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2020 AMC 10A Problem 15

2020 AMC 10A Problem 15AMC 10 Step by Step

Name: Date: October 28, 2026

Divisibility & Factors · difficulty 3 of 5 · about 4 min · position P11-15

A positive integer divisor of $12!$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?

A) $3$
B) $5$
C) $12$
D) $18$
E) $23$

Answer

Key insight

Write 12! as prime powers. A divisor chooses a power of each prime, and it is a square exactly when every chosen power is even.

Solution

Divisors are easiest to count once the number is broken into primes, so start there.

1. How many 2s are in $12!$ ? Among $1$ to $12$ : six numbers are even ( $2, 4, \ldots, 12$ ), giving one $2$ each. Three of them ( $4, 8, 12$ ) are multiples of $4$ and give an extra $2$ . One ( $8$ ) is a multiple of $8$ and gives one more. Total $6 + 3 + 1 = 10$ .
2. The other primes. Multiples of $3$ : $3, 6, 9, 12$ give $4$ threes, and $9$ gives one extra, so $5$ . Multiples of $5$ : $5, 10$ , so $2$ . Then $7$ and $11$ appear once each. So
$12! = 2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7 \cdot 11.$
3. Count all divisors. A divisor uses $2$ between $0$ and $10$ times ( $11$ choices), $3$ between $0$ and $5$ times ( $6$ choices), $5$ between $0$ and $2$ times ( $3$ choices), and $7$ and $11$ each $0$ or $1$ times ( $2$ choices each):
$11 \times 6 \times 3 \times 2 \times 2 = 792.$
4. Count square divisors. A number is a perfect square when every prime appears an even number of times. The even choices are: for $2$ , the powers $0, 2, 4, 6, 8, 10$ (six); for $3$ , $0, 2, 4$ (three); for $5$ , $0, 2$ (two); for $7$ and $11$ , only $0$ (one each):
$6 \times 3 \times 2 \times 1 \times 1 = 36.$
5. Probability. $\frac{36}{792} = \frac{1}{22}$ . So $m = 1$ , $n = 22$ , and $m + n = 23$ .

The answer is $\boxed{\textbf{(E)}\ 23}$ .

Why this works

Once a number is written as a product of prime powers, each divisor is just a choice of how many times to use each prime. Any rule about divisors that can be checked prime by prime (like "every power is even" for squares) is counted by multiplying the number of allowed choices for each prime.

The trap

Getting the power of 2 in 12! wrong (it is 10), or forgetting that power 0 is even, so 7 and 11 each still allow one choice for a square divisor.

Concise solution

First factor $12!$ . The exponent of $2$ is $\lfloor 12/2 \rfloor + \lfloor 12/4 \rfloor + \lfloor 12/8 \rfloor = 6 + 3 + 1 = 10$ ; of $3$ it is $4 + 1 = 5$ ; of $5$ it is $2$ ; and $7$ , $11$ appear once each:
$12! = 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11.$

A divisor picks an exponent for each prime from $0$ up to the exponent in $12!$ , so the number of divisors is
$11 \cdot 6 \cdot 3 \cdot 2 \cdot 2 = 792.$

A divisor is a perfect square exactly when every exponent it picks is even. The even choices are $0,2,\dots,10$ for $2$ (six), $0,2,4$ for $3$ (three), $0,2$ for $5$ (two), and only $0$ for $7$ and $11$ :
$6 \cdot 3 \cdot 2 \cdot 1 \cdot 1 = 36.$

The probability is $\frac{36}{792} = \frac{1}{22}$ , so $m + n = 1 + 22 = 23$ .

The answer is $\boxed{\textbf{(E)}\ 23}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques

Techniques

Set up the equation/formula and compute; no special trick needed