
content: $$ 3 $$ 
aoVal: C
content: $$ 4 $$ 
aoVal: D
content: $$ 5 $$ 
aoVal: E
content: $$ 7$$ 
Answer: C
Analysis: Let the first six terms of Kitty\textquotesingle s sequence be $$a$$, $$b$$, $$c$$, $$18$$ and $$29$$ respectively.  Then $$c+ 18= 29$$, so $$c= 11$$. Hence $$b+11= 18$$, so $$b=7$$.  Therefore, $$a+7=11$$, so $$a=4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Problem: \textbf{A set of 5,000 scores on a college readiness exam are known to be approximately normally distributed with a mean of 72 and a standard deviation of 6. To the nearest integer value, approximately how many scores are between 63 and 75.} 
Options:
aoVal: A
content: $$0.6247$$ 
aoVal: B
content: $$4115$$ 
aoVal: C
content: $$3650$$ 
aoVal: D
content: $$3123$$ 
aoVal: E
content: $$3227$$ 
Answer: D
Analysis: \textbf{P(63 \textless{} X \textless{} 75) calculator: normalcdf(63, 75, 72, 6) = 0.6247~}  \textbf{0.6247*5000 = 3123.3} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: If $$\frac{2\textbar x-2\textbar+1}{3}\textless{}1$$, the range of $x$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$-3\textless x\textless-1$$ 
aoVal: B
content: $$x\textgreater-1$$ or $$x\textless-3$$ 
aoVal: C
content: $$1\textless x\textless3$$ 
aoVal: D
content: $$x\textgreater3$$ or $$x\textless1$$ 
Answer: C
Analysis: $$\frac{2\textbar x-2\textbar+1}{3}\textless{}1$$  $$2\left\textbar x-2\right\textbar+1\textless3$$  $$2\left\textbar x-2\right\textbar\textless2$$  $$1\left\textbar x-2\right\textbar\textless1$$  $$-1\textless x-2\textless1$$  $$1\textless x\textless3$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: $$900+90+9+1=$$． 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$991$$ 
aoVal: C
content: $$1000$$ 
aoVal: D
content: $$9000$$ 
Answer: C
Analysis: $$900+90+9+1=999+1=1000$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Alicia and Emily agreed to meet at the cinema at 3.55pm. Emily left her house at 1.47pm but arrived at the cinema 17 minutes late. How long was Emily\textquotesingle s journey from her house to the cinema? 
Options:
aoVal: A
content: $$189$$ minutes 
aoVal: B
content: $$172$$ minutes 
aoVal: C
content: $$216$$ minutes 
aoVal: D
content: $$206$$ minutes 
aoVal: E
content: None of the above. 
Answer: E
Analysis: 3: 55pm - 1: 47om + 15 minutes = 145 minutes. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion
Source: 其它
Problem: $$55 + 55 + 55 = 44 + 44 +$$ . 
Options:
aoVal: A
content: $$33$$ 
aoVal: B
content: $$44$$ 
aoVal: C
content: $$66$$ 
aoVal: D
content: $$77$$ 
Answer: D
Analysis: $$55+55+55 = 44+11+44+11+55 = 44+44+(11+11+55)= 44+44+77$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: The average weight of dogs that come to a certain vet\textquotesingle s office is 55.6 lbs, with a standard deviation of 2.2 lbs. If the weights are normally distributed, what percent of dogs weight more than 60 lbs? 
Options:
aoVal: A
content: 66.8\% 
aoVal: B
content: 47.2\% 
aoVal: C
content: 33.4\% 
aoVal: D
content: 15.9\% 
aoVal: E
content: 2.28\% 
Answer: E
Analysis: $P(X\textgreater60) = P(Z\textgreater\frac{60-55.6}{2.2}) = p(Z\textgreater2) =1-0.9772 = 0.0228$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: There are four more girls than boys in Ms. Raub\textquotesingle s class of 28 students. What is the ratio of number of girls to the number of boys in her class? (2014 AMC 8, Question 7) 
Options:
aoVal: A
content: $3:4$ 
aoVal: B
content: $4:3$ 
aoVal: C
content: $3:2$ 
aoVal: D
content: $7:4$ 
aoVal: E
content: $2:1$ 
Answer: B
Analysis: We can set up an equation with $x$ being the number of girls in the class. The number of boys in the class is equal to $x-4$. Since the total number of students is equal to 28 , we get $x+x-4=28$. Solving this equation, we get $x=16$. There are $16-4=12$ boys in our class, and our answer is $16: 12=$ (B) $4: 3$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: There is a sequence of squares of some natural numbers: $1$, $4$, $9$, $16$, $\cdots$ One of the numbers in this sequence is $10^{8}$. What is the next number after $10^{8}$? ($2001$ Math Kangaroo Problem, Level $$11$$-$$12$$, Question \#$14$) 
Options:
aoVal: A
content: $\left (10^{4}+1\right )^{2}$ 
aoVal: B
content: $\left (10^{8}+1\right )^{2}$ 
aoVal: C
content: $\left (10^{5}\right )^{2}$ 
aoVal: D
content: $\left (10^{8}\right )^{2}$ 
aoVal: E
content: $\left (10^{4}\right )^{2}+1$ 
Answer: A
Analysis: As we know, $1$, $4$, $9$, $16$ are square numbers. $10^{8}=\left (10^{4}\right )^{2}$, so, the next one is $\left (10^{4}+1\right )^{2}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Application of Powers
Problem: Which is the solution of the following inequality:  $-11x-10\geq30-x$. 
Options:
aoVal: A
content: $x\leq-4$ 
aoVal: B
content: $x\geq-4$ 
aoVal: C
content: $x\leq-2$ 
aoVal: D
content: $x\geq-2$ 
Answer: A
Analysis: $-10x\leq40$  $x\leq-4$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: Mary went to the stationery store and she bought three pencils, a notebook, and a stationery box. A pencil costs $1$ dollar, a notebook costs $2$ dollars, a pencil case costs $3$ dollars, and Mary has a $1$ dollar discount coupon. How much did Mary spend in the end?~(adapted from $$2005$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$6$$) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$7$$ 
Answer: E
Analysis: Three pencils are three dollars. Notebook is two dollars. A three dollar pencil case and a one dollar free at last. So the last answer is $3+2+3-1=7$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition and Subtraction of Whole Numbers->Questions Involving Addition and Subtraction
Problem: Mike has $7y+1$ cans of coke. He gives his friends $$x$$ cans of coke. And he takes one out. How many cans of coke does he have left? 
Options:
aoVal: A
content: $$7y + 1$$ 
aoVal: B
content: $$7y + 1 -x$$ 
aoVal: C
content: $$7y-x$$ 
aoVal: D
content: $$7y$$ 
Answer: C
Analysis: $$7y+1-x-1=7y-x$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: The sum of the whole numbers from $$1$$ through $$100$$ is $$5050$$. What is the sum of the whole numbers from $$1$$ through $$200$$? 
Options:
aoVal: A
content: $$5150$$ 
aoVal: B
content: $$10100$$ 
aoVal: C
content: $$11050$$ 
aoVal: D
content: $$20100$$ 
Answer: D
Analysis: ($$1$$ to $$200$$)$$=$$($$1$$ to $$100$$)$$+[(100+1)+(100+2)+\cdots +(100+100)]=$$($$1$$ to $$100$$)$$+[(100\times 100)+$$($$1$$ to $$100$$)$$]=5050+[10000+5050]=20100$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: Which of the following is the same as 2018 cm? 
Options:
aoVal: A
content: 2 metres and 18 centimetres 
aoVal: B
content: 2 kilometres and 18 centimetres 
aoVal: C
content: 20 metres and 18 centimetres 
aoVal: D
content: 201 metres and 8 centimetres 
aoVal: E
content: None of the above 
Answer: C
Analysis: 1m = 100cm; 1km=1000m  A. 200 cm + 18 cm = 218 cm  B. 200 000cm + 18cm = 200 018cm  \textbf{C. 2000cm + 18cm = 2018 cm}  D. 20100cm + 8cm = 20108cm. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion
Source: 其它
Problem: What fraction of the integers from $$1$$ to $$1000$$ inclusive are cubes? ． 
Options:
aoVal: A
content: $$\frac{1}{50}$$ 
aoVal: B
content: $$\frac{1}{100}$$ 
aoVal: C
content: $$\frac{1}{200}$$ 
aoVal: D
content: $$\frac{1}{400}$$ 
aoVal: E
content: $$\frac{1}{800}$$ 
Answer: B
Analysis: As $$1=1^{3}$$ and $$1000 = 10^{3}$$, there are $$10$$ cubes from $$1$$ to $$1000$$. So the fraction of the integers from $$1$$ to $$1000$$ inclusive which are cubes is $$\frac{10}{1000}= \frac{1}{100}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers
Problem: After the final exam results came out, four students wanted to compare who had done best. Who had the best grades?~(adapted from $$2007$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$4$$)  Alice：$12+8$  Bob：$7+9$  Alan：$14+5$~  Tom：$6+8$  Susan: $6+15$ 
Options:
aoVal: A
content: Alice 
aoVal: B
content: Bob 
aoVal: C
content: Alan 
aoVal: D
content: Tom 
aoVal: E
content: Susan 
Answer: E
Analysis: Alice：$12+8=20$  Bob：$7+9=16$  Alan：$14+5=19$~  Tom：$6+8=14$  Susan: $6+15=21$  $21\textgreater20\textgreater19\textgreater16\textgreater14$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering->Comparing and Ordering the Value of Expressions
Problem: $3-(-4)=$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$-7$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$-1$$ 
Answer: A
Analysis: Subtract a negative number is the same as add a postive number with the same absolute value. $a-(-b)=a+b$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Negative Numbers->Operations on Negative numbers
Source: 其它
Problem: Kit played computer games for $225$ minutes. When he stopped gaming and decided to head to sleep, he checked the time and noticed that it was $11.20\text{p.m.}$ What time did he start playing computer games? 
Options:
aoVal: A
content: $7.35\text{p.m.}$ 
aoVal: B
content: $8.35\text{p.m.}$ 
aoVal: C
content: $8.55\text{p.m.}$ 
aoVal: D
content: $9.45\text{p.m.}$ 
Answer: A
Analysis: $225$ minutes $=3$ hours $45$ minutes.  He started playing at $7.35\text{p.m.}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Overseas In-curriculum->Knowledge Point->Measurement->Time->Time Calculation
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? (2013 Math Kangaroo Problem, Level 3 - 4, Question \#6) 
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: One tour bus can seat no more than $$50$$ people. What is the smallest number of buses needed to take $$160$$ people? (Adapted from 2000 Math Kangaroo Problem, Level 3-4, Question \#6) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: D
Analysis: $160\div50=3R10$, $3+1=4$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division->Division of Whole Numbers->Division with Remainders
Problem: If the degree measures of the angles of a convex quadrilateral are in the ratio~$3:4:5:6$, by how many degrees does the measure of the largest angle exceed the measure of the smallest angle? 
Options:
aoVal: A
content: $30^{\circ}$ 
aoVal: B
content: $45^{\circ}$ 
aoVal: C
content: $60^{\circ}$ 
aoVal: D
content: $75^{\circ}$ 
aoVal: E
content: $90^{\circ}$ 
Answer: C
Analysis: $\dfrac{6}{3+4+5+6}\times360^{\circ}-\dfrac{3}{3+4+5+6}\times360^{\circ}=\dfrac{360}{18}\left( 6-3\right)=20\times3=60^{\circ}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: There were five books on the shelf. Later, Jack took away three, and Mike took back four. Jim also took back three. How many books were there on the shelf?~(adapted from $$2005$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$5$$) 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$11$$ 
Answer: C
Analysis: $5-3+4+3=9$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition and Subtraction of Whole Numbers->Questions Involving Addition and Subtraction
Problem: How many different four-digit numbers can be formed by rearranging the four digits in $2004$?~(2004 AMC 8 Problem, Question \#2) 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$81$$ 
Answer: B
Analysis: $2 \times \_3P\_1 \times \_2C\_2=6$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The sum of the first $m$ positive odd integers is 212 more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$ ? 
Options:
aoVal: A
content: $$255$$ 
aoVal: B
content: $$256$$ 
aoVal: C
content: $$257$$ 
aoVal: D
content: $$258$$ 
aoVal: E
content: $$259$$ 
Answer: A
Analysis: The sum of the first $m$ odd integers is given by $m^{2}$. The sum of the first $n$ even integers is given by $n(n+1)$. Thus, $m^{2}=n^{2}+n+212$. Since we want to solve for $n$, rearrange as a quadratic equation: $n^{2}+n+\left(212-m^{2}\right)=0$. Use the quadratic formula: $n=\frac{-1+\sqrt{1-4\left(212-m^{2}\right)}}{2}$. Since $n$ is clearly an integer, $1-4\left(212-m^{2}\right)=4 m^{2}-847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer.  Let $x=\sqrt{4 m^{2}-847}$; note that this means $n=\frac{-1+x}{2}$. It can be rewritten as $x^{2}=4 m^{2}-847$, so $4 m^{2}-x^{2}=847$. Factoring the left side by using the difference of squares, we get $(2 m+x)(2 m-x)=847=7 \cdot 11^{2}$. Our goal is to find possible values for $x$, then use the equation above to find $n$. The difference between the factors is $(2 m+x)-(2 m-x)=2 m+x-2 m+x=2 x$. We have three pairs of factors, $847 \cdot 1,121 \cdot 7$, and $77 \cdot 11$. The differences between these factors are 846,114 , and 66 - those are all possible values for $2 x$. Thus the possibilities for $x$ are 423,57 , and 33 . Now plug in these values into the equation $n=\frac{-1+x}{2}$, so $n$ can equal 211,28 , or 16 , hence the answer is 255 . 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Unary Quadratic Equations
Source: 其它
Problem: 1. In the number 98, the digit "9" is in the ones place.~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: Yes 
aoVal: B
content: No 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the area of the triangle formed by the lines $y=5, y=1+x$, and $y=1-x$? (2019 AMC 8 Problems, Question \#21) 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$16$$ 
Answer: E
Analysis: $y=x+1$ and $y=-x+1$ have $y$-intercepts at $(0,1)$ and slopes of 1 and $-1$, respectively. Since the product of these slopes is $-1$, the two lines are perpendicular. From $y=5$, we see that $(-4,5)$ and $(4,5)$ are the other two intersection points, and they are 8 units apart. By symmetry, this triangle is a $45-45-90$ triangle, so the legs are $4 \sqrt{2}$ each and the area is $\frac{(4 \sqrt{2})^{2}}{2}=(\mathbf{E}) 16$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: 1. In the number 98, the digit "9" is in the ones place. 
Options:
aoVal: A
content: Yes 
aoVal: B
content: No 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它