Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[]

[ [ { "aoVal": "A", "content": "$$(14,9,10), (21,15,4)$$ " } ], [ { "aoVal": "B", "content": "$$(14,21,10), (9,15,4)$$ " } ], [ { "aoVal": "C", "content": "$$(14,4,10), (21,15,9)$$ " } ], [ { "aoVal": "D", "content": "$$(14,15,9), (21,10,4)$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->The Number of Zeros at the end of a Product" ]

[ "$$A$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

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

[ "When we divide by $$2$$, $$3$$, the remainder is $$1$$. Hence, if we first subtract $$1$$, the result will be divisible by $$2$$, $$3$$. $2\\times3+1=7$ " ]

[]

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

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

[ "$$87\\div 8=10$$ $\\text{R}$ $$7$$ $$10+1=11$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

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

[ "2 set of 2$\\times$5. " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

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

[ "Let the number be $x$. Since $x^{2} = 6x$, therefore $x=6$. " ]

[]

[ [ { "aoVal": "A", "content": "$$000$$ " } ], [ { "aoVal": "B", "content": "$$222$$ " } ], [ { "aoVal": "C", "content": "$$444$$ " } ], [ { "aoVal": "D", "content": "$$666$$ " } ] ]

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

[ "The product $$123 \\times124\\times125\\times126 \\times127$$ is a multiple of $$125$$; moreover, it also has a factor of $$2$$ three times, from $$124\\left( 2\\times 2\\times 31 \\right)$$ and from $$126\\left( =2\\times 63 \\right)$$. Therefore it is a multiple of $$125 \\times2\\times2\\times2= 1000$$, and so it must end in $$000$$. Alternatively, working from the options, it is easily seen that the product is a certainly an even multiple of $$5$$-so its unit digit is $$0$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$05$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$55$$ " } ], [ { "aoVal": "E", "content": "$$75$$ " } ] ]

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

[ "The result must be a multiple of $25.$ $$a=0(\\text{mod}25)$$ According to the divisibility rule of $4$, the remainder of the result divided by $4$ is $$1\\times 3\\times 1\\times 3\\times \\cdots \\times 1\\times 3\\times 1={{3}^{25}}\\times 1=3(\\text{mod}4)$$. When the result is $$75(\\text{mod}100)$$, it can be divisible by $25$ and have a remainder of $3$ when divided by $4.$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

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

[ "Note first that $$7632 =2\\times2\\times2\\times2\\times3\\times3\\times53$$. Therefore either the two-digit number $$de = 53$$ or the three-digit number $$abc$$ is a multiple of $$53$$. Since the multiplication uses each of the digits $$1$$ to $$9$$ once and $$7632$$ contains a $$3$$, the option $$de= 53$$ is not allowable. Hence we need to find a three-digit multiple of $$53$$ that does not share any digits with $$7632$$ and divides into $$7632$$ leaving an answer that also does not share any digits with $$7632$$. We can reject $$2 \\times 53 = 106$$ since it contains a $$6$$ but $$3 \\times 53 = 159$$ is a possibility. The value of $$7632\\div159$$ is $$2\\times2\\times2\\times2\\times3 = 48$$ which does not have any digits in common with $$7632$$ nor with $$159$$. We can also check that no other multiple of $$53$$ will work. Therefore the required multiplication is $$159 \\times 48 = 7632$$ and hence the value of $$b$$ is $$5$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$4$$ \\& $$18$$ " } ], [ { "aoVal": "B", "content": "$$5$$ \\&~$$25$$ " } ], [ { "aoVal": "C", "content": "$$6$$ \\&~$$33$$ " } ], [ { "aoVal": "D", "content": "$$8$$ \\&~$$35$$ " } ] ]

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

[ "The greatest common factors of the $$4$$ pairs of numbers are $$2$$, $$5$$, $$3$$, and $$1$$, respectively. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Common Multiples and Least Common Multiples->Least Common Multiple of Two Numbers" ]

[ "The least common multiple of $5$ and $7$ is $35$. " ]

[]

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

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

[ "$$11$$，$$13$$，$$17$$，$$31$$，$$37$$，$$71$$，$$73$$，$$79$$，$$97$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

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

[ "The total number of baseball cards is the least common multiple of $$2$$, $$3$$, and $$5$$, i.e. $$\\left[ 2,3,5\\right]=2\\times3\\times5=30$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2\\times 18$$ " } ], [ { "aoVal": "B", "content": "$$18\\times 18$$ " } ], [ { "aoVal": "C", "content": "$$3\\times 2$$ " } ], [ { "aoVal": "D", "content": "$$36\\times 36$$ " } ] ]

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

[ "$$\\sqrt {36}\\times \\sqrt {36}=6\\times 6=36 = 2\\times 18$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

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

[ "$8+8+5=9+9+3=10+10+1=21$, so after we removing $21$ cards, the number of the remaining cards can be divisible by $8, 9,$ and $10.$ " ]

[]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ], [ { "aoVal": "E", "content": "$$50$$ " } ] ]

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

[ "Each of five positive even numbers is $$1$$ larger than the corresponding odd number in $$1$$,$$3$$, $$5$$, $$7$$ and $$9$$, which we are told have a sum of $$25$$. So the sum of the first five positive even numbers is $$25 +5 =30$$. " ]

[ "其它" ]

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

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

[ "$$26\\div 5 = 5R1$$ Start from Tuesday, thus, it is Tuesday " ]

[]

[ [ { "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 " ]

[ "The sum of each numbers remainder can be divided by $$4$$ to get $$2$$: $$123\\div 4$$ R $$3$$；$$234\\div 4$$ R $$2$$；$$345\\div 4$$ R $$1$$; Therefore, $$\\left( 3+2+1 \\right)\\div 4$$ R $$2$$． " ]

[]

[ [ { "aoVal": "A", "content": "$$2350$$ " } ], [ { "aoVal": "B", "content": "$$2361$$ " } ], [ { "aoVal": "C", "content": "$$2470$$ " } ], [ { "aoVal": "D", "content": "$$2850$$ " } ] ]

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

[ "The sum of the squares of the first $$20$$ positive integers is $$2870$$. The sum of the squares of the first $$19$$ is $$2870-20^{2}=2870-400=2470$$. " ]

[ "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->Remainder Problems->Characteristics of Remainder " ]

[ "According to the additive and multiplicative properties of remainders, the remainder of this expression equals the remainder of $$(3\\times1+3\\times3)\\div 5=12\\div5=2 \\text{R} 2$$ " ]

[ "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->Division without Remainders->Divisibility Rules" ]

[ "The sums of the digits of $422$ and $971$ are not multiple of $3$. " ]

[]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$1000$$ " } ], [ { "aoVal": "C", "content": "$$10000$$ " } ], [ { "aoVal": "D", "content": "$$100000$$ " } ] ]

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

[ "The first perfect square is $$1^{2}$$, the $$2$$nd is $$2^{2}$$, and the $$3$$rd is $$3^{2}$$. With this pattern, the $$100$$th perfect square is $$100^{2} = 100\\times100= 10000$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

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

[ "$55=5\\times 11$ $100=2\\times 2\\times 5\\times 5$ Because $B$ is the factor both number contains, $B=5$ Thus, $A=11$, $C=20$. " ]

[]

[ [ { "aoVal": "A", "content": "$$ 8 $$ " } ], [ { "aoVal": "B", "content": "$$12 $$ " } ], [ { "aoVal": "C", "content": "$$16 $$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$10 $$ " } ] ]

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

[ "$$2\\times2 \\times2\\times2=16$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$101$$ " } ], [ { "aoVal": "B", "content": "$$111$$ " } ], [ { "aoVal": "C", "content": "$$99$$ " } ], [ { "aoVal": "D", "content": "$$113$$ " } ], [ { "aoVal": "E", "content": "None of the above. " } ] ]

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

[ "1,3,5,7,9 -\\/-\\/-\\/-\\/-\\/-\\/-\\/-\\/-\\/-⑤ 11,13,15,17,19-\\/-\\/-\\/-\\/-\\/-\\/-⑤ 31,33,35,37,39-\\/-\\/-\\/-\\/-⑤ 51,53,55,57,59-\\/-\\/-\\/-\\/-⑤ 71,73,75,77,79-\\/-\\/-\\/-\\/-⑤ 91,93,95,97,99-\\/-\\/-\\/-⑤ $$111$$ the 31st number is 111. " ]

[ "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->Remainder Problems->Characteristics of Remainder " ]

[ "According to the additive and multiplicative properties of remainders, the remainder of this expression equals the remainder of $$(3\\times1+3\\times3)\\div 5=12\\div5=2 \\text{R} 2$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$16$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$19$$ " } ], [ { "aoVal": "E", "content": "$$20$$ " } ] ]

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

[ "In tens place: there are ten $$5$$s: $$50\\sim59$$ In ones place: there are six $$5$$s: $$5, 15, 25, 35, 45, 55$ " ]

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

[]

[ [ { "aoVal": "A", "content": "$$5786$$ " } ], [ { "aoVal": "B", "content": "$$5787$$ " } ], [ { "aoVal": "C", "content": "$$5788$$ " } ], [ { "aoVal": "D", "content": "$$5789$$ " } ], [ { "aoVal": "E", "content": "$$5784$$ " } ] ]

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

[ "The number can be written as $$n+(n+1)=2n+1(n\\geqslant 1)$$ and $$x+(x+1)+(x+2)=3x+3$$. It must be a multiple of $3$ . $5790$ can be divisible by both $2$ and $3$, so it is $5790-3=5787$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$98765$$ " } ], [ { "aoVal": "B", "content": "$$98750$$ " } ], [ { "aoVal": "C", "content": "$$98755$$ " } ] ]

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

[ "The last three digits must be divisible by $125$, and the ones place must be divisible by $2$, so it can only be $98750$. " ]

[]

[ [ { "aoVal": "A", "content": "$$23\\times34$$ " } ], [ { "aoVal": "B", "content": "$$34\\times45$$ " } ], [ { "aoVal": "C", "content": "$$45\\times56$$ " } ], [ { "aoVal": "D", "content": "$$56\\times67$$ " } ], [ { "aoVal": "E", "content": "$$67\\times78$$ " } ] ]

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

[ "Of the options given, $$23\\times 34$$, $$56\\times 67$$ and $$67\\times 78$$ are all not divisible by $$5$$, so may be discounted. Also $$34$$ is not divisible by $$4$$ and $$45$$ is odd, so $$34\\times 45$$ may also be discounted as it is not divisible by $$4$$. The only other option is $$45\\times 56$$. As a product of prime factors, $$45\\times 56=2^{3}\\times3^{2}\\times5\\times7$$, so it is clear that it is divisible by all of the integers from $$1$$ to $$10$$ inclusive. " ]

[]

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

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

[ "$$(14\\times 9\\times 8)\\div \\left( 9\\times 7\\times 8 \\right)=14\\times 9\\times 8\\div 9\\div 7\\div 8=14\\div 7\\times (9\\div 9)\\times (8\\div 8)=14\\div 7=2$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$19$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

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

[ "$$10, 20, 30, 40, 50, 60, 70, 80, 90$$ " ]

[]

[ [ { "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. " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

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

[ "There are $$6$$ factors of $$5$$ when you prime factorise~ $$1\\times 2\\times 3\\times 4\\times 5\\times 6\\times \\cdots \\times 25$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$99$$ " } ], [ { "aoVal": "C", "content": "$$197$$ " } ], [ { "aoVal": "D", "content": "$$198$$ " } ] ]

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

[ "$$99 + 98 = 197$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

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

[ "Among $$3,4,5,6,8$$, only $$5$$ is a factor of $$70$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$101$$ " } ], [ { "aoVal": "B", "content": "$$102$$ " } ], [ { "aoVal": "C", "content": "$$103$$ " } ], [ { "aoVal": "D", "content": "$$104$$ " } ], [ { "aoVal": "E", "content": "$$105$$ " } ] ]

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

[ "The number of cards after adding $$2$$ is divisible by $$3$$, $$5$$, and $$7$$. Since the least common multiple of $$3$$, $$5$$, and $$7$$ is $$3\\times5\\times7=105$$, Jason has $$105-2=103$$ cards at least. " ]

[]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$30$$ " } ] ]

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

[ "$$10$$、$$30$$、$$50$$、$$80$$、$$90$$、$$18$$、$$38$$、$$58$$、$$98$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$49$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$51$$ " } ], [ { "aoVal": "D", "content": "$$59$$ " } ] ]

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

[ "There are $$60$$ whole numbers from $$0$$ to $$59$$. That\\textquotesingle s $$50$$ without $$0$$ to $$9$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2016$$ " } ], [ { "aoVal": "B", "content": "$$2015$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

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

[ "The sum of any even number of odd integers is always even. " ]

[]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$1$$ " } ] ]

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

[ "The only common factor of any two consecutive whole numbers is $$1$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2350$$ is a multiple of $$2$$, $$5$$ and $$10$$ " } ], [ { "aoVal": "B", "content": "$$1284$$ is a multiple of $$4$$ " } ], [ { "aoVal": "C", "content": "$$9972$$ is not a multiple of $$9$$ " } ], [ { "aoVal": "D", "content": "$$5742$$ is a multiple of $$6$$ and $$9$$ " } ] ]

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

[ "A: $$2350$$ is even, thus it is a multiple of $$2. 2350$$ ends with $$0$$, thus it is a multiple of $$5$$ and $$10$$; B: $$1284$$ ends with $$84$$, and $$84$$ is a multiple of $$4$$, thus $$1284$$ is a multiple of $$4$$; C: $$9 + 9 + 7 + 2 = 27$$ is a multiple of $$9$$, thus $$9972$$ is a multiple of $$9$$; D: $$5 + 7 + 4 + 2 = 18$$, it is a multiple of $$9$$ and $$3$$; since it ends with $$2$$, it is also a multiple of $$2$$, and thus a multiple of $$6$$; Hence the incorrect statement is C. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$70$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ] ]

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

[ "$77=7\\times 11$ $364=2\\times 2\\times 7\\times 13$ Because $B$ is the factor both number contains, $B=7$ Thus, $A=11$, $C=52$. " ]

[]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$63$$ " } ], [ { "aoVal": "C", "content": "$$72$$ " } ], [ { "aoVal": "D", "content": "$$96$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Common Multiples and Least Common Multiples->Least Common Multiple of Two Numbers" ]

[ "$$72$$ " ]

[]

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

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

[ "$$2000\\div1001$$ and $$2000\\div999$$ leave whole \\# remainders $$999$$ and $$2$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

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

[ "omitted PMC 2021 \\#7 " ]

[]

[ [ { "aoVal": "A", "content": "$$2940$$ " } ], [ { "aoVal": "B", "content": "$$2960$$ " } ], [ { "aoVal": "C", "content": "$$2980$$ " } ], [ { "aoVal": "D", "content": "$$3000$$ " } ], [ { "aoVal": "E", "content": "$$3020$$ " } ] ]

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

[ "$$M-N=3177$$ $$(30000+\\overline{abcd})-(10\\times \\overline{abcd}+3)=3177$$ $$30000-3177=9\\times \\overline{abcd}$$ $$\\overline{abcd}=2980$$. " ]

[]

[ [ { "aoVal": "A", "content": "＄$$2.40$$ " } ], [ { "aoVal": "B", "content": "＄$$3.80$$ " } ], [ { "aoVal": "C", "content": "＄$$4.40$$ " } ], [ { "aoVal": "D", "content": "＄$$5.20$$ " } ] ]

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

[ "The value of $$1$$ quarter, $$1$$ dime, and $$1$$ nickel is $$40$$￠. My coins must have a total value divisible by $$40$$, but ＄$$3.80$$ is not divisible by $$40$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

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

[ "$$6170$$ " ]

[]

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

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

[ "$$8+88+888+8888+88888=98760$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$47$$ " } ], [ { "aoVal": "C", "content": "$$49$$ " } ], [ { "aoVal": "D", "content": "$$51$$ " } ] ]

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

[ "$47$ is a prime number. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$29$$ " } ], [ { "aoVal": "C", "content": "$$31$$ " } ], [ { "aoVal": "D", "content": "$$37$$ " } ], [ { "aoVal": "E", "content": "$$14$$ " } ] ]

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

[ "If at least half the guesses are too low, the Norb\\textquotesingle s age must be greater than $34$. If two of the guesses are off by one, then his age is between two guesses whose difference is $2$. It could be $29$, $35$ or $37$, but because his age is greater than $34$, it can only be $35$ or $37$. Lastly, Norb\\textquotesingle s age is a prime number so the answer must be $37$. " ]

[ "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" ]

[ "None of them " ]

[]

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

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

[ "In division by $$4$$, the last $$2$$ digits determine the remainder, use $$22\\div4$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$29$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$33$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ], [ { "aoVal": "E", "content": "$$37$$ " } ] ]

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

[ "Because $$39$$ is an odd number, one of the two numbers must be an even prime number $$2$$. The other number is $$39-2=37$$, and their difference is $$35$$. So the answer is $$\\text{B}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$49$$ " } ], [ { "aoVal": "E", "content": "$$63$$ " } ] ]

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

[ "The prime factorization is $2016=2^{5} \\times 3^{2} \\times 7$. Since the problem is only asking us for the distinct prime factors, we have $2,3,7$. Their desired sum is then (B) 12 " ]

[]

[ [ { "aoVal": "A", "content": "$$4$$ and $$9$$ " } ], [ { "aoVal": "B", "content": "$$6$$ and $$27$$ " } ], [ { "aoVal": "C", "content": "$$27$$ and $$50$$ " } ], [ { "aoVal": "D", "content": "$$33$$ and $$100$$ " } ] ]

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

[ "The greatest common factors for the choices are $$1$$, $$3$$, $$1$$, $$1$$, respectively. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "April " } ], [ { "aoVal": "B", "content": "May " } ], [ { "aoVal": "C", "content": "June " } ], [ { "aoVal": "D", "content": "July " } ], [ { "aoVal": "E", "content": "August " } ] ]

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

[ "$$90\\div12=7R6$$ $$6$$ months afterFebruary will be August " ]

[]

[ [ { "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 smallest even factor is $$2$$; $$30\\div2= 15$$, the greatest odd factor. " ]

[]

[ [ { "aoVal": "A", "content": "Tulip\\textquotesingle s Reading Room " } ], [ { "aoVal": "B", "content": "Lily\\textquotesingle s Reading Room " } ], [ { "aoVal": "C", "content": "Both rooms " } ], [ { "aoVal": "D", "content": "None of the rooms " } ] ]

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

[ "There are two lamps on every table in Tulip\\textquotesingle s Reading Room, so the number of lamps in it is an even number. The total number of lamps in the two rooms is odd, and thus the number of lamps in Lily\\textquotesingle s Reading Room is an odd number. Since there are three lamps on every table in Lily\\textquotesingle s Reading Room, the number of tables in it is also an odd number. However, the total number of tables in the two rooms is odd, so the number of tables in Tulip\\textquotesingle s is even. Therefore, the answer is $$\\text{B}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

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

[ "A number divisible by $$12$$ \\& $$5$$ is divisible by a product of their factors, such as $$3\\times5$$. " ]

[]

[ [ { "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->Remainder Problems->Characteristics of Remainder " ]

[ "The remainder of $$A$$ $$\\div7$$ is $$5$$, and $$2A=A+A$$. Therefore the remainder of $$2A\\div7$$ is $$5+5=10$$. $$10= 7+3$$, therefore the remainder is $$3$$. " ]

[]

[ [ { "aoVal": "A", "content": "An odd number + An even number " } ], [ { "aoVal": "B", "content": "An even number + An even number " } ], [ { "aoVal": "C", "content": "An odd number + An odd number " } ] ]

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

[ "The sum of an odd number and an even number is always an odd number. The sum of two numbers which are both even or both odd is always an even number. " ]

[]

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

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

[ "$$23$$, $$32$$, and $$7$$ are positive; $$-56$$, $$-98$$ are negative. " ]

[]

[ [ { "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->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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$52$$ " } ], [ { "aoVal": "B", "content": "$$104$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$234$$ " } ] ]

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

[ "one back and forth $$156\\div3=52$$metres length of the lane$$52\\div2=26$$metres " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$19$$ " } ] ]

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

[ "The sum must be even, and it could be $$2+4+8=14$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$123$$ " } ], [ { "aoVal": "D", "content": "$$240$$ " } ], [ { "aoVal": "E", "content": "$$243$$ " } ] ]

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

[ "If the ping pong balls in the box are reduced by $3$, then there will be no extra balls when you count them $8$ by $8$, $10$ by $10$, or $12$ by $12$. It means that the number of the ping pong balls is a common multiple of $8$, $10$, and $12$ after being reduced by $3$. If you want to know how many ping pong balls there are at least, you can first find the least common multiple of $8$, $10$ and $12$, and then add $3$ to get the answer. $$\\begin{array}{l} {2\\left\\textbar{} \\underline{\\textasciitilde8\\textasciitilde\\textasciitilde10\\textasciitilde\\textasciitilde12}\\right. }\\textbackslash\\textbackslash{\\textasciitilde2\\left\\textbar{} \\underline{4\\textasciitilde\\textasciitilde\\textasciitilde5\\textasciitilde\\textasciitilde\\textasciitilde6}\\right. }\\textbackslash\\textbackslash{\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde2\\textasciitilde\\textasciitilde\\textasciitilde5\\textasciitilde\\textasciitilde\\textasciitilde3} \\end{array}$$ $2\\times2\\times2\\times5\\times3=120$ $120+3=123$ " ]

[]

[ [ { "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": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$44$$ " } ] ]

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

[ "$$3+2+2=7$$; $100\\div7=14R2$; $$14$$$\\times$$$3$$+$$2$$=$$44$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

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

[ "$55=5\\times 11$ $100=2\\times 2\\times 5\\times 5$ Because $B$ is the factor both number contains, $B=5$ Thus, $A=11$, $C=20$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "Such a representation is impossible. " } ] ]

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

[ "If so, $2$ must be one of the prime numbers, and the other must be $2003-2=2001$, which is not prime. Thus, such a representation is impossible. " ]

[]

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

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

[ "I wrote $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, and $$9$$ on my page. Adding $$1$$ to $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, and $$8$$ results in $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, and $$9$$. We cannot get a sum of $$1$$ or $$2$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$64$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

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

[ "$$\\sqrt {2\\times 4\\times 8}\\times \\sqrt {8\\times 8}=8\\times 8=64$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$321$$ " } ], [ { "aoVal": "B", "content": "$$489$$ " } ], [ { "aoVal": "C", "content": "$$644$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

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

[ "even numbers end with 0,2,4,6,8 " ]

[ "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" ]

[ "$0, 2, 4, 6,$ and $8$ " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

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

[ "The number of cards after adding $$2$$ is divisible by $$3$$, $$5$$, and $$7$$. Since the least common multiple of $$3$$, $$5$$, and $$7$$ is $$3\\times5\\times7=105$$, Jason has $$105-2=103$$ cards at least. " ]

[]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

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

[ "If $$1\\times2\\times3\\times4\\times5\\times6\\times7\\times$$$$8\\times9\\times10\\times11\\times12\\times13\\times14\\times15 = $$$$1307 674 368000$$, and we multiply each of these $$15$$ numbers by $$10$$, the new product will have an additional $$15$$ zeroes, and $$15+4=19$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$1239$$ " } ], [ { "aoVal": "E", "content": "$$1265$$ " } ] ]

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

[ "Note that $n \\equiv S(n) \\bmod 9$, so $S(n+1)-S(n) \\equiv n+1-n=1 \\bmod 9$. So, since $S(n)=1274 \\equiv 5 \\bmod 9 $, we have that $S(n+1) \\equiv 6 \\bmod 9$. Then, only one of the answer choices is congruent to $6 \\bmod 9$, which is $(D)=1239$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$48$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$125$$ " } ] ]

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

[ "$4\\times 4\\times3=48$ " ]

[]

[ [ { "aoVal": "A", "content": "$$13 $$ " } ], [ { "aoVal": "B", "content": "$$23 $$ " } ], [ { "aoVal": "C", "content": "$$31 $$ " } ], [ { "aoVal": "D", "content": "$$41 $$ " } ], [ { "aoVal": "E", "content": "$$43$$ " } ] ]

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

[ "First note that Arthur can write down three squares, namely $$16$$, $$25$$ and $$36$$. Also, he can write down four triangular numbers, namely $$15$$, $$21$$, $$36$$ and $$45$$. If he chooses $$16$$ and $$45$$ for the square and triangular number respectively, then the remaining digits are $$2$$ and $$3$$, the prime is $$23$$. If he chooses $$25$$ and $$36$$ then the remaining digits are $$1$$ and $$4$$, the prime is $$41$$. If he chooses $$36$$ for the square number, the remaining difits can be a prime. So the largest prime he write is $$41$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$11\\times99$$ " } ], [ { "aoVal": "B", "content": "$$44\\times33$$ " } ], [ { "aoVal": "C", "content": "$$55\\times22$$ " } ], [ { "aoVal": "D", "content": "$$88\\times66$$ " } ] ]

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

[ "The product is even $$except$$ when both numbers you multiply are odd. " ]

[]

[ [ { "aoVal": "A", "content": "$$456456456456456$$ " } ], [ { "aoVal": "B", "content": "$$678 678678678678$$ " } ], [ { "aoVal": "C", "content": "$$432432432432432$$ " } ], [ { "aoVal": "D", "content": "$$876876876 876876$$ " } ] ]

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

[ "To check for an odd quotient, divide only the last two digits by $$2$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "none of them " } ], [ { "aoVal": "B", "content": "statement 1 only " } ], [ { "aoVal": "C", "content": "statement 2 only " } ], [ { "aoVal": "D", "content": "statement 3 only " } ], [ { "aoVal": "E", "content": "statements 2 and 3 only " } ] ]

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

[ "3+5+8=16，$$3+5+6=14$$，$$5+6+8=19$$，$$16$$ is a multiple of 8, (1) wrong; No multiples of 3, (2) wrong, 14,16 is even, 19 is odd. (3) Wrong " ]

[]

[ [ { "aoVal": "A", "content": "$50^{}\\circ $ " } ], [ { "aoVal": "B", "content": "$120^{}\\circ $ " } ], [ { "aoVal": "C", "content": "$135^{}\\circ $ " } ], [ { "aoVal": "D", "content": "$150^{}\\circ $ " } ], [ { "aoVal": "E", "content": "$165^{}\\circ $ " } ] ]

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

[ "The smaller angle is $\\frac 5{12}$ of a full circle. A full circle has $360$ degrees, so the angle is $\\frac 5{12}\\times 360^{}\\circ =150^{}\\circ $. So, the answer is $\\rm D$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1.6$$ " } ], [ { "aoVal": "B", "content": "$$1.8$$ " } ], [ { "aoVal": "C", "content": "$$2.0$$ " } ], [ { "aoVal": "D", "content": "$$2.2$$ " } ], [ { "aoVal": "E", "content": "$$2.4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "After the first swap, we do casework on the next swap. Case 1: Silva swaps the two balls that were just swapped There is only one way for Silva to do this, and it leaves 5 balls occupying their original position. Case 2: Silva swaps one ball that has just been swapped with one that hasn\\textquotesingle t swapped There are two ways for Silva to do this, and it leaves 2 balls occupying their original positions. Case 3 : Silva swaps two balls that have not been swapped There are two ways for Silva to do this, and it leaves 1 ball occupying their original positions. Our answer is the average of all 5 possible swaps, so we get $$ \\frac{5+2 \\cdot 2+2 \\cdot 1}{5}=\\frac{11}{5}=(\\text { D) } 2.2 $$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "north " } ], [ { "aoVal": "B", "content": "north-east " } ], [ { "aoVal": "C", "content": "north-west " } ], [ { "aoVal": "D", "content": "south-east " } ], [ { "aoVal": "E", "content": "south-west " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "58 quarter-turn clockwise, 93 quarter-turn anti-clockwise, is the same thing as 35 and 1/4 quarter-turn counterclockwise, 35*1/4=8 turns more than 3/4, so it is facing southwest. " ]

[]

[ [ { "aoVal": "A", "content": "$$300$$ " } ], [ { "aoVal": "B", "content": "$$500$$ " } ], [ { "aoVal": "C", "content": "$$600$$ " } ], [ { "aoVal": "D", "content": "$$700$$ " } ], [ { "aoVal": "E", "content": "$$1000$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets" ]

[ "$(1200+1500)-(2006-6)=700$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$49200$$ " } ], [ { "aoVal": "B", "content": "$$94080$$ " } ], [ { "aoVal": "C", "content": "$$564480$$ " } ], [ { "aoVal": "D", "content": "$$1800$$ " } ], [ { "aoVal": "E", "content": "$$98400$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations" ]

[ "There are $3$ $E$s in total now with other $8$ letters remaining. But pay attention to $B$ and $P$: there are $3$ $B$s and $3$ $P$ here. There are $\\_8P\\_5 \\div \\_3P\\_3$ ways for us to arrange the $8$ letters\\textquotesingle{} positions. Then, we can put the $3$ $E$s in the $9$ intervals. So the answer is $\\_8P\\_5 \\div \\_3P\\_3 \\times \\_9C\\_3=94080$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$11$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Operational Problem" ]

[ "Start from either the left or the right and follow the steps until the basket in the middle has $$0$$ candy. And you can see that one side has $$5$$ pieces and the other side has $$6$$ pieces. So the largest number of pieces of candy is $$6$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$23$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "The prime factorization of $12 !$ is $2^{10} \\cdot 3^{5} \\cdot 5^{2} \\cdot 7 \\cdot 11$. This yields a total of $11 \\cdot 6 \\cdot 3 \\cdot 2 \\cdot 2$ divisors of $12 !$. In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in $12 !$. Thus, there are $6 \\cdot 3 \\cdot 2$ perfect squares. (For $2$ , you can choose $0,2,4,6,8,$ or $10$, etc. The probability that the divisor chosen is a perfect square is $$ \\frac{6 \\cdot 3 \\cdot 2}{11 \\cdot 6 \\cdot 3 \\cdot 2 \\cdot 2}=\\frac{1}{22} \\Longrightarrow \\frac{m}{n}=\\frac{1}{22} \\Longrightarrow m+n=1+22=\\text { (E) } 23 $$. " ]

[]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{2}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{1}{3}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{3}{4}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{1}{6}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "The probability that both show a green bean is $\\dfrac{1}{2}\\cdot \\dfrac{1}{3}=\\dfrac{1}{6}$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2015$$ " } ], [ { "aoVal": "B", "content": "$$2017$$ " } ], [ { "aoVal": "C", "content": "$$2097$$ " } ], [ { "aoVal": "D", "content": "$$2099$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "Each number in the second sum is $$18$$ greater than the corresponding number in the first sum. Thus the second sum is $$1935 +18 \\times9 = 2097$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$\\frac1{11}$ " } ], [ { "aoVal": "B", "content": "$\\frac4{11}$ " } ], [ { "aoVal": "C", "content": "$\\frac7{11}$ " } ], [ { "aoVal": "D", "content": "$\\frac{10}{11}$ " } ], [ { "aoVal": "E", "content": "$\\frac2{11}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability" ]

[ "There are $4$ numbers greater than $7$. Thus, the probability is $\\frac4{11}$. " ]

[]

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

[ "Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations" ]

[ "$3\\times 2\\times 1=6$ There are six different ways for three kids to line up. " ]

[]

[ [ { "aoVal": "A", "content": "$$13$$ minutes " } ], [ { "aoVal": "B", "content": "$$23$$ minutes " } ], [ { "aoVal": "C", "content": "$$39$$ minutes " } ], [ { "aoVal": "D", "content": "$$47$$ minutes " } ], [ { "aoVal": "E", "content": "$$73$$ minutes " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Using Formulas" ]

[ "The rider travels for $$13$$ hours over $$60$$ rides, which is anaverage time of $$\\frac{13}{60}$$ of an hour per ride, hence $$13$$ minutes. " ]

[]

[ [ { "aoVal": "A", "content": "$$120$$ " } ], [ { "aoVal": "B", "content": "$$240$$ " } ], [ { "aoVal": "C", "content": "$$285$$ " } ], [ { "aoVal": "D", "content": "$$320$$ " } ], [ { "aoVal": "E", "content": "$$400$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers" ]

[ "The smallest \\textquotesingle V-number\\textquotesingle{} is $$101$$ and the largest \\textquotesingle V-mumber\\textquotesingle{} is $$989$$. Consider the tens digits. The smallest tens digit is $$0$$ and the largest tens digit is $$8$$. If the tens digit is $$0$$, the hundreds digit can be $$1$$ to $$9$$, and the units digit can be $$1$$ to $$9$$, giving $$9 \\times 9$$ possible \\textquotesingle V-numbers\\textquotesingle. If the tens digit is $$1$$, then the hundreds digit can be $$2$$ to $$9$$ and the units digit can be $$2$$ to $$9$$, giving $$8 \\times 8$$ possible \\textquotesingle V-numbers\\textquotesingle. If the tens digit is $$d$$, where $$d$$ can be any digit from $$0$$ to $$8$$, the hundreds digit can be $$(d + 1)$$ to $$9$$ and the units digit can be $$(d + 1)$$ to $$9$$, giving $$(9-d)\\times (9-d)$$ possible \\textquotesingle V-numbers\\textquotesingle. The greatest value of $$d$$ is $$8$$. In this case, the hundreds digit can only be $$9$$ and the units digit can only be $$9$$, which gives just $$1 \\times 1$$ possibilities. This gives the total number of possible \\textquotesingle V-numbers\\textquotesingle{} to be $$9\\times9+8\\times8+\\cdots +1\\times1 = 285$$, which is the sum of the squares from $$1$$ to $$9$$ inclusive. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$42$$ " } ], [ { "aoVal": "B", "content": "$$52$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$46$$ " } ], [ { "aoVal": "E", "content": "$$56$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle" ]

[ "In $31+28+31=90$ days, they could visit the grandpa in $[90\\div4]+[90\\div5]+[90\\div6]-[90\\div20]-[90\\div12]-[90\\div30]+[90\\div60]=42$ days. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration" ]

[ "The possible groups of three integers with product $$36$$ are $$(1,1,36)$$, $$(1,2,18)$$, $$(1,3,12)$$, $$(1,4,9)$$, $$(1,6,6)$$, $$(2,2,9)$$, $$(2,3,6)$$ and $$(3,3,4)$$ with sums $$38$$, $$21$$, $$16$$, $$14$$, $$13$$, $$13$$, $$11$$ and $$10$$ respectively. The only value for the sum that occurs twice is $$13$$. Hence, since Topaz does not know what the three integers chosen are, the sum of Harriet\\textquotesingle s three integers is $$13$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$55$$ " } ], [ { "aoVal": "C", "content": "$$65$$ " } ], [ { "aoVal": "D", "content": "$$75$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle" ]

[ "E " ]