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Divisibility & Factors

Divisors, number/sum of divisors, factorization, divisibility rules

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Divisibility & FactorsAMC 10 Step by Step

Name: Date: October 28, 2026

20 problems, easiest first

What you need to know

Fundamental Theorem of Arithmetic. Every whole number $n > 1$ is a product of primes in exactly one way (apart from order). We write $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ , where the $p_i$ are different primes and the $a_i$ are their exponents. Example: $360 = 2^3\cdot3^2\cdot5$ , and no other primes multiply to $360$ .
Counting divisors. Write $d(n)$ for the number of positive divisors of $n$ . Then $d(n) = (a_1+1)(a_2+1)\cdots(a_k+1)$ , because a divisor uses anywhere from $0$ to $a_i$ copies of each prime $p_i$ : that is $a_i+1$ choices. Example: $72 = 2^3\cdot3^2$ has $4\cdot3 = 12$ divisors.
Adding divisors. Write $\sigma(n)$ (Greek "sigma") for the sum of the positive divisors. Multiply one sum per prime: $\sigma(n) = (1+p_1+\dots+p_1^{a_1})\cdots(1+p_k+\dots+p_k^{a_k})$ . Example: $\sigma(72) = (1+2+4+8)(1+3+9) = 15\cdot13 = 195$ .
Divisors come in pairs $d$ and $\frac{n}{d}$ . So $n$ has an odd number of divisors exactly when $n$ is a perfect square, since only then does one divisor, $\sqrt n$ , pair with itself. Example: $36$ pairs up as $1\cdot36$ , $2\cdot18$ , $3\cdot12$ , $4\cdot9$ , $6\cdot6$ , giving $9$ divisors.
Divisibility rules. Last digit for $2$ , $5$ , $10$ ; last two digits for $4$ , $25$ ; last three digits for $8$ ; digit sum for $3$ , $9$ ; alternating digit sum (add and subtract the digits in turn) for $11$ . Example: $7425$ is divisible by $25$ (ends in $25$ ), by $9$ ( $7+4+2+5 = 18$ ), and by $11$ ( $7-4+2-5 = 0$ ).
Counting multiples. From $1$ to $N$ there are $\lfloor N/k \rfloor$ multiples of $k$ , where $\lfloor\ \rfloor$ (the floor) means round down. Example: $\lfloor 100/7 \rfloor = 14$ multiples of $7$ up to $100$ , the last being $98$ .
Combining. If $a$ divides both $b$ and $c$ , then $a$ divides $bx+cy$ for any integers $x$ and $y$ . Example: $7$ divides $21$ and $35$ , so it divides $3\cdot21+2\cdot35 = 133$ .

How AMC 10 tests it

Early problems (1–8) factor the contest year and ask for a sum or a digit fact about its factors.
Middle problems (10–18) ask "how many positive divisors of $N$ are …" (even, perfect squares, multiples of $6$ ). That is $d(n)$ with limits on the exponents.
"How many integers up to $N$ are divisible by $a$ or $b$ but not $c$ ?" combines floor counting with the Principle of Inclusion-Exclusion (PIE).
Products of consecutive integers: which number must divide $n(n+1)(n+2)$ ? One counterexample rules a choice out.
Late problems go backwards: given $d(n)$ or $\sigma(n)$ , find the smallest such $n$ or count the possibilities.

Standard approaches

Prime factorize right away; almost every divisor question becomes a question about exponents.
To count divisors with a property, list the allowed exponents for each prime and multiply. Example: an even divisor of $2^3\cdot3^2$ has $1$ to $3$ twos and $0$ to $2$ threes: $3\cdot3 = 9$ .
For the smallest $n$ with a given number of divisors, write that number as a product $(a_1+1)(a_2+1)\cdots$ , give the largest exponents to the smallest primes, and compare every such product.
Count multiples with floors, and handle overlaps with the lcm: numbers divisible by both $a$ and $b$ are the multiples of $\text{lcm}(a,b)$ .
For a "must divide" claim, test one small example such as $5\cdot6\cdot7$ .

Worked example

Let $N$ be the smallest positive integer with exactly $20$ positive divisors. What is the sum of the digits of $N$ ?

(A) $4$ (B) $6$ (C) $8$ (D) $9$ (E) $12$

Step 1: the shapes. The divisor count is a product of (exponent $+1$ ) factors. So write $20$ as a product of factors that are each at least $2$ , largest first: $20$ , $10\cdot2$ , $5\cdot4$ , $5\cdot2\cdot2$ .

Step 2: smallest number for each shape. Subtract $1$ from each factor to get the exponents, and give the biggest exponent to $2$ , the next to $3$ , then $5$ :

$20$ gives $2^{19} = 524288$ .
$10\cdot2$ gives $2^9\cdot3 = 1536$ .
$5\cdot4$ gives $2^4\cdot3^3 = 432$ .
$5\cdot2\cdot2$ gives $2^4\cdot3\cdot5 = 240$ .

Step 3: pick the smallest. $N = 240$ , and its digit sum is $2+4+0 = 6$ .

The answer is $\boxed{\textbf{(B)}\ 6}$ .

Pitfalls

Forgetting the $+1$ : $2^4\cdot3^3$ has $5\cdot4 = 20$ divisors, not $4\cdot3 = 12$ .
Counting $1$ and $n$ inconsistently. "Proper divisors" leaves out $n$ itself, so check the wording. Example: $12$ has $6$ divisors but $5$ proper divisors.
Assuming the shape with the most factors always gives the smallest $n$ ; compare them all. Example: for $8$ divisors, $2\cdot2\cdot2$ gives $2\cdot3\cdot5 = 30$ , but $4\cdot2$ gives $2^3\cdot3 = 24$ .
Using $ab$ instead of $\text{lcm}(a,b)$ in PIE when $a$ and $b$ share a factor. Example: up to $100$ , the numbers divisible by both $4$ and $6$ are the $8$ multiples of $12$ , not the $4$ multiples of $24$ .

Traps that recur

Counting only the 15 primes as failures (answer 35) and missing that n = 4 also fails, or forgetting that n = 1 works. (2019 AMC 10A #25)
Counting only the n with d(n) = 6 (eight of them) and missing 36 and 48, whose f_1 values 18 and 20 also feed into the fixed point 12. (2021 AMC Fall 10A #23)
Trying to compute C(52,10) exactly, or using divisibility by 11 (which every choice satisfies) and getting nowhere. (2020 AMC 10B #19)
Answering 121 by forgetting the word 'distinct', or subtracting only the two obvious cases 1 and 10^10 to get 119. (2019 AMC 10B #19)

1.2021 AMC 10B #5
The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give $24$ , while the other two multiply to $30$ . What is the sum of the ages of Jonie's four cousins?
1. A) $21$
2. B) $22$
3. C) $23$
4. D) $24$
5. E) $25$
2.2019 AMC 10A #2
What is the hundreds digit of $(20!-15!)?$
1. A) $0$
2. B) $1$
3. C) $2$
4. D) $4$
5. E) $5$
3.2008 AMC 10A #3
For the positive integer $n$ , let $\langle n\rangle$ denote the sum of all the positive divisors of $n$ with the exception of $n$ itself. For example, $\langle 4\rangle=1+2=3$ and $\langle 12 \rangle =1+2+3+4+6=16$ . What is $\langle\langle\langle 6\rangle\rangle\rangle$ ?
1. A) $6$
2. B) $12$
3. C) $24$
4. D) $32$
5. E) $36$
4.2000 AMC 10 #1
In the year $2001$ , the United States will host the International Mathematical Olympiad. Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$ ?
1. A) $23$
2. B) $55$
3. C) $99$
4. D) $111$
5. E) $671$
5.2024 AMC 10A #7
The product of three integers is $60$ . What is the least possible positive sum of the three integers?
1. A) $2$
2. B) $3$
3. C) $5$
4. D) $6$
5. E) $13$
6.2024 AMC 10B #8
Let $N$ be the product of all the positive integer divisors of $42$ . What is the units digit of $N$ ?
1. A) $0$
2. B) $2$
3. C) $4$
4. D) $6$
5. E) $8$
7.2024 AMC 10B #6
A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?
1. A) $160$
2. B) $180$
3. C) $222$
4. D) $228$
5. E) $390$
8.2023 AMC 10A #12
How many three-digit positive integers $N$ satisfy the following properties?
- The number $N$ is divisible by $7$ .
- The number formed by reversing the digits of $N$ is divisible by $5$ .
1. A) $13$
2. B) $14$
3. C) $15$
4. D) $16$
5. E) $17$
9.2022 AMC 10B #8
Consider the following $100$ sets of $10$ elements each: $\begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*}$ How many of these sets contain exactly two multiples of $7$ ?
1. A) $40$
2. B) $42$
3. C) $43$
4. D) $49$
5. E) $50$
10.2021 AMC Fall 10B #6
The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$ , where $m$ and $k$ are integers and $6$ is not a divisor of $m$ . What is $m+k?$
1. A) $47$
2. B) $58$
3. C) $59$
4. D) $88$
5. E) $90$
11.2020 AMC 10A #9
A single bench section at a school event can hold either $7$ adults or $11$ children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N?$
1. A) $9$
2. B) $18$
3. C) $27$
4. D) $36$
5. E) $77$
12.2017 AMC 10A #16
There are 10 horses, named Horse 1, Horse 2, $\ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S>0$ , in minutes, at which all 10 horses will again simultaneously be at the starting point is $S=2520$ . Let $T>0$ be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of $T$ ?
1. A) $2$
2. B) $3$
3. C) $4$
4. D) $5$
5. E) $6$
13.2014 AMC 10B #12
The largest divisor of $2,014,000,000$ is itself. What is its fifth largest divisor?
1. A) $125,875,000$
2. B) $201,400,000$
3. C) $251,750,000$
4. D) $402,800,000$
5. E) $503,500,000$
14.2013 AMC 10B #9
Three positive integers are each greater than $1$ , have a product of $27000$ , and are pairwise relatively prime. What is their sum?
1. A) $100$
2. B) $137$
3. C) $156$
4. D) $160$
5. E) $165$
15.2012 AMC 10B #10
How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$
1. A) $6$
2. B) $7$
3. C) $8$
4. D) $9$
5. E) $10$
16.2011 AMC 10B #5
In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161$ . What is the correct value of the product of $a$ and $b$ ?
1. A) $116$
2. B) $161$
3. C) $204$
4. D) $214$
5. E) $224$
17.2011 AMC 10A #10
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $\$17.71$ . What was the cost of a pencil in cents?
1. A) $7$
2. B) $11$
3. C) $17$
4. D) $23$
5. E) $77$
18.2010 AMC 10B #8
A ticket to a school play cost $x$ dollars, where $x$ is a whole number. A group of 9th graders buys tickets costing a total of $\$48$ , and a group of 10th graders buys tickets costing a total of $\$64$ . How many values for $x$ are possible?
1. A) $1$
2. B) $2$
3. C) $3$
4. D) $4$
5. E) $5$
19.2009 AMC 10B #6
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?
1. A) $10$
2. B) $12$
3. C) $16$
4. D) $18$
5. E) $24$
20.2008 AMC 10A #9
Suppose that $\frac{2x}{3}-\frac{x}{6}$ is an integer. Which of the following statements must be true about $x$ ?
1. A) $\text{It is negative.}$
2. B) $\text{It is even, but not necessarily a multiple of 3.}$
3. C) $\text{It is a multiple of 3, but not necessarily even.}$
4. D) $\text{It is a multiple of 6, but not necessarily a multiple of 12.}$
5. E) $\text{It is a multiple of 12.}$

Divisibility & Factors — answersAMC 10 Step by Step

Name: Date: October 28, 2026

Keep this page separate from the problems.

Answers

#	Answer	The idea it turns on
1	B	Using single digits, 30 can only be 5 × 6. Since 6 is now taken, 24 has to be 3 × 8, not 4 × 6.
2	A	15! already contains three 5s and plenty of 2s, so it is a multiple of 1000. So is 20!, and so is their difference.
3	A	The divisors of 6 other than 6 are 1, 2 and 3, which add up to 6. So each step turns 6 back into 6.
4	E	Break 2001 into primes: 3, 23 and 29. To make the sum big, make two of the numbers as small as possible, 1 and 3.
5	B	The word is integers, so negatives are allowed. Two negative numbers keep the product positive but make the sum small: (−1)(−6)(10) = 60 with sum 3.
6	D	The divisors of 42 pair up so each pair multiplies to 42. Four pairs make N equal to 42 to the 4th power, which ends in 6.
7	B	The perimeter is smallest when the two sides are as close as possible. Since 2024 is one less than 45 squared, the sides 44 and 46 work.
8	B	Reversing N moves its first digit to the end. A multiple of 5 ends in 0 or 5, and N cannot start with 0, so N is in the 500s.
9	B	Every set of 10 holds one or two multiples of 7. There are 142 multiples in 100 sets, so 142 − 100 = 42 sets hold two.
10	B	2021 = 43 times 47, so use exponents 46 and 42, the bigger one on 2. The number is 16 times 6 to the 42.
11	B	Cut a bench into 77 pieces: an adult uses 11 and a child 7, so each adult-child pair uses 18. So 77 times N must be a multiple of 18.
12	B	Horse k is at the start at multiples of k minutes. So T is the smallest number divisible by at least five of 1 to 10, which is 12.
13	C	Divisors come in pairs that multiply to N, so the fifth largest divisor is N divided by the fifth smallest. The smallest divisors are 1, 2, 4, 5, 8.
14	D	Relatively prime numbers share no prime, so all the 2s go in one number, all the 3s in another, all the 5s in the third: 8, 27, 125.
15	D	Cross-multiply to get M times N = 36. Every divisor of 36 chosen as M gives exactly one N, and 36 has 9 divisors.
16	E	Break 161 into factors: it is 7 times 23. The reversed number has two digits, so it was 23. That makes the real number 32 and the other number 7.
17	B	Factor 1771 cents as 7 × 11 × 23. A majority of 30 is 16 to 30 students, so 23 bought; each bought 7 pencils at 11 cents.
18	E	The ticket price must divide both 48 and 64 evenly. The prices that do are 1, 2, 4, 8 and 16, the numbers that divide 16.
19	D	The twins share an age, so 128 is Kiana's age times the twins' age twice. Try the few possible twin ages and keep the one where Kiana is younger.
20	B	Over a common bottom of 6 the expression becomes x over 2. So x must be even, and x = 2 shows it need not be a multiple of 3.

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.