Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$33$$ " } ], [ { "aoVal": "C", "content": "$$35$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

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

[ "$21=3\\times 7$ $60=2\\times 2\\times 3\\times 5$ Because $B$ is the factor both number contains, $B=3$ Thus, $A=7$, $C=20$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$6$ and $10$ " } ], [ { "aoVal": "B", "content": "$5$ and $12$ " } ], [ { "aoVal": "C", "content": "$10$ and $12$ " } ], [ { "aoVal": "D", "content": "$12$ and $16$ " } ] ]

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

[ "A: The least common multiple of $6$ and $10$ is $30$. B: Two numbers are relatively prime. C: The least common multiple is $60$, and the greatest common factor is $2$. D: The least common multiple is $48$, and the greatest common factor is $4$. " ]

[]

[ [ { "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 books 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 $$\\rm A$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$89$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$101$$ " } ] ]

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

[ "$47+53=100$ " ]

[ "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 $$156\\div3=52$$meters length of the lane$$52\\div2=26$$meters " ]

[]

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

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

[ "$$360={{2}^{3}}\\times {{3}^{2}}\\times5$$, so the number of its factors would be $$\\left( 3+1 \\right)\\times \\left( 2+1 \\right)\\times\\left( 1+1 \\right)=24$$. Among them, there are $$2\\times 2\\times(1+1)=8$$ factors which have $${{2}^{2}}\\times 3$$ as its factors. " ]

[]

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

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

[ "Since $$10$$ is a factor of this product, the ones digit is $$0$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$1.01$$ " } ], [ { "aoVal": "C", "content": "$$1\\frac1{111}$$ " } ], [ { "aoVal": "D", "content": "$$1.009$$ " } ] ]

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

[ "$$0.\\overline{234} +0.\\overline{342} +0.\\overline{432} =1.\\overline{009} $$$$=1\\frac 1{111}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$62$$ " } ], [ { "aoVal": "D", "content": "$$72$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$4$$ tens $$5$$ ones: $$45$$ $$2$$ tens $$7$$ ones: $$27$$ Greater than: $$45 + 27 = 72$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$152$$ " } ], [ { "aoVal": "D", "content": "$$383$$ " } ] ]

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

[ "Since $$231=3\\times7\\times11$$, the sum of its prime factors is $$3 +7+11=21$$. " ]

[]

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

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

[ "Multiply the three smallest primes: $$2\\times3 \\times5=30$$, whose $$8$$ divisors are $$1$$, $$2$$, $$3$$, $$5$$, $$6$$, $$10$$, $$15$$, and $$30$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ], [ { "aoVal": "E", "content": "$$32$$ " } ] ]

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

[ "The digit $3$ appears in the ones place: $3$, $13$, $23$, The digit $3$ appears in the tens place: $30, 31, 32, 33, 34, 35$ In total, $3 + 6 = 9$. " ]

[]

[ [ { "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 $$9$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$63$$ " } ], [ { "aoVal": "B", "content": "$$77$$ " } ], [ { "aoVal": "C", "content": "$$84$$ " } ], [ { "aoVal": "D", "content": "$$91$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "A multiple of $$3$$ and $$7$$ also a multiple of $$21$$ $$21\\times3=63$$ $$21\\times4=even$$ $$21\\times5=105$$, exceed Thus, only $$63$$ " ]

[ "其它" ]

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

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

[ "$7756, 3728,$ and $8064$ can be divisible by $4$. " ]

[]

[ [ { "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. $$\\begin{array}{l} {2\\left\\textbar{} \\underline{\\textasciitilde8\\textasciitilde\\textasciitilde10\\textasciitilde\\textasciitilde12}\\right. }\\textbackslash\\textbackslash{\\textasciitilde2\\left\\textbar{} \\underline{4\\textasciitilde\\textasciitilde\\textasciitilde5\\textasciitilde\\textasciitilde\\textasciitilde6}\\right. }\\textbackslash\\textbackslash{\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde2\\textasciitilde\\textasciitilde\\textasciitilde5\\textasciitilde\\textasciitilde\\textasciitilde3} \\end{array}$$ $2\\times2\\times2\\times5\\times3=120$ $120+3=123$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Monday " } ], [ { "aoVal": "B", "content": "Wednesday " } ], [ { "aoVal": "C", "content": "Thursday " } ], [ { "aoVal": "D", "content": "Saturday " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$30-1+1=30$$ $$30\\div7=4R2$$ Tuesday -\\/-\\textgreater~Wednesday. " ]

[]

[ [ { "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" ]

[ "$$123123123$$ is divisible by $$3$$, since the sum of its digits is $18$ which is divisible by $3$. " ]

[]

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

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

[ "The product of two odd numbers is odd, so there is only one even product, $$2\\times 11$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "All prime numbers are odd numbers. " } ], [ { "aoVal": "B", "content": "In every $3$ consecutive numbers, there must be a composite number. " } ], [ { "aoVal": "C", "content": "All even numbers are composite numbers. " } ], [ { "aoVal": "D", "content": "The sum of two different odd number must be a composite number. " } ] ]

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

[ "A: Any non-$3$ multiple of 3 are composite B: $1$, $2$, $3$ does not have composite number C: $2$ is prime number D: $1+$any odd number is multiple of $2$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$407$$ " } ], [ { "aoVal": "B", "content": "$$408$$ " } ], [ { "aoVal": "C", "content": "$$418$$ " } ], [ { "aoVal": "D", "content": "$$419$$ " } ], [ { "aoVal": "E", "content": "$$428$$ " } ] ]

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

[ "$$ \\left\\textbackslash{\\begin{array}{l} a(b+1)=1443=3 \\times 13 \\times 37=37 \\times 39 \\textbackslash\\textbackslash{} b(a+1)=1444=38^{2} \\end{array}\\right. $$ Compare and $a=37, b=38 .$ So $10 a+b=370+38=408$. " ]

[ "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" ]

[ "In total, there were $3+4=7$ marbles left from both Ringo and Paul.We know that $7 \\equiv 1(\\bmod 6)$. This means that there would be $1$ marble leftover. " ]

[]

[ [ { "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": "$$55$$ " } ], [ { "aoVal": "B", "content": "$$71$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$66$$ " } ] ]

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

[ "$$6\\times 11+5=71$$． " ]

[ "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": "$$10$$ and $$20 $$ " } ], [ { "aoVal": "B", "content": "$$20$$ and $$30$$ " } ], [ { "aoVal": "C", "content": "$$30$$ and $$40$$ " } ], [ { "aoVal": "D", "content": "$$40$$ and $$50$$ " } ] ]

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

[ "The $$3$$ prime numbers between $$40$$ and $$50$$ are $$41$$, $$43$$, and $$47$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

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

[ "$$3+2+B+9$$ = 14+B is divisble by 3. Max single digit possible is 9. Then 14+9 = 23. Range is between 15 to 23. Multiple of 3 - 3 x 7=21, 3 x 6 =18, 3 x 5 = 15. Since, it should be even number, 18-14=4. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

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

[ "The result must hold for any three-digit number with hundreds digit being 2 more than the units digit. $301$ is such a number. Evaluating, we get $301-103=198$. Thus, the units digit in the final result is $(\\mathbf{E}) 8$ " ]

[ "其它" ]

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

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

[ "$30=5 \\times 6$. " ]

[]

[ [ { "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": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$19$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

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

[ "$$2$$, $$12$$, $$20-29$$, $$32$$, $$42$$, $$52$$, $$62$$, $$72$$, $$82$$, and $$92$$ use a $$2$$; that\\textquotesingle s $$19$$ numbers. " ]

[]

[ [ { "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 have $${{2}^{2}}$$ as its factors. " ]

[]

[ [ { "aoVal": "A", "content": "$$000$$ " } ], [ { "aoVal": "B", "content": "$$222$$ " } ], [ { "aoVal": "C", "content": "$$444$$ " } ], [ { "aoVal": "D", "content": "$$666$$ " } ], [ { "aoVal": "E", "content": "$$888$$ " } ] ]

[ "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": "$$15$$ " } ], [ { "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": "$$43$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ], [ { "aoVal": "E", "content": "$$3$$ " } ] ]

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

[ "According to information $(3)$ and $(4)$, we know $3$ must be in the ones place and the digit in the tens place is an even number. Thus, it should be one of the numbers from $23$, $43$, $63$ and $83$. According to information $(1)$, it can be $23$, $43$, and $83$. According to information $(2)$, it can only be $23$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$43$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ], [ { "aoVal": "E", "content": "$$3$$ " } ] ]

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

[ "According to information $(3)$ and $(4)$, we know $3$ must be in the ones place and the digit on the tens place is an even number. Thus, it should be one of the numbers from $23$, $43$, $63$ and $83$. According to information $(1)$, it can be $23$, $43$, and $83$. According to information $(2)$, it can only be $23$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4451$$ " } ], [ { "aoVal": "B", "content": "$$4453$$ " } ], [ { "aoVal": "C", "content": "$$4450$$ " } ], [ { "aoVal": "D", "content": "$$4452$$ " } ] ]

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

[ "Odd + Even = Odd~~ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$Yes.$$ " } ], [ { "aoVal": "B", "content": "$$No.$$ " } ] ]

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

[ "If there are odd number of odd page number, the sum of page numbers is odd. Therefore, it could be 953. " ]

[]

[ [ { "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": "$$69$$ " } ], [ { "aoVal": "B", "content": "$$67$$ " } ], [ { "aoVal": "C", "content": "$$63$$ " } ], [ { "aoVal": "D", "content": "$$61$$ " } ] ]

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

[ "omitted " ]

[]

[ [ { "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->Numbers" ]

[ "The thousands\\textquotesingle{} digit is $$2$$ and the tens\\textquotesingle{} digit is $$4$$. The sum is $$6$$. " ]

[]

[ [ { "aoVal": "A", "content": "　one " } ], [ { "aoVal": "B", "content": "one hundred " } ], [ { "aoVal": "C", "content": "one thousand " } ], [ { "aoVal": "D", "content": "one million " } ] ]

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

[ "One million $$= 1000000$$. Adding, $$1+0+0+0+0+0+0 = 1$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ], [ { "aoVal": "E", "content": "$$18$$ " } ] ]

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

[ "$6\\times8\\times10 \\div(2\\times2\\times2)=60.$ $60=2^{2}\\times3\\times5$, which means it has $3\\times2\\times2=12$ factors. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ], [ { "aoVal": "E", "content": "$$48$$ " } ] ]

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

[ "Obsever that $3\\times27=9^{2}$ and $12\\times27=24^{2}$ are both squares. Hence, their product is also a perfect square and the fifth number must be $6$. " ]

[]

[ [ { "aoVal": "A", "content": "$$108$$ is a multiple of $$3$$ and $$9$$ " } ], [ { "aoVal": "B", "content": "$$100$$ is a multiple of $$10$$, but it is not a multiple of $$5$$ " } ], [ { "aoVal": "C", "content": "$$164$$ is a multiple of $$4$$ " } ], [ { "aoVal": "D", "content": "$$132$$ is a multiple of $$6$$ " } ] ]

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

[ "$$\\rm A$$: $$1+0+8=9$$, it is a multiple of $$3$$ and $$9$$; $$\\rm B$$: $$100$$ ends with $$0$$, thus it is a multiple of $$10$$ and $$5$$; $$\\rm C$$: $$164$$ ends with $$64$$, and $$64$$ is a multiple of $$4$$, thus $$164$$ is also a multiple of $$4$$; $$\\rm D$$: $$132$$ is a multiple of $$2$$ and $$3$$, thus it is a multiple of $$6$$. So we choose $$\\rm B$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$19$$ " } ] ]

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

[ "As $$3$$ and $$5$$ are coprime, the squares that have more than one mark are multiples of both $$3$$ and $$5$$, $$\\left( {} \\right.$$multiples of $$15$$$$\\left. {} \\right)$$; or multiples of both $$3$$ and $$7$$, $$\\left( {} \\right.$$multiples of $$21$$$$\\left. {} \\right)$$; or multiples of $$5$$ and $$7$$, $$\\left( {} \\right.$$multiples of $$35$$$$\\left. {} \\right)$$; or multiples of $$3$$, $$5$$ and $$7$$, $$\\left( {} \\right.$$multiples of $$105$$$$\\left. {} \\right)$$. However, the latter will be included in all of the first three categories. Between $$1$$ and $$150$$ inclusive, there are ten multiples of $$15$$, seven multiples of $$21$$ and four multiples of $$35$$, making a total of $$21$$ multiples. However, there is one multiple of $$3$$, $$5$$ and $$7$$ between $$1$$ and $$150$$, namely $$105$$. So $$105$$ has been counted three times in those $$21$$ multiples, but corresponds to exactly one marked square. Therefore the total number of marked squares is $$21−2=19$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Yes " } ], [ { "aoVal": "B", "content": "No " } ] ]

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

[ "$$722-12-385=325$$ " ]

[]

[ [ { "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": "$$45$$ " } ], [ { "aoVal": "B", "content": "$$46$$ " } ], [ { "aoVal": "C", "content": "$$47$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ], [ { "aoVal": "E", "content": "$$49$$ " } ] ]

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

[ "Let the two digits be $$a$$ and $$b$$. The correct score was $$10a+b$$. Clara misinterpreted it as $$10b+a$$. The difference between the two is $$\\left\\textbar{} 9a-9b\\right\\textbar$$ which factors into $$\\left\\textbar{} 9(a-b)\\right\\textbar$$. Therefore, since the difference is a multiple of $$9$$, the only answer choice that is a multiple of $$9$$ is $$45$$ . ( $$2013$$ AMC $$8$$ Problem, Question \\#$$13$$) " ]

[]

[ [ { "aoVal": "A", "content": "$$97=1\\times97$$ " } ], [ { "aoVal": "B", "content": "$$85=5\\times 17$$ " } ], [ { "aoVal": "C", "content": "$$64=8\\times 8$$ " } ], [ { "aoVal": "D", "content": "$$52=1\\times 2\\times 2\\times 13$$ " } ] ]

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

[ "①A prime number is a natural number greater than $$1$$ that has no other factors than $$1$$ and itself. ②Decomposing a composite number into the product of several prime factors is called decomposing prime factors. ③$$1$$ is neither a prime number nor a composite number. So $$\\text{D}$$ is not correct. The $$\\text{A}$$ option is also incorrect. $$8$$ is not a prime number, so $$\\text{C}$$ is incorrect. " ]

[]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

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

[ "As an example, $$42$$ is divisible by $$6$$ and by $$14$$ but not by $$12$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$2530\\sim2540$ " } ], [ { "aoVal": "B", "content": "$2540\\sim2550$ " } ], [ { "aoVal": "C", "content": "$2550\\sim2560$ " } ], [ { "aoVal": "D", "content": "$2570\\sim2590$ " } ], [ { "aoVal": "E", "content": "$2580\\sim2600$ " } ] ]

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

[ "If he distributes them among $12$ children evenly, there will be $11+12\\times4=59$ apples left. If he distributes them among $13$ children evenly, there will be $7+13\\times4=59$ apples left.~ If he distributes them among $14$ children evenly, there will be $3+14\\times 4=59$ apples left. So at least he takes $59+12\\times 13\\times 14=2243$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2:1$$ " } ], [ { "aoVal": "B", "content": "$$100:1$$ " } ], [ { "aoVal": "C", "content": "$$101:1$$ " } ], [ { "aoVal": "D", "content": "$$1001:1$$ " } ], [ { "aoVal": "E", "content": "It depends on his number " } ] ]

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

[ "Let Hasan\\textquotesingle s two-digit number be \\textquotesingle$$ab$$\\textquotesingle, which is equal to $$10 a + b$$. The four-digit number he forms is therefore \\textquotesingle$$abab$$\\textquotesingle, which is equal to $$1000 a + 100 b + 10 a + b$$ and hence to $$100 (10 a + b)+ 10 a + b = 101 \\times (10 a + b)$$. Therefore the ratio of his four-digit number to his two-digit number is $$101:1$$. " ]

[]

[ [ { "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 " ]

[ "$$123\\div 4$$ R $$3$$；$$234\\div 4$$ R $$2$$；$$345\\div 4$$ R $$1$$; Therefore, $$\\left( 3+2+1 \\right)\\div 4$$ R $$2$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "All prime numbers are odd numbers " } ], [ { "aoVal": "B", "content": "In every $3$ consecutive numbers, there must be a composite number " } ], [ { "aoVal": "C", "content": "All even numbers are composite numbers " } ], [ { "aoVal": "D", "content": "The sum of two different odd number must be a composite number " } ] ]

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

[ "A: Any non-$3$ multiple of 3 are composite B: $1$, $2$, $3$ does not have composite number C: $2$ is prime number D: $1+$any odd number is multiple of $2$. " ]

[]

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

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

[ "The number whose square is $$16$$ is $$4$$. The square root of $$4$$ is $$2$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$232$$ " } ], [ { "aoVal": "B", "content": "$$168$$ " } ], [ { "aoVal": "C", "content": "$$247$$ " } ], [ { "aoVal": "D", "content": "$$94$$ " } ], [ { "aoVal": "E", "content": "$$132$$ " } ] ]

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

[ "Odd number $+$ Odd number $=$ Even number Odd number $-$ Even number $=$ Odd number " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$14$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

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

[ "If the sum of $$2$$ whole numbers is $$12$$, and the product is $$32$$, then the numbers are $$8$$ and $$4$$; $$8-4 = 4$$. " ]

[]

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

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

[ "If the sum of $$7$$ whole numbers is even, there must be an even number of odd numbers. The total number of odd numbers could be $$6$$. " ]

[ "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. " ]

[ "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" ]

[ "$131$, $137$, $139$, $149$ " ]

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

[]

[ [ { "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->Remainder Problems->Characteristics of Remainder ->The additive property of the remainder" ]

[ "$$1001\\div a=b~~\\text {R} 5$$, $a$ is a one-digit number bigger than $$5$$, so $a$ =6; $$1005\\div a=c~ \\text {R} 9$$; $$2006\\div a=d \\text {R} 14$$; $$14=2\\times6 + 2$$, so the remainder is $$2$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$57$$ " } ], [ { "aoVal": "E", "content": "$$58$$ " } ] ]

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

[ "We compute $2^{8}+1=257$. We\\textquotesingle re all familiar with what $6^{3}$ is, namely, which is too small. The smallest cube greater than it is $7^{3}=343$. $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^{6})^{3}=64^{3}$, which therefore clearly will be the largest cube less than $2^{18}+1$. Therefore, the required number of cubes is $64-7+1=58$. " ]

[]

[ [ { "aoVal": "A", "content": "$$140$$ " } ], [ { "aoVal": "B", "content": "$142.8$ " } ], [ { "aoVal": "C", "content": "$142.86$ " } ], [ { "aoVal": "D", "content": "$143$ " } ], [ { "aoVal": "E", "content": "$$100$$ " } ] ]

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

[ "$$142.857$$ rounded to nearest: hundred($100$), ten($140$), one($143$), tenth($142.9$) and hundredth($142.86$). " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ] ]

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

[ "$32$ can not be divided by $12$. " ]

[ "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": "January " } ], [ { "aoVal": "B", "content": "February " } ], [ { "aoVal": "C", "content": "March " } ], [ { "aoVal": "D", "content": "April " } ], [ { "aoVal": "E", "content": "May " } ] ]

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

[ "$$100\\div12=8R4$$ $$4$$ months after December will be April. " ]

[ "其它" ]

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

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples->Common Factors and the Greatest Common Factors->The Greatest Common Factor of Multiple Numbers" ]

[ "The greatest common factor of $10$, $28$ and $30$ is $2$. " ]

[]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

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

[ "Divisibility by $$6$$ \\& $$20$$ does not promise divisibility by $$7$$, $$13$$, or $$8$$. $$\\rm A$$. $$12 = 4\\times3$$; $$\\rm B$$. $$14 = 2\\times7$$; $$\\rm C$$. $$26 = 2\\times13$$; $$\\rm D$$. $$120 = 8\\times15$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$143$$ " } ], [ { "aoVal": "D", "content": "$$116$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

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

[ "Note that $n \\equiv S(n) \\bmod 9$, so $S(n+2)-S(n) \\equiv n+2-n\\equiv2 \\bmod 9$. So, since $S(n)=691 \\equiv 7 \\bmod 9 $, we have that $S(n+2) \\equiv 9 \\equiv 0\\bmod 9$. Then, only one of the answer choices is congruent to $0 \\bmod 9$, which is $(A)=9$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$39$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ], [ { "aoVal": "E", "content": "$$48$$ " } ] ]

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

[ "Common multiple of $$12$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5\\rm{cm}$$ " } ], [ { "aoVal": "B", "content": "$$6\\rm{cm}$$ " } ], [ { "aoVal": "C", "content": "$$7\\rm{cm}$$ " } ], [ { "aoVal": "D", "content": "$$8\\rm{cm}$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$7\\times7=49\\rm{cm}^{2}$$ " ]

[ "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" ]

[ "$35=5\\times 7$ $84=2\\times 2\\times 3\\times 7$ Because $B$ is the factor both number contains, $B=7$ Thus, $A=5$, $C=12$. " ]

[]

[ [ { "aoVal": "A", "content": "$$14 $$ " } ], [ { "aoVal": "B", "content": "$$20 $$ " } ], [ { "aoVal": "C", "content": "$$36 $$ " } ], [ { "aoVal": "D", "content": "$$45 $$ " } ], [ { "aoVal": "E", "content": "$$50$$ " } ] ]

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

[ "The prime digits are $$2$$, $$3$$, $$5$$ and $$7$$. So the largest two-digit integer whose digits are both prime is $$77$$. However, $$77$$ is not prime, nor is $$75$$, but $$73$$ is prime. So Emily writes down $$73$$. The smallest two-digit integer whose digits are both prime is $$22$$. However, $$22$$ is not prime, but $$23$$ is prime. So Krish writes down $$23$$. Therefore the answer which Kirsten obtains is $$73 - 23 = 50$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "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 $(\\rm B)$ $3$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$25$$ " } ] ]

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

[ "It takes three days for one plant to turn to $8$ plants. " ]

[]

[ [ { "aoVal": "A", "content": "$$35$$ " } ], [ { "aoVal": "B", "content": "$$72$$ " } ], [ { "aoVal": "C", "content": "$$111$$ " } ], [ { "aoVal": "D", "content": "$$152$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "$$35+37+39+41=152$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$5555555$$ " } ], [ { "aoVal": "B", "content": "$$6666666$$ " } ], [ { "aoVal": "C", "content": "$$7777777$$ " } ], [ { "aoVal": "D", "content": "$$8888888$$ " } ], [ { "aoVal": "E", "content": "$$9999999$$ " } ] ]

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

[ "$$(2+3+4+5+6+7+8)\\times1111111\\div5$$ $$=7777777 " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$33$$ " } ], [ { "aoVal": "E", "content": "$$34$$ " } ] ]

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

[ "There are $$33$$ from $$1$$ to $$500$$. Remove number \"$$15$$\" and \"$$30$$\", thus $$33-2=31$$ " ]

[]

[ [ { "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" ]

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

[]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$33$$ " } ] ]

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

[ "We note that $$31\\times 31 = 961$$ and $$32\\times 32 = 1024$$, hence there are $31$ perfect squares greater than $$0$$ and less than $$1000$$. " ]

[ "其它" ]

[ [ { "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->Prime Factorization" ]

[ "Since $2020=2^{2} \\cdot 5 \\cdot 101$, we can simply list its factors: $$ 1,2,4,5,10,20,101,202,404,505,1010,2020 . $$ There are $12$ of these; only $1,2,4,5,101$ (i.e. $5$ of them) don\\textquotesingle t have over $3$ factors, so the remaining $12-5=(\\mathbf{B}) 7$ factors have more than $3$ factors. " ]

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[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

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

[ "The prime is $${}\\textless33$$, so it could be $$2$$, $$3$$, $$5$$, $$7$$, $$11$$, $$13$$, $$17$$, $$19$$, $$23$$, $$29$$, or $$31$$. " ]

[]

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

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

[ "$$36={{2}^{2}}\\times {{3}^{2}}$$, so the number of its factors would be $$\\left( 2+1 \\right)\\times \\left( 2+1 \\right)=9$$. But $$\\textasciitilde N+2$$ cannot be $$1$$ or $$2$$, so the number of possible values of $$N$$ is$$\\textasciitilde9-2=7$$. " ]

[]

[ [ { "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->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 $4$, we can ignore the hundreds. The remainder when $87$ is divided by $4$ is $3$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "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 units digit of $$123456789\\times8$$ is $$2$$, since $$9\\times 8=72$$ . So, if the statement in the question is correct then the two digits which are in a different order are $$1$$ and $$2$$, whose sum is $$3$$. As a check, $$123456789\\times8$$ is indeed $$987654312$$ . " ]

[]

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

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

[ "Since $$8=8\\times 1$$. We choose $$\\rm A$$. " ]

[]

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

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

[ "omitted " ]

[]

[ [ { "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->Factors and Multiples" ]

[ "$$16$$：$$1$$, $$2$$, $$4$$, $$8$$, $$16$$ $$17$$：$$1$$, $$17$$ $$18$$：$$1$$, $$2$$, $$3$$, $$6$$, $$9$$, $$18$$ $$19$$：$$1$$, $$19$$ $$20$$：$$1$$, $$2$$, $$4$$, $$5$$, $$10$$, $$20$$ " ]

[]

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

[ "Among $64, 154$\\ldots~the smallest possible $$N$$ that satisfies the two conditions is $$64$$, and $$64 \\div 11\\rm R9$$. " ]

[]

[ [ { "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" ]

[ "The product includes the factor $$5\\times2=10$$, so the digit in the ones place is $$0$$. " ]

[]

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

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

[ "The only positive whole number whose square is equal to its square root is $$1$$, so the answer is $$\\text{B}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$4150\\sim4160$ " } ], [ { "aoVal": "B", "content": "$3950\\sim3960$ " } ], [ { "aoVal": "C", "content": "$4500\\sim4600$ " } ], [ { "aoVal": "D", "content": "$7920\\sim7960$ " } ], [ { "aoVal": "E", "content": "$7970\\sim7980$ " } ] ]

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

[ "$[4, 8, 9, 10, 11]=3960$, so the possible number of coins could be $3962$ or $7922$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$33$$ " } ] ]

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

[ "Odd numbers from $$1$$ to $$19$$: $$1, 3, 5, 7, 9, 11, 13, 15, 17, 19$$ Even numbers from $$2$$ to $$14$$: $$2, 4, 6, 8, 10, 12, 14$$ $$10+7=17$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$99$$ " } ], [ { "aoVal": "B", "content": "$$25900$$ " } ], [ { "aoVal": "C", "content": "$$35904$$ " } ], [ { "aoVal": "D", "content": "$$34589$$ " } ], [ { "aoVal": "E", "content": "$$39804$$ " } ] ]

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

[ "$99\\div3=33$ $32+33+34=99$ $32\\times33\\times34=35904$ " ]

[]

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

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

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

[ "Only two prime numbers: $$127$$, $$131$$. Therefore, we choose $$\\rm C$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$29$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$77$$ " } ], [ { "aoVal": "D", "content": "$$78$$ " } ] ]

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

[ "Connie counts from $$1$$ to $$20$$. The sum of the prime numbers she counts is $$2+3+5+7+11+13+17+19=77$$. (Note: $$1$$ is not prime.) " ]