Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers" ]

[ "$97$, $101$, $103$, $107$, $109$ " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "$$36={{2}^{2}}\\times {{3}^{2}}$$, so the number of its factors would be $$\\left( 2+1 \\right)\\times \\left( 2+1 \\right)=9$$, Among them, there are $$\\left( 2+1 \\right)\\times 1=3$$ factors which has $${{2}^{2}}$$ as its factors. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers" ]

[ "For any $$2$$ such primes, the factors are $$1$$, the primes, and their product. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder " ]

[ "The smallest possible $$N$$ is $$73$$, and $$73 \\div 11\\rm R7$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$44$$ " } ], [ { "aoVal": "D", "content": "$$88$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders " ]

[ "$2^{10}=1024\\equiv24($mod$100)$ $\\left (2^{10}\\right )^{2}=1024^{2}\\equiv76($mod$100)$ $\\left (2^{10}\\right )^{3}\\equiv 76\\times24\\equiv24($mod$100)$ $\\left (2^{10}\\right )^{4}\\equiv76($mod$100)$ $\\cdots\\cdots$ $$2^{2018201}-8=2\\left (2^{2018200}\\right )-8 \\equiv 2\\left (76\\right )-8 \\equiv 44($$mod$$100)$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder " ]

[ "The remainder of $$A$$ $$\\div9$$ is $$7$$, and $$2A=A+A$$. Therefore the remainder of $$2A\\div9$$ is $$7+7=14$$. $$14= 9+5$$, therefore the remainder is $$5$$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "$$36={{2}^{2}}\\times {{3}^{2}}$$, so the number of its factors would be $$\\left( 2+1 \\right)\\times \\left( 2+1 \\right)=9$$, Among them, there are $$\\left( 2+1 \\right)\\times 1=3$$ factors which has $${{2}^{2}}$$ as its factors. " ]

[]

[ [ { "aoVal": "A", "content": "♦ $$-\\textasciitilde3$$ " } ], [ { "aoVal": "B", "content": "☺ $$+$$ ♦ " } ], [ { "aoVal": "C", "content": "☺ $$\\times$$ ☺ " } ], [ { "aoVal": "D", "content": "♦ $$\\times$$~♦ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers" ]

[ "♥ $$\\times$$ ☺ $$=$$ ♦ Since ☺ is an even number,~♦ must also be an even number. ♦ $$-\\textasciitilde3$$ is the only option to given an odd answer because even $$-$$ odd $$=$$ odd. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Congruence" ]

[ "We note that $19^{2} \\equiv 1$ $( \\text{mod} \\textbackslash; 4)$ and hence: $19^{2021} \\equiv 1^{1010} \\times 3 \\equiv 3$ $( \\text{mod} \\textbackslash; 3)$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1001$$ " } ], [ { "aoVal": "B", "content": "$$1100$$ " } ], [ { "aoVal": "C", "content": "$$1999$$ " } ], [ { "aoVal": "D", "content": "$$2001$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers" ]

[ "If $$1$$ more than a $$3$$-digit number is a $$4$$-digit number, then the numbers are $$999$$ and $$1000$$ and their sum is $$1999$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$5 $$ " } ], [ { "aoVal": "B", "content": "$$6 $$ " } ], [ { "aoVal": "C", "content": "$$ 7 $$ " } ], [ { "aoVal": "D", "content": "$$8 $$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders" ]

[ "For the number to be as small as possible, we need the number of digits to be as small as possible. For instance, $$111\\cdots 1 (2022\\textasciitilde1\\text{s})$$ has a digit sum of $$2022$$, but it is a much larger number than $$333\\cdots 3 (674\\textasciitilde3\\text{s})$$, which also has a digit sum of $$2022$$. Clearly, to reduce the number of digits in the number, we need to make as many as possible of the digits in the number equal to $$9$$. Now $$2022 \\div9 = 224$$ remainder $$6$$, so the smallest positive integer with digit sum of $$2022$$ is $$699\\cdots 9 (224\\textasciitilde9\\text{s})$$. Its last digit is $$9$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "The factors of $$30$$ are $$1$$, $$30$$, $$2$$, $$15$$, $$3$$, $$10$$, $$5$$, $$6$$, so the greatest odd factor of $$30$$ is $$15$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ], [ { "aoVal": "E", "content": "$$22$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "We have $$ \\begin{aligned} 16383 \\& =2^{14}-1 \\textbackslash\\textbackslash{} \\& =\\left(2^{7}+1\\right)\\left(2^{7}-1\\right) \\textbackslash\\textbackslash{} \\& =129 \\cdot 127 \\end{aligned} $$ Since 129 is composite, 127 is the largest prime divisible by 16383 . The sum of 127 \\textquotesingle s digits is $$ 1+2+7=\\text { (C) } 10 $$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$20$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "If we start off with the first digit, we know that it can\\textquotesingle t be $9$ since $9$ is not a factor of $120$ . We go down to the digit $8$ , which does work since it is a factor of $120$ . Now, we have to know what digits will take up the remaining four spots. To find this result, just divide $\\frac{120}{8}=15$. The next place can be $5$ , as it is the largest factor, aside from $15$ . Consequently, our next three values will be $3,1$ and $1$ if we use the same logic. Therefore, our five-digit number is $85311$ , so the sum is $8+5+3+1+1=18 \\Longrightarrow($ D) 18 . " ]

[]

[ [ { "aoVal": "A", "content": "$$111$$ " } ], [ { "aoVal": "B", "content": "$$121$$ " } ], [ { "aoVal": "C", "content": "$$132$$ " } ], [ { "aoVal": "D", "content": "$$115$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders" ]

[ "$$8\\times 13+7=111$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2288$$ " } ], [ { "aoVal": "B", "content": "$$2290$$ " } ], [ { "aoVal": "C", "content": "$$2294$$ " } ], [ { "aoVal": "D", "content": "$$2098$$ " } ], [ { "aoVal": "E", "content": "$$2300$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "The number can be written as $$4n+2$$ and $$3x$$. It must be a multiple of $3$ but leaves a remainder of $2$ when divided by $4$. $2300$ can be divisible by both $2$ and $3$, so it is $2098$. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples" ]

[ "The LCM of $6$, $7$, $9$ and $10$ is $2 \\times 3^{2}\\times 5 \\times7= 630$. $180 000=285 \\times 630 + 450$, ∴$$$$the$$$$ smallest number is $286 \\times 630=180 180$, Sum of the last 4 digits, $A + B + C+ D=9$. " ]

[]

[ [ { "aoVal": "A", "content": "$213$ " } ], [ { "aoVal": "B", "content": "$214$ " } ], [ { "aoVal": "C", "content": "$216$ " } ], [ { "aoVal": "D", "content": "$219$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "$$\\text{A}$$. $$2+1+3=6=2\\times3$$; $$\\text{B}$$. $$2+1+4=7$$, and $$7$$ is not a multiple of $$3$$; $$\\text{C}$$. $$2+1+6=9=3\\times3$$; $$\\text{D}$$. $$2+1+9=12=4\\times3$$. We choose $$\\text{B}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$23\\times24$$ " } ], [ { "aoVal": "B", "content": "$$24\\times35$$ " } ], [ { "aoVal": "C", "content": "$$42\\times53$$ " } ], [ { "aoVal": "D", "content": "$$53\\times45$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers" ]

[ "The product is odd if and only if every factor is odd. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "$$36={{2}^{2}}\\times {{3}^{2}}$$, so the number of its factors would be $$\\left( 2+1 \\right)\\times \\left( 2+1 \\right)=9$$, Among them, there are $$\\left( 2+1 \\right)\\times 1=3$$ factors which has $${{2}^{2}}$$ as its factors. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "The counting numbers that leave a remainder of 4 when divided by 6 are $4,10,16,22,28,34, \\cdots$ The counting numbers that leave a remainder of 3 when divided by 5 are $3,8,13,18,23,28,33, \\cdots$ So 28 is the smallest possible number of coins that meets both conditions. Because $4 \\cdot 7=28$, there are (A) 0 coins left when they are divided among seven people. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers" ]

[ "Only $5$ " ]

[]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$54$$ " } ], [ { "aoVal": "C", "content": "$$75$$ " } ], [ { "aoVal": "D", "content": "$$152$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders" ]

[ "The number when added by $$2$$ is divisible by $$7$$ and by $$11$$. Hence, the smallest value is $$7\\times 11-2=75$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers" ]

[ "First of all $${{70}^{2}}=4900$$, So the number is between $$70$$ and $$80$$. According to the last digit, the number should be $$72 $$ or $$78 $$. $$\\because {{72}^{2}}=5184$$，$${{78}^{2}}=6084$$． $$\\therefore a+b=1+8=9$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$33$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$66$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "We can divide the $$99$$ numbers into $$3$$ groups according to their remainder when divided by $$3$$. $$1$$, $$4$$, $$7\\_\\cdots$$ they belong to the group $$A$$, since they leave a remainder of $$1$$ when divided by $$3$$. $$2$$, $$5$$, $$8\\_\\cdots$$ they belong to the group $$B$$, since they leave a remainder of $$2$$ when divided by $$3$$. $$3$$, $$6$$, $$9\\_\\cdots$$ they belong to the group $$C$$, since they leave a remainder of $$0$$ when divided by $$3$$. As we know, the members in group $$C$$ cannot get together. Also, when the member of group $$A$$ and the member of group $$B$$ get together, they will create the sum which is a multiple of $$3$$. So the $$33$$ members of group $$C$$ will be in $$33$$ groups. Group $$A$$ and group $$B$$ can join in the $$33$$ groups however they want. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "$$2+8+8=18$$, $$18$$ is multiple of $$3$$ Thus, $$X$$ itself must be a multiple of $$3$$, or $$0$$. We have $$3, 6, 9$$ and $$0$$. Four possible value. " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ], [ { "aoVal": "E", "content": "$$22$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "It can be shown that a positive integer has an odd number of factors if and only if it is square. ~The smallest three-digit square number is $${{10}^{2}}=100$$ and the largest is $${{31}^{2}}=961$$. Hence there are $$31-9=22$$ three-digit numbers which have an odd number of factors. " ]

[]

[ [ { "aoVal": "A", "content": "♦ $$-\\textasciitilde7$$ " } ], [ { "aoVal": "B", "content": "☺ $$+$$ ♦ " } ], [ { "aoVal": "C", "content": "☺ $$\\times$$ ☺ " } ], [ { "aoVal": "D", "content": "♦ $$\\times$$~♦ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers" ]

[ "♥ $$\\times$$ ☺ $$=$$ ♦ Since ☺ is an even number,~♦ must also be an even number. ♦ $$-\\textasciitilde7$$ is the only option to given an odd answer because even $$-$$ odd $$=$$ odd. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "We know that a number is divisible by $11$ if the odd numbers added together minus the even numbers added together(or vice versa) is a multiple of $11$ . So, we have $1+2-A=$ a multiple of $11$ . The only multiple that works here is $0$ , as $11 \\cdot 0=0$. Thus, $A=(D)$ " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)" ]

[ "$$2016=7\\times288=8 \\times252=9\\times224$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$321$ " } ], [ { "aoVal": "B", "content": "$354$ " } ], [ { "aoVal": "C", "content": "$720$ " } ], [ { "aoVal": "D", "content": "$360$ " } ], [ { "aoVal": "E", "content": "$240$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem" ]

[ "$8\\times9\\times10\\div2=360$ $360-6=354$ " ]

[]

[ [ { "aoVal": "A", "content": "Correct " } ], [ { "aoVal": "B", "content": "Wrong " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers" ]

[ "Her calculation was wrong. The answer is supposed to be an even number. $$117\\times32$$，$$19\\times12$$ are even numbers and $$133\\times11$$ is an odd number. So in that case, we can think of the equation as: odd + even + even - even + odd = even Hence, the answer should be an even number! However, Chloe\\textquotesingle s answer is an odd number. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "use divisibility rule. 3+ 7+8+R = divisible by 3. 18 + R = divisible by 3. smallest possible number for R is 0, max possible amount for R is 9 3x6=18 - 18 =0 (can be divisible by 3) 3x7=21 -18 = 3 3x 8= 24 -18 =6 3x 9= 27 - 18 = 9 Total = 4 ways. " ]

[]

[ [ { "aoVal": "A", "content": "$$2014$$ " } ], [ { "aoVal": "B", "content": "$$2015$$ " } ], [ { "aoVal": "C", "content": "$$2016$$ " } ], [ { "aoVal": "D", "content": "$$2017$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem" ]

[ "The total number of stamps when divided by $$32$$, $$9$$, and $$7$$ leaves remainders of $$30$$, $$7$$, and $$5$$, respectively. In other words, the number when added by $$2$$ is divisible by $$32$$, $$9$$, and $$7$$. Hence, the smallest such number is $$32\\times9\\times7-2=2014$$. The answer is $$\\rm A$$. " ]

[]

[ [ { "aoVal": "A", "content": "$${{2}^{3}}\\times {{3}^{2}}\\times 5\\times 7$$ " } ], [ { "aoVal": "B", "content": "$${{2}^{4}}\\times {{3}^{2}}\\times 5$$ " } ], [ { "aoVal": "C", "content": "$${{2}^{4}}\\times {{3}^{2}}\\times 5\\times 7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers" ]

[ "$$24\\times 105={{2}^{3}}\\times {{3}^{2}}\\times 5\\times 7$$ " ]

[]

[ [ { "aoVal": "A", "content": "$\\left (8231\\right )\\_5$ " } ], [ { "aoVal": "B", "content": "$\\left (2001\\right )\\_5$ " } ], [ { "aoVal": "C", "content": "$\\left (4341\\right )\\_7$ " } ], [ { "aoVal": "D", "content": "$\\left (2345\\right )\\_5$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Properties and Applications of Number Bases" ]

[ "In quinary, the base number must be $5$, and all the digits in the parentheses must be less than $5$. " ]

[]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number" ]

[ "$$2010-15=1995=3\\times 5\\times 7\\times 19$$ $$\\left( 1+1 \\right)\\times \\left( 1+1 \\right)\\times \\left( 1+1 \\right)\\times \\left( 1+1 \\right)=16$$ factors. Remove $$1$$, $$3$$, $$5$$, $$7$$, and $$15$$ from them, we can get $$16-5=11$$ possible values. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "Let the original number be ab, and reverse it to be ba, calculated by the place value principle, ba-ab=72 10b+a-(10a+b)=72, 9b-9a=72, b-a=8, b=9 and a=1. So the original number is 19, last digit is 9. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$67$$ " } ], [ { "aoVal": "C", "content": "$$69$$ " } ], [ { "aoVal": "D", "content": "$$76$$ " } ], [ { "aoVal": "E", "content": "$$77$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "Let $k$ be any positive integer so that $13 k$ is a multiple of 13 . For the smallest three-digit number, $13 k\\textgreater100$ and $k\\textgreater\\frac{100}{13} \\approx 7.7$. For the greatest three-digit number, $13 k\\textless999$ and $k\\textless\\frac{999}{13} \\approx 76.8$. The number $k$ can range from 8 to 76 so there are $(\\mathbf{C}) 69$ threedigit numbers. " ]

[]

[ [ { "aoVal": "A", "content": "Only $179$ " } ], [ { "aoVal": "B", "content": "$179$ and $$187$$ " } ], [ { "aoVal": "C", "content": "$179$ and $157$ " } ], [ { "aoVal": "D", "content": "$179$, $157$, and $187$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers" ]

[ "$129=3\\times43$. $187=11\\times17$ " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$64$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "The square root of $$16$$ is $$4$$ and the square root of $$4$$ is $$2$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$\\text{milk}$$ " } ], [ { "aoVal": "B", "content": "$$\\text{juice}$$ " } ], [ { "aoVal": "C", "content": "$$\\text{coke}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems" ]

[ "$$15\\div 4=3\\text{R}3$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ], [ { "aoVal": "E", "content": "$$17$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "$2000=2^{3}\\times5^{3}$, the number of its factors would be $(4+1)(3+1)=20$, but $1$, $2$, and $5$ do not meet the condition. " ]

[]

[ [ { "aoVal": "A", "content": "$$12=2\\times2\\times3$$ " } ], [ { "aoVal": "B", "content": "$$51=3\\times17$$ " } ], [ { "aoVal": "C", "content": "$$8=2\\times2\\times2$$ " } ], [ { "aoVal": "D", "content": "$$45=5\\times9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "$$45=5\\times3\\times3$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$32$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples" ]

[ "$5\\times 8=40$. " ]

[]

[ [ { "aoVal": "A", "content": "$$55$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$25$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ], [ { "aoVal": "E", "content": "$$65$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "$30+25=55$ " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems" ]

[ "We can use the equation: divisor $$\\times$$ quotient $$=$$ dividend $$-$$ remainder, so here we can get divisor $$\\times$$ quotient $$=28-4=24$$. Therefore, the only possibilities are $$1$$ and $$24$$, $$2$$ and $$12$$, $$3$$ and $$8$$, and $$4$$ and $$6$$ for a total of four possible combinations. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "$$\\sqrt{10^{2} - 6^{2}} + \\sqrt{3^{2} + 4^{2}}=\\sqrt{64} + \\sqrt{25} = 8 + 5 = 13$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$23$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$29$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "$45=3\\times 3\\times 5$ $50=2\\times 5\\times 5$ Because $B$ is the factor both number contains, $B=5$ Thus, $A=9$, $C=10$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder " ]

[ "$$539\\div 4$$ R $$3$$；$$142\\div 4$$ R $$2$$；$$539-142$$ divided by $$4$$, the remainder is 3-2=1. " ]

[]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$49$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders" ]

[ "$$56 = 8 \\times 7$$, $$56 = 7 \\times 8$$, so we choose $$\\rm A$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$a$ can only be $1$ " } ], [ { "aoVal": "B", "content": "$a$ can only be $37$ " } ], [ { "aoVal": "C", "content": "$a$ can be 1 or $37$ " } ], [ { "aoVal": "D", "content": "$a$ can be any number " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers" ]

[ "\"$$a$$ is a factor of $37$\" means $37$ is divisible by $a$. " ]

[]

[ [ { "aoVal": "A", "content": "$$49$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers" ]

[ "The $$3$$ such whole numbers are $$16$$, $$32$$, and $$64$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "A multiple of $$3$$, $$4$$, and $$5$$ is a multiple of $$60$$, and $$16\\times60\\textless{}1000 \\textless{}17\\times60$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "$15=3\\times 5$ $35=5\\times 7$ Because $B$ is the factor both number contains, $B=5$ Thus, $A=3$, $C=7$, $A+B+C=15$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems" ]

[ "We can use the equation: divisor $$\\times$$ quotient $$=$$ dividend $$-$$ remainder, so here we can get divisor $$\\times$$ quotient $$=28-4=24$$. Therefore, the only possibilities are $$1$$ and $$24$$, $$2$$ and $$12$$, $$3$$ and $$8$$, and $$4$$ and $$6$$ for a total of four possible combinations. " ]

[]

[ [ { "aoVal": "A", "content": "$$89$$ " } ], [ { "aoVal": "B", "content": "$$92$$ " } ], [ { "aoVal": "C", "content": "$$95$$ " } ], [ { "aoVal": "D", "content": "$$97$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers" ]

[ "Because $$99$$ is an odd number, one of these two prime numbers must be an even prime number. The only choice is $$2$$. So the other is $$99-2 = 97$$. So, the difference between the two numbers is $$95$$. Therefore, we choose $$\\rm C$$ . " ]

[]

[ [ { "aoVal": "A", "content": "$$402$$ " } ], [ { "aoVal": "B", "content": "$$403$$ " } ], [ { "aoVal": "C", "content": "$$404$$ " } ], [ { "aoVal": "D", "content": "$$410$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders " ]

[ "First, we need to remove all the integers which are multiples of $5$, otherwise the last digit of the products is $0$ or $5$. Hence, $403$ integers need to be removed. Next, note that the last digit of each of the products below is $6$. $1\\times2\\times3\\times4\\times6\\times7\\times8\\times9$, $11\\times12\\times13\\times14\\times16\\times17\\times18\\times19$, $\\cdots\\cdots$ $2001\\times2002\\times2003\\times2004\\times2006\\times2007\\times2008\\times2009$, and the last digit of the product $2011\\times2012\\times2013\\times2014\\times2016\\times2017\\times2018$ is $4$. Hence, we need to remove one more \"$2$\"and the last digit of the product will be $2$. So the answer is $404$. " ]

[]

[ [ { "aoVal": "A", "content": "odd " } ], [ { "aoVal": "B", "content": "even " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers" ]

[ "$88-29+1=60$ $60\\div2=30$ Odd numbers all paired. " ]

[]

[ [ { "aoVal": "A", "content": "$$10\\textbackslash\\%$$ " } ], [ { "aoVal": "B", "content": "$$20\\textbackslash\\%$$ " } ], [ { "aoVal": "C", "content": "$$40\\textbackslash\\%$$ " } ], [ { "aoVal": "D", "content": "$$80\\textbackslash\\%$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Properties and Applications of Number Bases->Mixed Operations of Number Bases", "Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Word Problems Involving Fractions and Percentages->Finding the Percentage Given a Part and a Whole" ]

[ "50-10=40；$$40\\div$$ 50=0.8；0.8=80\\% " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->Finding Factors Given the Product" ]

[ "$a^{2}b=2000=2^{4}\\times5^{3}$. Thus, the value of $a$ can be $1$, $2$, $4$, $5$, $10$, and $20$. " ]

[]

[ [ { "aoVal": "A", "content": "$$385$$ " } ], [ { "aoVal": "B", "content": "$$1155$$ " } ], [ { "aoVal": "C", "content": "$$77$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples" ]

[ "Since $$5=1\\times5$$, $$7=1\\times7$$, and $$11=1\\times11$$, we have one $$5$$, one $$7$$, and one $$11$$, thus the least common multiple for $$5$$, $$7$$, and $$11$$ is $$5\\times7\\times11=385$$. We choose $$\\rm A$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$35\\times {{10}^{10}}$$ " } ], [ { "aoVal": "B", "content": "$$3.5\\times {{10}^{11}}$$ " } ], [ { "aoVal": "C", "content": "$$3.5\\times {{10}^{12}}$$ " } ], [ { "aoVal": "D", "content": "$$3.5\\times {{10}^{19}}$$ " } ], [ { "aoVal": "E", "content": "$$3.5\\times {{10}^{18}}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->The Relationship between Exponents and the Number of Factors" ]

[ "$$7\\times {{10}^{9}}\\times 5\\times {{10}^{2}}=3.5\\times {{10}^{12}}$$ " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders " ]

[ "We can find the last (units) digit by looking at the units digits of the four parts. We can quickly note that the units digits of $$0^{2016}$$ and $$1^{2016}$$ and $$6^{2016}$$are $$0$$, $$1$$ and$$6$$ respectively. Now then we have to look at powers of $$2:{{2}^{1}}=2$$, $${{2}^{2}}=4$$, $${{2}^{3}}=8$$, $${{2}^{4}}=16$$, and after this the units digits repeat ($${{2}^{5}}=3\\underline{2}$$, $${{2}^{6}}=6\\underline{4}$$, $${{2}^{7}}=12\\underline{8}$$, $${{2}^{8}}=25\\underline{6}$$, $$\\cdots $$). We notice that when the indices are multiples of 4 (eg. $${{2}^{4}}=16$$, $${{2}^{8}}=256$$, $$\\cdots $$) the units digit of the power of $$2$$ is $$6$$. Hence the units digit of $$2^{2016}+0^{2016}+1^{2016}+6^{2016}$$ is the same as the units digit of $$6 + 0 + 1 + 6$$, that is $$3$$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "We use the property that the digits of a number must sum to a multiple of $9$ if it is divisible by $9$. This means $2+0+1+8+U$ must be divisible by $9$. The only possible value for U then must be $7$. Since we are looking for the remainder when divided by $8$, we can ignore the thousands. The remainder when $187$ is divided by $8$ is $(\\rm B)$ $3$. " ]

[]

[ [ { "aoVal": "A", "content": "$${{2}^{6}}\\times {{3}^{7}}\\times {{5}^{9}}$$ " } ], [ { "aoVal": "B", "content": "$${{2}^{2}}\\times {{3}^{3}}\\times {{5}^{20}}$$ " } ], [ { "aoVal": "C", "content": "$${{2}^{2}}\\times {{3}^{3}}\\times {{5}^{9}}$$ " } ], [ { "aoVal": "D", "content": "$${{2}^{8}}\\times {{3}^{12}}\\times {{5}^{20}}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->The Relationship between Exponents and the Number of Factors" ]

[ "Answer $$={{2}^{2\\times 4}}\\times {{3}^{3\\times 4}}\\times {{5}^{5\\times 4}}={{2}^{8}}\\times {{3}^{12}}\\times {{5}^{20}}$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$52$$ " } ], [ { "aoVal": "B", "content": "$$60$$ " } ], [ { "aoVal": "C", "content": "$$62$$ " } ], [ { "aoVal": "D", "content": "$$72$$ " } ], [ { "aoVal": "E", "content": "$$84$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number" ]

[ "The sum of the factors is $$(3^{0}+3^{1})$$$$\\times (2^{0}+2^{1}+2^{2}+2^{3})=60$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2 $$ " } ], [ { "aoVal": "B", "content": "$$$$prime " } ], [ { "aoVal": "C", "content": "$$$$odd " } ], [ { "aoVal": "D", "content": "$$$$even " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers" ]

[ "Add two numbers from: $$1$$, $$3$$, $$5$$, $$7$$, $$9$$. The result is always even. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ], [ { "aoVal": "E", "content": "None of these. " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Questions involving Square Numbers" ]

[ "Using the difference of two squares, we can get: $$\\left(2^{2}-1\\right)\\cdot \\left(3^{2}-1\\right)\\cdot \\left(4^{2}-1\\right)\\cdots \\left(n^{2}-1\\right) = 1\\times 3\\times 2\\times 4\\times 3\\times 5\\times \\cdots \\times (n-1) \\times (n+1)$$ you will notice that this is equal to: $$1\\times 2\\times 3^{2} \\times 4^{2} \\times \\cdots \\times (n-1)^{2} \\times n \\times (n+1)$$ So this will only be a square if $2n(n+1)$ is a square, which happens when $n=8$ as $2n(n+1) = 144=12^{2}$. Answer: $B$ " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "Five consecutive numbers can be: $$12$$, $$13$$, $$14$$, $$15$$, and $$16$$ (without carrying) or $$17$$, $$18$$, $$19$$, $$20$$, $$21$$ (with carrying in the tens place). Without carrying, among each of the $$5$$ consecutive integers, we can find one whose sum of digits is a multiple of $$5$$. So we need to carry. To make the number of integers the greatest, we can start from a number whose remainder is $$1$$ when divided by five. And the most important thing is, after we write the fourth number, the carry appears. For example: $$56$$, $$57$$, $$58$$, $$59$$, $$60$$, $$61$$, $$62$$, $$63$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "Already divisible by $$2$$, so we must check it divides by $$9$$ by adding the digits together: $$9+5+x+9+4+7+7+5+9+9+8=54+x$$, this must be a multiple of $$9$$ so $$x$$ is $$0$$ or $$9$$. It should also be divisible by $$11$$, so the alternating sum of digits should too. $$9-5+x-9+4-7+7-5+9-9+8=2+x$$ is divisible by $$11$$. Hence $$x$$ is $$9$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$76$$ " } ], [ { "aoVal": "B", "content": "$$160$$ " } ], [ { "aoVal": "C", "content": "$$202$$ " } ], [ { "aoVal": "D", "content": "$$328$$ " } ], [ { "aoVal": "E", "content": "$$412$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples" ]

[ "The number of total price can be divided by $3$ and $8$. It can be divided by $8$ which means the tens digit must be $3$ or $7$. And when the tens digit is $7$, the number can be divided by $3$. Thus, the total price is $7872$ dollars, and each of the machine is $328$ dollars. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$75$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples" ]

[ "*and* = 2 x 3 = 6 Multiple of 6 with 2 digits = 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96. " ]

[]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$46$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$91$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers" ]

[ "Add $$10$$ to $$1$$, $$3$$, $$5$$, $$7$$, $$\\cdots $$, $$87$$, and $$89$$. None of these sums is more than $$99$$. There are $$45$$ such sums. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->The Relationship between Exponents and the Number of Factors" ]

[ "$$\\frac{{{2}^{2019}}}{{{2}^{1964}}-{{2}^{1963}}}$$ $$={{2}^{2019-1963}}$$ $$={{2}^{56}}$$， ∴$${{2}^{56}}:6$$， $$6\\div 5:1$$． " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "All four expressions are perfect squares: $$1^{3} + 2^{3} = 1 + 8 = 9 = 3^{2}$$; $$1^{3}+2^{3}+3^{3}=1+8+27=36=6^{2}$$; $$1^{3}+2^{3}+3^{3}+4^{3}=1+8+27+64=100=10^{2}$$; $$1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} = 1 + 8 + 27 + 64 + 125 = 225 = 15^{2}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "$$42$$ is not divisible by $$4$$, so neither is $$135 798 642$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$4^{2}$$ " } ], [ { "aoVal": "B", "content": "$$6^{2}$$ " } ], [ { "aoVal": "C", "content": "$$12^{2}$$ " } ], [ { "aoVal": "D", "content": "$$33^{2}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "$$3^{2}+3^{2}+3^{2}+3^{2}=9+9+9+9=36=6^{2}$$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)" ]

[ "$$2016=7\\times288=8 \\times252=9\\times224$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers" ]

[ "The smallest prime factors of $250$ are $2$ and $5$. Thus, the sum is $2 + 5 = 7$. " ]

[]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder " ]

[ "Since $$16\\div 4$$ has a remainder of $$0$$, the remainder is $$0+0+0+0=0$$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value" ]

[ "We can think of the number as $$10a+b$$, where $$a$$ and $$b$$ are digits. Since the number is equal to the product of the digits $$(a\\cdot b)$$ plus the sum of the digits $$(a+b)$$, we can say that $$10a+b=a\\cdot b+a+b$$. We can simplify this to $$10a=a\\cdot b+a$$, and factor to $$(10)a=(b+1)a$$. Dividing by $$a$$, we have that $$b+1=10$$. Therefore, the units digit, $$b$$, is $$9$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$107$$ " } ], [ { "aoVal": "B", "content": "$$184$$ " } ], [ { "aoVal": "C", "content": "$$534$$ " } ], [ { "aoVal": "D", "content": "$$641$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders" ]

[ "Each number on Lee\\textquotesingle s list must have the same remainder when divided by $$7$$. Divide each choice by $$7$$. The respective remainders are $$2$$, $$2$$, $$2$$, and $$4$$. Thus, $$107$$, $$184$$, and $$534$$ work. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "Since $1001$ is a multiple of $13$, $111111 = 111 \\times 1001$ is also a multiple of $13$. It follows that both $666666$ and $444444$ are both multiples of $26$. $666666XY 444444 = 66666600000000 + XY 000000 + 444444$ $\\Rightarrow XY$ must be divisible by $13$. Smallest $X+Y=1+3=4$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ], [ { "aoVal": "E", "content": "$$14$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem" ]

[ "The number is a multiple of $5$. When it is divided by $6$, it will have a remainder of $1$; when it is divided by $7$, it will have a remainder of $2$. Thus, if we add another $5$ to the number, it can be multiple of all the three numbers, which is $5\\times6\\times7=210$ at least. But the number is more than $400$, so we can get $210\\times2-5=415.$ " ]

[]

[ [ { "aoVal": "A", "content": "odd　 " } ], [ { "aoVal": "B", "content": "even　 " } ], [ { "aoVal": "C", "content": "$$1995$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers" ]

[ "even number $$+$$ odd number $$=$$ odd number. " ]

[]

[ [ { "aoVal": "A", "content": "$$120$$ " } ], [ { "aoVal": "B", "content": "$$150$$ " } ], [ { "aoVal": "C", "content": "$$180$$ " } ], [ { "aoVal": "D", "content": "$$240$$ " } ], [ { "aoVal": "E", "content": "$$300$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Word Problems Involving Factors and Multiples->Multiples Word Problems" ]

[ "$6=2\\times3, 9=3\\times3, 2\\times3\\times3\\times5=90$. The number of students is a multiple of 90. $180\\div90=2$. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples" ]

[ "The LCM of $6$, $7$, $9$ and $10$ is $2 \\times 3^{2}\\times 5 \\times7= 630$. $180 000=285 \\times 630 + 450$, ∴$$$$the$$$$ smallest number is $286 \\times 630=180 180$, Sum of the last 4 digits, $A + B + C+ D=9$. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$13$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Casting Out Nines" ]

[ "No matter how we write the three-digit number, the sum of its three digits is always the same. Therefore, the remainders of the two numbers divided by $$9$$ are also the same. The difference will always be a number that is divisible by $$9$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$23$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$29$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "$45=3\\times 3\\times 5$ $50=2\\times 5\\times 5$ Because $B$ is the factor both number contains, $B=5$ Thus, $A=9$, $C=10$. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders" ]

[ "$$104\\div7=14R6$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$222$$ " } ], [ { "aoVal": "B", "content": "$$444$$ " } ], [ { "aoVal": "C", "content": "$$888$$ " } ], [ { "aoVal": "D", "content": "$$1000$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders" ]

[ "If the sum is twice the difference, one \\# is triple the other. The sum is divisible by $$4$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$65$$ " } ], [ { "aoVal": "B", "content": "$$572$$ " } ], [ { "aoVal": "C", "content": "$$715$$ " } ], [ { "aoVal": "D", "content": "$$2860$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers" ]

[ "C " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$76$$ " } ], [ { "aoVal": "B", "content": "$$160$$ " } ], [ { "aoVal": "C", "content": "$$202$$ " } ], [ { "aoVal": "D", "content": "$$328$$ " } ], [ { "aoVal": "E", "content": "$$412$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples" ]

[ "The number of total price can be divided by $3$ and $8$. It can be divided by $8$ which means the tens digit must be $3$ or $7$. And when the tens digit is $7$, the number can be divided by $3$. Thus, the total price is $7872$ dollars, and each of them is $328$ dollars. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder " ]

[ "When $$10$$ students or$$13$$ students are in a group, there are always $$3$$ students remained. It means that if we subtract $$3$$ from the total number, the result is a common multiple of $$10$$ and $$13$$. Therefore when there are $$130$$ in one group, there are still$$\\textasciitilde3$$ students remained. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$27$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$51$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems" ]

[ "The Lowest Common Multiple of $$6$$ and $$8$$ is $$24$$. $$24+3=27$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$1:16$ " } ], [ { "aoVal": "B", "content": "$1:15$ " } ], [ { "aoVal": "C", "content": "$1:14$ " } ], [ { "aoVal": "D", "content": "$1:8$ " } ], [ { "aoVal": "E", "content": "$1:3$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "Prime factorizing $N$, we see $N=2^{3} \\cdot 3^{5} \\cdot 5 \\cdot 7 \\cdot 17^{2}$. The sum of $N$ \\textquotesingle s odd divisors are the sum of the factors of $N$ without 2 , and the sum of the even divisors is the sum of the odds subtracted by the total sum of divisors. The sum of odd divisors is given by $$ a=\\left(1+3+3^{2}+3^{3}+3^{4}+3^{5}\\right)(1+5)(1+7)\\left(1+17+17^{2}\\right) $$ and the total sum of divisors is $$ (1+2+4+8)\\left(1+3+3^{2}+3^{3}+3^{4}+3^{5}\\right)(1+5)(1+7)\\left(1+17+17^{2}\\right)=15 a . $$ Thus, our ratio is $$ \\frac{a}{15 a-a}=\\frac{a}{14 a}=\\text { (C) } 1: 14 \\text {. } $$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples" ]

[ "The remainder of $$A$$ $$\\div9$$ is $$7$$, and $$2A=A+A$$. Therefore the remainder of $$2A\\div9$$ is $$7+7=14$$. $$14= 9+5$$, therefore the remainder is $$5$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$55$$ " } ], [ { "aoVal": "B", "content": "$$57$$ " } ], [ { "aoVal": "C", "content": "$$58$$ " } ], [ { "aoVal": "D", "content": "$$59$$ " } ], [ { "aoVal": "E", "content": "$$61$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers" ]

[ "The smallest prime factor is $2$, and $58$ is the only multiple of $2$ among these five numbers. " ]

[]

[ [ { "aoVal": "A", "content": "$$9 + 6$$ " } ], [ { "aoVal": "B", "content": "$$9 - 6$$ " } ], [ { "aoVal": "C", "content": "$$2 \\times 9$$ " } ], [ { "aoVal": "D", "content": "$$2 \\times 9 + 1$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "$$\\text{A}$$: corresponding to $$\\text{a}+\\text{b}$$; $$\\text{B}$$: corresponding to $$\\text{a}-\\text{b}$$; $$\\text{C}$$: $$\\text{na}$$, where $$\\text{n}$$\\emph{\\emph{~}}is an integer. We choose $$\\text{D}$$. " ]