AMC 10 · 2020 · #15

Grade 7 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

factorialprime-factorizationdivisor-countperfect-squareslegendre-formulaprobability-basic identify-subproblemssystematic-enumeration ↑ Prerequisites: prime-factorizationdivisor-count

📏 Long solution 💡 3 insights

Problem

A divisor of 12! is chosen at random (each divisor equally likely). The probability that the chosen divisor is a perfect square equals

\frac{m}{n}

for relatively prime positive integers m, n. Find m + n.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): break the question into four cleanly separate pieces — (1) prime factorization of 12!, (2) total number of divisors, (3) number of perfect-square divisors, (4) probability and m + n. Each piece is a small standalone calculation. Tool #2 (Systematic List): list the primes ≤ 12 in order (2, 3, 5, 7, 11) and, for each prime, list the legal exponent choices and the smaller "must be even" sub-list. Tool #3 (Eliminate): match m + n to the answer choices.

1STEP 1

Legendre's formula counts each prime's exponent inside 12!, giving 2¹⁰ · 3⁵ · 5² · 7¹ · 11¹.

12! = 2¹⁰ · 3⁵ · 5² · 7¹ · 11¹

2STEP 2

Each divisor picks every exponent independently, so the total divisor count is 11 · 6 · 3 · 2 · 2 = 792.

#divisors = (10+1)(5+1)(2+1)(1+1)(1+1) = 11 · 6 · 3 · 2 · 2 = 792

3STEP 3

A perfect-square divisor needs every exponent even, so the even choices multiply: 6 · 3 · 2 · 1 · 1 = 36.

#square divisors = 6 · 3 · 2 · 1 · 1 = 36

4STEP 4

The probability is $\frac{36}{792}$ , which reduces to $\frac{1}{22}$ , so m = 1 and n = 22.

\frac{36}{792}

\frac{1}{22}

, gcd(1, 22) = 1

5STEP 5

Add them: m + n = 1 + 22 = 23.

m + n = 1 + 22 = 23

6STEP 6

Only 23 matches, which is choice (E); the others come from dropping a factor, mis-reducing, or stopping early.

m + n = 23 → (E)

Answer

Quick sanity:

\frac{1}{22}

is the rough chance, ≈ 4.5%. There's a clean conceptual reason — each odd-exponent prime contributes a factor (⌈ (a_i + 1)/2 ⌉)/(a_i + 1) to the ratio. For our exponents (10, 5, 2, 1, 1), that's

\frac{6}{11}

\frac{3}{6}

\frac{2}{3}

\frac{1}{2}

\frac{1}{2}

\frac{6 · 3 · 2}{11 · 6 · 3 · 2 · 2}

\frac{1}{22}

. ✓ Matches. m + n = 23.

💡Key takeaway

This AMC 10 problem only needs Grade 7 counting you already know! Factor 12! = 2¹⁰ · 3⁵ · 5² · 7 · 11. Total divisors: 11 · 6 · 3 · 2 · 2 = 792. Perfect-square divisors need every exponent even, so count even choices in each range: 6 · 3 · 2 · 1 · 1 = 36. Probability = $\frac{36}{792}$ = $\frac{1}{22}$ , so m + n = 1 + 22 = 23, answer (E).

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