Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "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 $$2+1=3$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ] ]

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

[ "$$5$$ apples = $$16$$ pineapples $$5\\times3$$ apples = $15$ apples $$16\\times3$$ pineapples = $48$ pineapples " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "Fibonacci number. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$-19$ " } ], [ { "aoVal": "B", "content": "$-18$ " } ], [ { "aoVal": "C", "content": "$18$ " } ], [ { "aoVal": "D", "content": "$0$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Negative Numbers" ]

[ "The opposite of $-18$ is $18$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$11$$ " } ] ]

[ "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 $$\\frac{x+6 x}{3 x-2x}=\\frac{7 x}{x}=\\text { (C) } 7$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

[ "$(12 + 4) \\div 2 = 8$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$15.5$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$16.5$$ " } ], [ { "aoVal": "E", "content": "$$17$$ " } ] ]

[ "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 $\\frac{8+18+7}{2}= 16.5$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$24$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

[ "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 $(\\mathbf{A}) 2$. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers" ]

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$420$$ " } ], [ { "aoVal": "C", "content": "$$3600$$ " } ], [ { "aoVal": "D", "content": "$$7200$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions" ]

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

[]

[ [ { "aoVal": "A", "content": "$$3800$$ " } ], [ { "aoVal": "B", "content": "$$3900$$ " } ], [ { "aoVal": "C", "content": "$$3945$$ " } ], [ { "aoVal": "D", "content": "$$3960$$ " } ], [ { "aoVal": "E", "content": "$$4000$$ " } ] ]

[ "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 $$4000$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ], [ { "aoVal": "E", "content": "$$15$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

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

[]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Proportions" ]

[ "$\\frac{2}{3}=\\frac{8}{12}$ " ]

[]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

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

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$57$$ " } ], [ { "aoVal": "C", "content": "$$58$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ], [ { "aoVal": "E", "content": "$$61$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

[ "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) $60$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$53$$ " } ], [ { "aoVal": "B", "content": "$$54$$ " } ], [ { "aoVal": "C", "content": "$$69$$ " } ], [ { "aoVal": "D", "content": "$$70$$ " } ] ]

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

[ "B " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

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

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

[]

[ [ { "aoVal": "A", "content": "$$401$$ " } ], [ { "aoVal": "B", "content": "$$455$$ " } ], [ { "aoVal": "C", "content": "$$428$$ " } ], [ { "aoVal": "D", "content": "$$431$$ " } ], [ { "aoVal": "E", "content": "$$530$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers" ]

[ "$$531-103=428$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$55$$ " } ], [ { "aoVal": "C", "content": "$$79$$ " } ], [ { "aoVal": "D", "content": "$$2022$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

[ "$55\\div 5 =11$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$2$$ " } ], [ { "aoVal": "E", "content": "$$1$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions" ]

[ "$$8\\times \\frac{1}{4}=2$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$171$$ " } ], [ { "aoVal": "B", "content": "$$167$$ " } ], [ { "aoVal": "C", "content": "$$164$$ " } ], [ { "aoVal": "D", "content": "$$160$$ " } ] ]

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

[ "$$7+4\\times (41-1)=167$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\overset{\\frown}{Rounds}=-0.6442+22.94\\textbackslash{} Minutes$ " } ], [ { "aoVal": "B", "content": "$\\overset{\\frown}{Rounds}=22.94+0.5466\\textbackslash{} Minutes$ " } ], [ { "aoVal": "C", "content": "$\\overset{\\frown}{Minutes}=22.94+2.866\\textbackslash{} Rounds$ " } ], [ { "aoVal": "D", "content": "$\\overset{\\frown}{Minutes}=22.94-0.6442\\textbackslash{} Rounds$ " } ], [ { "aoVal": "E", "content": "$\\overset{\\frown}{Minutes}=-0.6442+0.5466\\textbackslash{} Rounds$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

[ "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). " ]

[]

[ [ { "aoVal": "A", "content": "$$\\frac{11}{6250}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{23}{12500}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{6}{3125}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{13}{6250}$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Complex Fractions" ]

[ "$$\\frac{1}{5^{5}+1}+ \\frac{1}{5^{5}+5}+ \\frac{1}{5^{5}+5^{2}}+ \\cdots + \\frac{1}{5^{5}+5^{8}}+ \\frac{1}{5^{5}+5^{9}}+ \\frac{1}{5^{5}+5^{10}}$$ $$=\\left ( \\frac{1}{5^{5}+1}+ \\frac{1}{5^{5}+5^{10}}\\right )+\\left ( \\frac{1}{5^{5}+5}+ \\frac{1}{5^{5}+5^{9}}\\right )$$$$+\\left ( \\frac{1}{5^{5}+5^{2}}+ \\frac{1}{5^{5}+5^{8}}\\right )+\\cdots+\\left (\\dfrac{1}{5^{5}+5^{5}}\\right )$$ $$=\\left ( \\frac{5^{5}+1}{5^{5}\\left (5^{5}+1\\right )}\\right )+\\left ( \\frac{5^{4}+1}{5^{5}(5^{4}+1)}\\right )$$$$+\\left ( \\frac{5^{3}+1}{5^{5}(5^{3}+1)}\\right )+ \\cdots +$$ $$\\left (\\frac{1}{2\\left (5\\right )^{5}}\\right )$$ $$= \\left (\\frac{5}{5^{5}}\\right )+\\left ( \\frac{1}{2(5)^{5}}\\right )= \\frac{11}{6250}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\frac{9}{5}(0.4)+32$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{9}{5}(0.4)$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{9}{5}(0.4)^{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{9}{5}^{2}(0.4)$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{9}{5}^{2}(0.4)^{2}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$39$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$210$$ " } ], [ { "aoVal": "D", "content": "$$400$$ " } ], [ { "aoVal": "E", "content": "$$401$$ " } ] ]

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

[ "On day $20$, she sold $1+2\\times(20-1)=39$ Sum of $20$ days: $(1+39)\\times 20\\div2=400$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{462}$ " } ], [ { "aoVal": "B", "content": "$\\frac{1}{231}$ " } ], [ { "aoVal": "C", "content": "$\\frac{1}{132}$ " } ], [ { "aoVal": "D", "content": "$\\frac{2}{213}$ " } ], [ { "aoVal": "E", "content": "$\\frac{1}{22}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$\\frac13$ " } ], [ { "aoVal": "B", "content": "$\\frac34$ " } ], [ { "aoVal": "C", "content": "$$\\frac57$$ " } ], [ { "aoVal": "D", "content": "$\\frac79$ " } ], [ { "aoVal": "E", "content": "$\\frac8{11}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions" ]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$31$$ " } ], [ { "aoVal": "B", "content": "$$41$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$51$$ " } ], [ { "aoVal": "E", "content": "$$60$$ " } ] ]

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

[ "$$27+11+13$$ $$=27+13+11$$ $$=40+11$$ $$=51$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$249$$ " } ], [ { "aoVal": "B", "content": "$$250$$ " } ], [ { "aoVal": "C", "content": "$$251$$ " } ], [ { "aoVal": "D", "content": "$$252$$ " } ], [ { "aoVal": "E", "content": "$$253$$ " } ] ]

[ "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, $a=253$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "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 $9$. " ]

[ "其它" ]

[ [ { "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": "$\\frac{1}{2001}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Fractions" ]

[ "$1\\div 1\\frac{2}{2001}\\div 1\\frac{2}{2003}\\cdots \\div 1\\frac{2}{2023}$ $=1\\div \\frac{2003}{2001}\\div \\frac{2005}{2003}\\cdots \\div \\frac{2025}{2023}$ $=1\\times \\frac{2001}{2003}\\times \\frac{2003}{2005}\\cdots \\times \\frac{2023}{2025}$ $=\\frac{2001}{2025}$ " ]

[]

[ [ { "aoVal": "A", "content": "$$22222$$ " } ], [ { "aoVal": "B", "content": "$2222^{2}$ " } ], [ { "aoVal": "C", "content": "$222^{22}$ " } ], [ { "aoVal": "D", "content": "$22^{222}$ " } ], [ { "aoVal": "E", "content": "$2^{2222}$ " } ] ]

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

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$50$$ " } ], [ { "aoVal": "E", "content": "$$60$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$1+1= 2$$, $$1+2 = 3$$, $$\\cdots$$, $$5+8 = 13$$, $$8+13 = 21$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$6a+4$ " } ], [ { "aoVal": "B", "content": "$2a+7$ " } ], [ { "aoVal": "C", "content": "$6a+12$ " } ], [ { "aoVal": "D", "content": "$32a+34$ " } ] ]

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

[ "$3\\times (2a+4)=3(2a+4)=6a+12$ " ]

[]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ] ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$-5$ " } ], [ { "aoVal": "B", "content": "$\\frac{11}{2}$ " } ], [ { "aoVal": "C", "content": "$10.5$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

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

[ "$2 a+10-4 x=-4 x+21$ $2 a=11 $ $a=\\frac{11}{2}$ " ]

[ "其它" ]

[ [ { "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 " } ] ]

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

[ "15-9=6, 29+7=36 " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$42$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "E " ]

[]

[ [ { "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": "$$\\frac{50}{99}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas" ]

[ "$$=\\frac{2\\times 2}{(2+1)\\times (2-1)}\\times \\frac{3\\times 3}{(3+1)\\times (3-1)}\\times \\frac{4\\times 4}{(4+1)\\times (4-1)}\\times \\cdots \\times \\frac{98\\times 98}{(98+1)\\times (98-1)}\\times \\frac{99\\times 99}{(99+1)\\times (99-1)}$$ $$=\\frac{2\\times 2}{3\\times 1}\\times \\frac{3\\times 3}{4\\times 2}\\times \\frac{4\\times 4}{5\\times 3}\\times \\frac{5\\times 5}{6\\times 4}\\times \\cdots \\times \\frac{98\\times 98}{99\\times 97}\\times \\frac{99\\times 99}{100\\times 98}$$ $$=\\frac{2}{1}\\times \\frac{2}{3}\\times \\frac{3}{2}\\times \\frac{3}{4}\\times \\frac{4}{3}\\times \\frac{4}{5}\\times \\cdots \\times \\frac{98}{97}\\times \\frac{98}{99}\\times \\frac{99}{98}\\times \\frac{99}{100}$$ $$=\\frac{2}{1}\\times \\frac{99}{100}$$ $$=\\frac{99}{50}$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$25$$ " } ] ]

[ "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 $k=22 \\rightarrow(\\mathbf{B})22$. " ]

[]

[ [ { "aoVal": "A", "content": "$$707$$ " } ], [ { "aoVal": "B", "content": "$$907$$ " } ], [ { "aoVal": "C", "content": "$$916$$ " } ], [ { "aoVal": "D", "content": "$$1000$$ " } ], [ { "aoVal": "E", "content": "$$1001$$ " } ] ]

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

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

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "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 $$10$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ], [ { "aoVal": "E", "content": "$$-3$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Calculation Modules" ]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

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

[ "$$36 \\div 5 = 7R1$$, so the answer is $$5$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$-11$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$-2$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Mathematical Thoughts->Absolute Value" ]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "William " } ], [ { "aoVal": "B", "content": "Mark " } ], [ { "aoVal": "C", "content": "Diana " } ], [ { "aoVal": "D", "content": "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$28$$ " } ], [ { "aoVal": "B", "content": "$$29$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$31$$ " } ], [ { "aoVal": "E", "content": "$$32$$ " } ] ]

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

[ "August have $31$ days. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning" ]

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

[ "其它" ]

[ [ { "aoVal": "A", "content": "Yes, he has the winning strategy. " } ], [ { "aoVal": "B", "content": "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "every month has 28 days. " ]

[]

[ [ { "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 " } ] ]

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

[ "Pablo is the name of his son. " ]

[]

[ [ { "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 " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning" ]

[ "nil " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning" ]

[ "March, May, July, August, October and December. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "Yes, it can " } ], [ { "aoVal": "B", "content": "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Ivy has the winning strategy. " } ], [ { "aoVal": "B", "content": "Cathy has the winning strategy. " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Game 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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Mathematical Thoughts->Absolute Value" ]

[ "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 $x=0$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ], [ { "aoVal": "E", "content": "$$18$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles" ]

[ "$11\\times13\\times7=1001$ $1+3+7+0=11$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$190$$ " } ], [ { "aoVal": "B", "content": "$$188$$ " } ], [ { "aoVal": "C", "content": "$$186$$ " } ], [ { "aoVal": "D", "content": "$$142$$ " } ], [ { "aoVal": "E", "content": "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.} " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$63$$ " } ], [ { "aoVal": "B", "content": "$$47$$ " } ], [ { "aoVal": "C", "content": "$$57$$ " } ], [ { "aoVal": "D", "content": "$$59$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Patterns of Figures" ]

[ "1, 3, 7, 15, 31, \\cdots .. 2. 4. 8.~ 16.~ 32 31+32 = 64 " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ] ]

[ "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. $$16\\div (1+2)=5 \\textbackslash{} \\text{R} 1$$ 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. " ]

[]

[ [ { "aoVal": "A", "content": "Lynn gives Sara $5$ coins, and Molly gives Sara $5$ coins. " } ], [ { "aoVal": "B", "content": "Chris gives Molly $5$ coins, and Chris gives Lynn $10$ coins. " } ], [ { "aoVal": "C", "content": "Chris gives Molly $5$ coins, and Chris gives Sara $5$ coins. " } ], [ { "aoVal": "D", "content": "Chris gives Lynn $10$ coins. " } ], [ { "aoVal": "E", "content": "Chris gives Sara $10$ coins. " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$625$$ " } ], [ { "aoVal": "B", "content": "$$620$$ " } ], [ { "aoVal": "C", "content": "$$602$$ " } ], [ { "aoVal": "D", "content": "$$260$$ " } ], [ { "aoVal": "E", "content": "$$206$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles" ]

[ "Start from the ones digit, $$4+7=11$$, $$0+1=1$$, so $$C=0$$. In the tens place, $$8+1+1=10$$, $$0+2=2$$, so $$B=2$$. In the hundreds place, $$8+5+1=14$$, $$14+6=20$$, so $$A=6$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$SURGEON$$ " } ], [ { "aoVal": "B", "content": "$$HARBORS$$ " } ], [ { "aoVal": "C", "content": "$SWEATER$ " } ], [ { "aoVal": "D", "content": "$MESSAGE$ " } ] ]

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

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Magic Square" ]

[ "$A=s^{2}$ $s^{2}=16$ $s=4$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$5$$ " } ] ]

[ "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: $$AA\\xrightarrow{BB}$$ $$AA\\xleftarrow{B}B$$ $$B\\xrightarrow{AA}B$$ $$B\\xleftarrow{B}AA$$ $$\\xrightarrow{BB}AA$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Yes, she can take $5$ balls from group $A$ at first. " } ], [ { "aoVal": "B", "content": "Yes, she can take $3$ balls from group $B$ at first. " } ], [ { "aoVal": "C", "content": "Yes, she can take $6$ balls from group $A$ at first. " } ], [ { "aoVal": "D", "content": "Yes, she can take $5$ balls from group $B$ at first. " } ], [ { "aoVal": "E", "content": "No, she doesn\\textquotesingle t have winning strategy. " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Game 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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Yes, he has. " } ], [ { "aoVal": "B", "content": "No, he doesn\\textquotesingle t. " } ] ]

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

[ "$124\\div(1+5)=20R4$. 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. " ]

[]

[ [ { "aoVal": "A", "content": "$$2015$$ " } ], [ { "aoVal": "B", "content": "$$2016$$ " } ], [ { "aoVal": "C", "content": "$$2017$$ " } ], [ { "aoVal": "D", "content": "$$2018$$ " } ], [ { "aoVal": "E", "content": "$$2019$$ " } ] ]

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

[ "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 $$2017$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Tuesday " } ], [ { "aoVal": "B", "content": "Wednesday " } ], [ { "aoVal": "C", "content": "Thursday " } ], [ { "aoVal": "D", "content": "Friday " } ], [ { "aoVal": "E", "content": "Sunday " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "$25\\div7=3R4$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Tuesday " } ], [ { "aoVal": "B", "content": "Wednesday " } ], [ { "aoVal": "C", "content": "Thursday " } ], [ { "aoVal": "D", "content": "Saturday " } ], [ { "aoVal": "E", "content": "Sunday " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "$28\\div7=4$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$10:20$ A.M. " } ], [ { "aoVal": "B", "content": "$11:00$ A.M. " } ], [ { "aoVal": "C", "content": "$11:10$ A.M. " } ], [ { "aoVal": "D", "content": "$11:20$ A.M. " } ], [ { "aoVal": "E", "content": "$11:30$ A.M. " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "$$45+15+45+15=120$$ minutes = $2$ hours So, he will finish at $$11:00$$ A.M. " ]

[]

[ [ { "aoVal": "A", "content": "$$23$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ], [ { "aoVal": "E", "content": "$$72$$ " } ] ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$1$ hour $40$ minutes " } ], [ { "aoVal": "B", "content": "$2$ hours $10$ minutes " } ], [ { "aoVal": "C", "content": "$2$ hours $20$ minutes " } ], [ { "aoVal": "D", "content": "$1$ hour $50$ minutes " } ], [ { "aoVal": "E", "content": "$50$ minutes " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "$6:40$ - $4:30$ = $2$ hours $10$ minutes " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "Sunday " } ], [ { "aoVal": "B", "content": "Wednesday " } ], [ { "aoVal": "C", "content": "Thursday " } ], [ { "aoVal": "D", "content": "Friday " } ], [ { "aoVal": "E", "content": "Saturday " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning" ]

[ "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$10$ " } ], [ { "aoVal": "B", "content": "$8$ " } ], [ { "aoVal": "C", "content": "$6$ " } ], [ { "aoVal": "D", "content": "$4$ " } ], [ { "aoVal": "E", "content": "$2$ " } ] ]

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

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "Yes, it is possible. " } ], [ { "aoVal": "B", "content": "No, it is impossible. " } ], [ { "aoVal": "C", "content": "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 $7$ 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. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

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

[ "We can create three drawers: $(1, 4, 7, 10, 13, 16)$. $(2, 5, 8, 11, 14, 17)$. $(3, 6, 9, 12, 15)$. At least $3\\times3+1=10$ numbers should be chosen. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$0$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning" ]

[ "$$5+1-4=2$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$2$$ " } ] ]

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

[ "NA " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "7 x 12 " } ], [ { "aoVal": "B", "content": "7 x 2 x 12 " } ], [ { "aoVal": "C", "content": "2 x 7 x 2 x 12 " } ], [ { "aoVal": "D", "content": "(7+7) x 12 " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "Conversion of unit. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "6 pm " } ], [ { "aoVal": "B", "content": "7 pm " } ], [ { "aoVal": "C", "content": "8 pm " } ], [ { "aoVal": "D", "content": "9 pm " } ], [ { "aoVal": "E", "content": "10 pm " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "$$300\\div20-300\\div30=5$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12:00$$ " } ], [ { "aoVal": "B", "content": "$$12:10$$ " } ], [ { "aoVal": "C", "content": "$$12:15$$ " } ], [ { "aoVal": "D", "content": "$$12:20$$ " } ], [ { "aoVal": "E", "content": "$$11:50$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation" ]

[ "$12:30$ - $40$ min = $11:50$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10 min$$ " } ], [ { "aoVal": "B", "content": "$$15 min$$ " } ], [ { "aoVal": "C", "content": "$$20 min$$ " } ], [ { "aoVal": "D", "content": "$$25 min$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "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 $$A$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$8$ hours and $5$ minutes " } ], [ { "aoVal": "B", "content": "$$7$$ hours and $30$ minutes " } ], [ { "aoVal": "C", "content": "$8$ hours " } ], [ { "aoVal": "D", "content": "$7$ hours " } ], [ { "aoVal": "E", "content": "$6$ hours " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "Cindy " } ], [ { "aoVal": "B", "content": "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "3 brothers and 4 sisters " } ], [ { "aoVal": "B", "content": "2 brothers and 3 sisters " } ], [ { "aoVal": "C", "content": "3 brothers and 3 sisters " } ], [ { "aoVal": "D", "content": "2 brothers and 4 sisters " } ], [ { "aoVal": "E", "content": "3 brothers and 2 sisters " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning" ]

[ "NA " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle" ]

[ "There are $12$ months. Thus, $42 \\div 12 = 3R6$, $3+1 = 4$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$81$$ " } ], [ { "aoVal": "B", "content": "$$82$$ " } ], [ { "aoVal": "C", "content": "$$83$$ " } ], [ { "aoVal": "D", "content": "$$84$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math" ]

[ "Let first 10 numbers be $${{a}\\_{1}}$$、$${{a}\\_{2}}$$、$${{a}\\_{3}}$$、$${{a}\\_{4}}$$、$$\\ldots \\ldots $$、$${{a}\\_{10}}$$. $${{a}\\_{1}}+{{a}\\_{2}}+{{a}\\_{3}}={{a}\\_{2}}+{{a}\\_{3}}+{{a}\\_{4}}$$，$${{a}\\_{1}}={{a}\\_{4}}$$. Therefore, $${{a}\\_{1}}={{a}\\_{4}}={{a}\\_{7}}={{a}\\_{10}}=16$$，$${{a}\\_{8}}=100-{{a}\\_{10}}-{{a}\\_{9}}=100-16-{{a}\\_{9}}=84-{{a}\\_{9}}$$. The maximum value of $${{a}\\_{8}}$$ is $$83$$．$$For$$ example：$$16$$、$$83$$、$$1$$、$$16$$、$$83$$、$$1$$、$$16$$、$$83$$、$$1$$、$$16$$． " ]

[]

[ [ { "aoVal": "A", "content": "John " } ], [ { "aoVal": "B", "content": "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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Time Problem" ]

[ "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). " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Martin " } ], [ { "aoVal": "B", "content": "Adam " } ], [ { "aoVal": "C", "content": "Susan " } ], [ { "aoVal": "D", "content": "Dana " } ], [ { "aoVal": "E", "content": "Lucy " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning" ]

[ "L \\textgreater{} D\\textgreater{} M \\textgreater{} A \\textgreater{} S " ]