math-eval/TAL-SCQ5K huggingface dataset:
Problem: Suppose it is now the month of December. What month will it be $$100$$ calendar months from now? 
Options:
aoVal: A
content: January 
aoVal: B
content: February 
aoVal: C
content: March 
aoVal: D
content: April 
aoVal: E
content: May 
Answer: D
Analysis: $$100\div12=8R4$$  $$4$$ months after December will be April. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems
Source: 其它
Problem: There are three ropes. One of them is $10$ inches long, the other two are $28$ and $30$ inches respectively. If we cut those ropes into small pieces with nothing left and each piece has the same length, how long is each piece at most? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$10$$ 
Answer: A
Analysis: The greatest common factor of $10$, $28$ and $30$ is $2$. 
Knowledge points: 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
Source: 其它
Problem: A number divisible by both $$6$$ and $$20$$ must also be divisible by. 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$14$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$120$$ 
Answer: A
Analysis: Divisibility by $$6$$ \& $$20$$ does not promise divisibility by $$7$$, $$13$$, or $$8$$.  $$\rm A$$. $$12 = 4\times3$$; $$\rm B$$. $$14 = 2\times7$$; $$\rm C$$. $$26 = 2\times13$$; $$\rm D$$. $$120 = 8\times15$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: Every positive integer is congruent modulo $9$ to the sum of its decimal digits.  Now, let $S(n)$ equal the sum of the digits of positive integer $n$. For example, $S(1507)=13$. For a particular positive integer $n, S(n)=691$. Which of the following could be the value of $S(n+2)$? (Adapted From 2017 AMC 12A Problems, Question \#18) 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$17$$ 
aoVal: C
content: $$143$$ 
aoVal: D
content: $$116$$ 
aoVal: E
content: $$6$$ 
Answer: A
Analysis: 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 $(A)=9$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Congruence
Source: 其它
Problem: What is the number between $$37$$ and $$49$$ is exactly divisible by both $$3$$ and $$4$$? 
Options:
aoVal: A
content: $$39$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$42$$ 
aoVal: D
content: $$45$$ 
aoVal: E
content: $$48$$ 
Answer: E
Analysis: Common multiple of $$12$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: What is the length of the largest square that can be made from $$50$$ one-centimetre square tile? 
Options:
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 
Answer: C
Analysis: $$7\times7=49\rm{cm}^{2}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers
Source: 其它
Problem: There are three whole numbers $A$, $B$ and $C$ ($B\neq1$). $A\times B=35$, $B\times C=84$. $A+B+C=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$23$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$29$$ 
Answer: C
Analysis: $35=5\times 7$  $84=2\times 2\times 3\times 7$  Because $B$ is the factor both number contains, $B=7$  Thus, $A=5$, $C=12$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: Emily writes down the largest two-digit prime such that each of its digits is prime.  Krish writes down the smallest two-digit prime such that each of its digits is prime.  Kirsten subtracts Krish\textquotesingle s number from Emily\textquotesingle s number.  What answer does Kirsten obtain? ． 
Options:
aoVal: A
content: $$14 $$ 
aoVal: B
content: $$20 $$ 
aoVal: C
content: $$36 $$ 
aoVal: D
content: $$45 $$ 
aoVal: E
content: $$50$$ 
Answer: E
Analysis: 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 $$73 - 23 = 50$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers
Problem: The $5$-digit number "$2018U$" is divisible by $9$, where $$U$$ is the ones digit. What is the remainder when this number is divided by $8$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: B
Analysis: 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 $(\rm B)$ $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: A kind of water plant grows so fast that it doubles every day. If one plant is put into the pond on the first day, it will turn to two plants on the second day, and on the $26$\textsuperscript{th} days, they can fill the pond. If $8$ water plants are put into the pond on the first day, how many days it will take to fill the pond? 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$25$$ 
Answer: C
Analysis: It takes three days for one plant to turn to $8$ plants. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: What is the total value of all the odd numbers between $$34$$ and $$42$$? 
Options:
aoVal: A
content: $$35$$ 
aoVal: B
content: $$72$$ 
aoVal: C
content: $$111$$ 
aoVal: D
content: $$152$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: $$35+37+39+41=152$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers
Problem: $$(2345678+3456782+4567823+5678234+6782345+7823456+8234567)\div5$$=~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$5555555$$ 
aoVal: B
content: $$6666666$$ 
aoVal: C
content: $$7777777$$ 
aoVal: D
content: $$8888888$$ 
aoVal: E
content: $$9999999$$ 
Answer: C
Analysis: $$(2+3+4+5+6+7+8)\times1111111\div5$$  $$=7777777 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: How many numbers can be divided by both $$3$$ and $$5$$ from $$50$$ to $$500$$? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$33$$ 
aoVal: E
content: $$34$$ 
Answer: B
Analysis: There are $$33$$ from $$1$$ to $$500$$.  Remove number "$$15$$" and "$$30$$", thus $$33-2=31$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Source: 其它
Problem: A whole number is a perfect square if it can be expressed as the product of two equal whole numbers. How many perfect squares are greater than $$0$$ and less than $$1000$$? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$31$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$33$$ 
Answer: B
Analysis: We note that $$31\times 31 = 961$$ and $$32\times 32 = 1024$$, hence there are $31$ perfect squares greater than $$0$$ and less than $$1000$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: How many positive integer factors of $2020$ have more than $3$ factors? (2020 AMC 8 Problems, Question \#17) 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$10$$ 
Answer: B
Analysis: Since $2020=2^{2} \cdot 5 \cdot 101$, we can simply list its factors:  $$ 1,2,4,5,10,20,101,202,404,505,1010,2020 . $$  There are $12$ of these; only $1,2,4,5,101$ (i.e. $5$ of them) don\textquotesingle t have over $3$ factors, so the remaining $12-5=(\mathbf{B}) 7$ factors have more than $3$ factors. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它
Problem: How many whole numbers between $$1$$ and $$100$$ are $$3$$ times a prime? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$12$$ 
Answer: C
Analysis: The prime is $${}\textless33$$, so it could be $$2$$, $$3$$, $$5$$, $$7$$, $$11$$, $$13$$, $$17$$, $$19$$, $$23$$, $$29$$, or $$31$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: For how many positive integer values of $$N$$ is the expression $$\frac{36}{N+2}$$ an integer? ($$1994$$ AHSME Problem, Question \#$$10$$) 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$12$$ 
Answer: A
Analysis: $$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$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number
Problem: The $$5$$-digit number $$2018U$$ is divisible by $$9$$. What is the remainder when this number is divided by $$4$$? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: D
Analysis: We use the property that the digits of a number must sum to a multiple of $9$ if it is divisible by $9$. This means $2+0+1+8+U$ must be divisible by $9$. The only possible value for U then must be $7$. Since we are looking for the remainder when divided by $4$, we can ignore the hundreds. The remainder when $87$ is divided by $4$ is $3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: The result of the calculation $$123456789\times8$$ is almost the same as $$987654321$$ except that two of the digits are in a different order. What is the sum of these two digits? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$15$$ 
Answer: A
Analysis: The units digit of $$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$$ . 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders
Problem: Which of the followings is a multiple of $$8$$? 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
Answer: A
Analysis: Since $$8=8\times 1$$.  We choose $$\rm A$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: The product of $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$ ends withzeros． 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: C
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->The Number of Zeros at the end of a Product
Problem: How many numbers in the number set $16,17,18,19$ and $20$ have more than $3$ factors? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: C
Analysis: $$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$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Problem: $$N$$ is a two$$-$$digit number. When $$N$$ is divided by $$9$$, the remainder is $$1$$. When $$N$$ is divided by $$10$$, the remainder is $$4$$. What is the remainder when $$N$$ is divided by $$11$$? (Adapted from $$2016$$ AMC $$8$$ Problem, Question \#$$5$$) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$9$$ 
Answer: E
Analysis: Among $64, 154$\ldots~the smallest possible $$N$$ that satisfies the two conditions is $$64$$, and $$64 \div 11\rm R9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: Find the digit in the ones place for $$2\times2\times5\times5\times5$$. ． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$5$$ 
Answer: A
Analysis: The product includes the factor $$5\times2=10$$, so the digit in the ones place is $$0$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Problem: I am thinking of a whole number greater than $$0$$ whose square equals its square root. How many such numbers are there? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$4$$ 
Answer: B
Analysis: The only positive whole number whose square is equal to its square root is $$1$$, so the answer is $$\text{B}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Some pirates found lots of gold coins. They divided all the coins equally into $5$ groups. Then, the leader of each group would divide the coins he got with his teammates. The $5$ groups had $4,$ $8,$ $9,$ $10,$ and $11$ pirates, respectively. In addition, all the leaders found that after they divided the coins evenly, there were always $2$ coins left. Which of the following could be the possible range where the number of coins of each group was in? 
Options:
aoVal: A
content: $4150\sim4160$ 
aoVal: B
content: $3950\sim3960$ 
aoVal: C
content: $4500\sim4600$ 
aoVal: D
content: $7920\sim7960$ 
aoVal: E
content: $7970\sim7980$ 
Answer: D
Analysis: $[4, 8, 9, 10, 11]=3960$, so the possible number of coins could be $3962$ or $7922$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Problem: On one side of Long Street the houses are numbered with the consecutive odd numbers from $$1$$ to $$19$$. On the other side of that street, the houses are numbered with the consecutive even numbers from $$2$$ to $$14$$. How many houses are there on Long Street? (2006 Math Kangaroo Problem, Level 3-4, Question \#9) 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$33$$ 
Answer: C
Analysis: Odd numbers from $$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$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: The sum of three $2-$digit consecutive numbers is the largest $2-$digit number. What is their product? 
Options:
aoVal: A
content: $$99$$ 
aoVal: B
content: $$25900$$ 
aoVal: C
content: $$35904$$ 
aoVal: D
content: $$34589$$ 
aoVal: E
content: $$39804$$ 
Answer: C
Analysis: $99\div3=33$  $32+33+34=99$  $32\times33\times34=35904$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Problem: How many $$0$$\textquotesingle s does the product of $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$ end with? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: C
Analysis: There are $$6$$ factors of $$5$$ when you prime factorise~ $$1\times 2\times 3\times 4\times 5\times 6\times \cdots \times 25$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Problem: Among the eleven numbers from $$121$$ to $$131$$, how many prime numbers are there? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$1$$ 
Answer: C
Analysis: Only two prime numbers: $$127$$, $$131$$. Therefore, we choose $$\rm C$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: Connie counts from $$1$$ to $$20$$. What is the sum of the prime numbers she counts? 
Options:
aoVal: A
content: $$29$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$77$$ 
aoVal: D
content: $$78$$ 
Answer: C
Analysis: Connie counts from $$1$$ to $$20$$. The sum of the prime numbers she counts is $$2+3+5+7+11+13+17+19=77$$. (Note: $$1$$ is not prime.) 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: Divide $$4$$、$$9$$、$$10$$、$$14$$、$$15$$ and $$21$$ into 2 groups with 3 numbers in each group to make the product of numbers in each group the same. How can we divide the numbers? 
Options:
aoVal: A
content: $$(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)$$ 
Answer: A
Analysis: $$A$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->The Number of Zeros at the end of a Product
Problem: Which of the following numbers leaves a remainder of $$1$$ when divided by $$2$$ and $$3$$? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$7$$ 
Answer: D
Analysis: When we divide by $$2$$, $$3$$, the remainder is $$1$$. Hence, if we first subtract $$1$$, the result will be divisible by $$2$$, $$3$$.  $2\times3+1=7$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Problem: Teacher Angel has $$87$$ apples. She want to pack the apples into container so she can keep them nicely in the fridge. Each fo the container can only hold $$8$$ apples. What is the least number of containers needed so that teacher Angel can pack all the apples in the fridge? 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
Answer: D
Analysis: $$87\div 8=10$$ $\text{R}$ $$7$$  $$10+1=11$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders
Problem: How many 0s are there at the end of the product $$2\times3\times5\times2\times5\times3\times5\times5$$. ． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$5$$ 
Answer: C
Analysis: 2 set of 2$\times$5. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Problem: A square of a positive number is $$500\textbackslash\%$$ greater than that number. What number is it? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$10$$ 
Answer: B
Analysis: Let the number be $x$. Since $x^{2} = 6x$, therefore $x=6$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: What are the last three digits of the answer to the calculation below?  $$123\times 124\times 125\times 126\times 127$$ 
Options:
aoVal: A
content: $$000$$ 
aoVal: B
content: $$222$$ 
aoVal: C
content: $$444$$ 
aoVal: D
content: $$666$$ 
Answer: A
Analysis: The product $$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$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders
Problem: What are the last two digits of the result of $$1\times 3\times 5\times 7\times \cdots \times 101$$? 
Options:
aoVal: A
content: $$05$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$55$$ 
aoVal: E
content: $$75$$ 
Answer: E
Analysis: The result must be a multiple of $25.$  $$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)$$, it can be divisible by $25$ and have a remainder of $3$ when divided by $4.$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Problem: The multiplication $$abc\times de=7632$$ uses each of the digits $$1$$ to $$9$$ exactly once. What is the value of $$b$$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$9$$ 
Answer: C
Analysis: Note first that $$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$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization
Problem: The greatest common factor is smallest for which of the following pairs of numbers? 
Options:
aoVal: A
content: $$4$$ \& $$18$$ 
aoVal: B
content: $$5$$ \&~$$25$$ 
aoVal: C
content: $$6$$ \&~$$33$$ 
aoVal: D
content: $$8$$ \&~$$35$$ 
Answer: D
Analysis: The greatest common factors of the $$4$$ pairs of numbers are $$2$$, $$5$$, $$3$$, and $$1$$, respectively. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples