math-eval/TAL-SCQ5K huggingface dataset:
Problem: How many positive factors of $$36$$ are also multiples of $$4$$? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: B
Analysis: $$36={{2}^{2}}\times {{3}^{2}}$$, so the number of its factors would be $$\left( 2+1 \right)\times \left( 2+1 \right)=9$$. Among them, there are $$\left( 2+1 \right)\times 1=3$$ factors which have $${{2}^{2}}$$ as its factors. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
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: $$666$$ 
aoVal: E
content: $$888$$ 
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: If $$1\times2\times3\times4\times5\times6\times7\times8\times$$$$9\times10\times11\times12\times13\times$$$$14\times15 = 1307 674 368000$$, how many times does the digit "$$0$$" appear in the product $$10\times20\times30\times40\times50\times60\times$$$$70\times80\times90\times100\times110\times120\times130\times140\times150$$? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$19$$ 
aoVal: D
content: $$20$$ 
Answer: B
Analysis: 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 $$15+4=19$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization
Problem: Malcolm wants to visit Isabella after school today and he knows the street where she lives but does not know her house number. She tells him,~"My house number has two digits, and all of the following four statements about it are true."  $(1)$ It is a prime number.  $(2)$ It is less than $40$.  $(3)$ One of its digits is $3$.  $(4)$ Another digit is an even number.  This information allows Malcolm to determine Isabella\textquotesingle s house number. What is the house number? (adapted from 2017 AMC 8 problem, Question \#8) 
Options:
aoVal: A
content: $$43$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$32$$ 
aoVal: E
content: $$3$$ 
Answer: C
Analysis: 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 $23$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: Malcolm wants to visit Isabella after school today and he knows the street where she lives but doesn't know her house number. She tells him, ``My house number has two digits, and all of the following four statements about it are true.''  $(1)$ It is a prime number.  $(2)$ It is less than $40$.  $(3)$ One of its digits is $3$.  $(4)$ Another digit is an even number.  This information allows Malcolm to determine Isabella's house number. What is the house number? (adapted from 2017 AMC 8 problem, Question \#8) 
Options:
aoVal: A
content: $$43$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$32$$ 
aoVal: E
content: $$3$$ 
Answer: C
Analysis: 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 $23$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: Without calculating, can you quickly choose the correct answer?  $$1236 + 3217$$ = ? 
Options:
aoVal: A
content: $$4451$$ 
aoVal: B
content: $$4453$$ 
aoVal: C
content: $$4450$$ 
aoVal: D
content: $$4452$$ 
Answer: B
Analysis: Odd + Even = Odd~~ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: There is a book with 650 pages. Henry tears 31 paper from the book, each paper contains two pages. Is it possible that the sum of their page number equals to 953? 
Options:
aoVal: A
content: $$Yes.$$ 
aoVal: B
content: $$No.$$ 
Answer: A
Analysis: If there are odd number of odd page number, the sum of page numbers is odd. Therefore, it could be 953. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Source: 其它
Problem: What is the smallest prime number greater than $$59$$? 
Options:
aoVal: A
content: $$69$$ 
aoVal: B
content: $$67$$ 
aoVal: C
content: $$63$$ 
aoVal: D
content: $$61$$ 
Answer: D
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers
Problem: The sum of the thousands\textquotesingle{} digit and the tens\textquotesingle{} digit of $$12345$$ is． 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
Answer: B
Analysis: The thousands\textquotesingle{} digit is $$2$$ and the tens\textquotesingle{} digit is $$4$$. The sum is $$6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers
Problem: What is the sum of the digits in the number one million? 
Options:
aoVal: A
content: 　one 
aoVal: B
content: one hundred 
aoVal: C
content: one thousand 
aoVal: D
content: one million 
Answer: A
Analysis: One million $$= 1000000$$. Adding, $$1+0+0+0+0+0+0 = 1$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: Vivian bakes a cuboid cake with a dimension of $6\times8\times10$ and cuts it into many small cubical cakes with a dimension of $2\times2\times2$. She wants to equally divide the cake without remainder among $x$ people so that each person can get at least one small cake. How many possible values of $x$ are there? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$16$$ 
aoVal: E
content: $$18$$ 
Answer: C
Analysis: $6\times8\times10 \div(2\times2\times2)=60.$ $60=2^{2}\times3\times5$, which means it has $3\times2\times2=12$ factors. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: Four numbers are chosen from the set $3$, $6$, $12$, $27$, $48$ so that the product is a perfect square. What is the number that was not chosen? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$27$$ 
aoVal: E
content: $$48$$ 
Answer: B
Analysis: 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 $6$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers
Source: 其它
Problem: Which of the following statements in incorrect? 
Options:
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 $$6$$ 
Answer: B
Analysis: $$\rm A$$: $$1+0+8=9$$, 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 $$\rm B$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples
Problem: The positive integers from $$1$$ to $$150$$ inclusive are placed in a $$10$$ by $$15$$ grid so that each cell contains exactly one integer. Then the multiples of $$3$$ are given a red mark, the multiples of $$5$$ are given a blue mark, and the multiples of $$7$$ are given a green mark.  How many cells have more than $$1$$ mark? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$19$$ 
Answer: E
Analysis: 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 $$21−2=19$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples
Problem: Is $$12722385$$ divisible by $$13$$? 
Options:
aoVal: A
content: Yes 
aoVal: B
content: No 
Answer: A
Analysis: $$722-12-385=325$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders
Source: 其它
Problem: $$123+234+345$$ divided by $$4$$ has a remainder of~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: C
Analysis: 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 $$2$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of a two-digit score. By which of the following might her incorrect sum have differed from the correct one? 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$46$$ 
aoVal: C
content: $$47$$ 
aoVal: D
content: $$48$$ 
aoVal: E
content: $$49$$ 
Answer: A
Analysis: 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 \#$$13$$) 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: Which of the following prime factorization below is correct? 
Options:
aoVal: A
content: $$97=1\times97$$ 
aoVal: B
content: $$85=5\times 17$$ 
aoVal: C
content: $$64=8\times 8$$ 
aoVal: D
content: $$52=1\times 2\times 2\times 13$$ 
Answer: B
Analysis: 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. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Problem: A whole number divisible by $$6$$ and by $$14$$ need \emph{not} be divisible by． 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$3$$ 
Answer: B
Analysis: As an example, $$42$$ is divisible by $$6$$ and by $$14$$ but not by $$12$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders
Problem: Allen has some apples, and the number of apples is more than $2200$ but less than $2300$. If he distributes them among $12$ children evenly, there will be $11$ apples left. If he distributes them among $13$ children evenly, there will be $7$ apples left.~ If he distributes them among $14$ children evenly, there will be $3$ apples left. Which is the correct range of the number of pens? 
Options:
aoVal: A
content: $2530\sim2540$ 
aoVal: B
content: $2540\sim2550$ 
aoVal: C
content: $2550\sim2560$ 
aoVal: D
content: $2570\sim2590$ 
aoVal: E
content: $2580\sim2600$ 
Answer: A
Analysis: 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 $59+12\times 13\times 14=2243$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem
Source: 其它
Problem: Hasan writes down a two-digit number. He then writes the same two-digit number next to his original number to form a four-digit number. What is the ratio of his four-digit number to his two-digit number? 
Options:
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 
Answer: C
Analysis: 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 $$101:1$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: $$123+234+345$$ divided by $$4$$ has a remainder of~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: C
Analysis: $$123\div 4$$ R $$3$$；$$234\div 4$$ R $$2$$；$$345\div 4$$ R $$1$$; Therefore, $$\left( 3+2+1 \right)\div 4$$ R $$2$$． 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder
Problem: Which of the following statement is correct? 
Options:
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 
Answer: D
Analysis: 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 $2$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: What is the square root of the number whose square is $$16$$? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$\sqrt{{8}}$$ 
aoVal: D
content: $$16$$ 
Answer: A
Analysis: The number whose square is $$16$$ is $$4$$. The square root of $$4$$ is $$2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
Problem: Teacher divides $$365$$ coins among $$3$$ kids. One of them gets an odd number of coins, and another gets also an odd number of coins. Which of the following would be the possible number of coins that the last kid gets? 
Options:
aoVal: A
content: $$232$$ 
aoVal: B
content: $$168$$ 
aoVal: C
content: $$247$$ 
aoVal: D
content: $$94$$ 
aoVal: E
content: $$132$$ 
Answer: C
Analysis: Odd number $+$ Odd number $=$ Even number  Odd number $-$ Even number $=$ Odd number 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers
Problem: One day Randy was bored and he wrotre all the whole numbers from $49$ to $97$ on his paper. How many times did he write the digit \textquotesingle\textquotesingle$8$\textquotesingle\textquotesingle{} on his paper? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$13$$ 
aoVal: E
content: None of the above 
Answer: E
Analysis: $$14$$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers->Understanding Numbers and Digits
Source: 其它
Problem: The sum of two whole numbers is $$12$$, and the product is $$32$$. Find the difference between these numbers. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$0$$ 
Answer: B
Analysis: If the sum of $$2$$ whole numbers is $$12$$, and the product is $$32$$, then the numbers are $$8$$ and $$4$$; $$8-4 = 4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization
Problem: If the sum of $$7$$ whole numbers is even, at mostof the numbers could be odd. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$1$$ 
Answer: A
Analysis: 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 $$6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers
Problem: If the four-digit number 3P78 is divisible by 3, how many possible values are there for P? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: 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. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules
Source: 其它
Problem: How many prime numbers are there between $130$ and $150$?  $$\textasciitilde$$  $$\textasciitilde$$ 
Options:
aoVal: A
content: $3$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: $131$, $137$, $139$, $149$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: What is the sum of the two smallest prime factors of $250$? (2007 AMC 8 Problems, Question \#3) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$12$$ 
Answer: C
Analysis: The two smallest prime factors of $250$ are $2$ and $5$. Thus, the sum is $2 + 5 = 7$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers
Source: 其它
Problem: When $$1001$$ is divided by a certain one-digit number, the remainder is $$5$$. What is the remainder when the same one-digit number divides $$2006$$? ． 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: A
Analysis: $$1001\div a=b~~\text {R} 5$$, $a$ 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 $$2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder ->The additive property of the remainder
Problem: How many perfect cubes lie between $2^{8}+1$ and $2^{18}+1$, inclusive? (2018 AMC 8 Problem, Question \#25) 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$57$$ 
aoVal: E
content: $$58$$ 
Answer: E
Analysis: 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 $64-7+1=58$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules
Source: 其它
Problem: When I rounded $$142.857$$ to the nearest hundred, ten, one, tenth and hundredth,was not one of my rounded numbers. 
Options:
aoVal: A
content: $$140$$ 
aoVal: B
content: $142.8$ 
aoVal: C
content: $142.86$ 
aoVal: D
content: $143$ 
aoVal: E
content: $$100$$ 
Answer: B
Analysis: $$142.857$$ rounded to nearest: hundred($100$), ten($140$), one($143$), tenth($142.9$) and hundredth($142.86$). 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value
Problem: Which of the following is not a multiple of $12$? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$36$$ 
aoVal: D
content: $$48$$ 
Answer: B
Analysis: $32$ can not be divided by $12$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples
Source: 其它
Problem: Let $N=34 \cdot 34 \cdot 63 \cdot 270$. What is the ratio of the sum of the odd divisors of $N$ to the sum of the even divisors of $N$ ? 
Options:
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 
Answer: C
Analysis: 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 $$ \frac{a}{15 a-a}=\frac{a}{14 a}=\text { (C) } 1: 14 \text {. } $$ 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization
Source: 其它