AMC 10 · 2023 · #15

Grade 8 arithmetic
perfect-squaresprime-factorizationfactorialparityexponents identify-subproblemspattern-recognitioncasework ↑ Prerequisites: prime-factorizationperfect-squaresfactorial
📏 Long solution 💡 3 insights
Problem
Find the smallest positive integer m that makes m · 2! · 3! · 4! · 5! … 16! a perfect square.

Pick an answer.

(A)
30
(B)
30030
(C)
70
(D)
1430
(E)
1001

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Tool #7 (Identify Subproblems) splits the work: (a) reduce N to a form that exposes a perfect-square chunk, (b) find the leftover non-square chunk, (c) determine which primes in the leftover have odd exponents — those are exactly what m must contain. Tool #5 (Pattern) catches the pairing trick: (2k)! · (2k+1)! = (2k+1) · [(2k)!]², so consecutive pairs of factorials peel off a square. Tool #16 (Change Focus) shifts the question from "is m N a perfect square?" (hard) to "which primes in N have odd exponents?" (a clean parity check) — the complement of the perfect-square structure. Tool #2 (Systematic List) of primes 2, 3, 5, 7, 11, 13 via Legendre's formula on 16! does the bookkeeping.

1STEP 1

Pair adjacent factorials: each pair (2k)! · (2k+1)! = (2k+1) · [(2k)!]², an odd factor times a perfect square. 16! stays alone.

(2k)! · (2k+1)! = (2k+1) · [(2k)!]²
2STEP 2

All 7 pairs give leftover odds 3 · 5 · 7 · 9 · 11 · 13 · 15 times a perfect-square block, so N = that odd product · square · 16!.

N = (3 · 5 · 7 · 9 · 11 · 13 · 15) · square · 16!
3STEP 3

Factor the odds (9 = 3², 15 = 3 · 5): K = 3⁴ · 5² · 7 · 11 · 13, so 7, 11, 13 have odd exponents.

K = 3⁴ · 5² · 7¹ · 11¹ · 13¹
4STEP 4

Legendre on 16! gives odd exponents for 2, 5, 11, 13 (E₂=15, E₅=3, E₁₁=1, E₁₃=1) and even for 3, 7.

E₂ = 8+4+2+1 = 15 (odd); E₃ = 5+1 = 6 (even); E₅ = 3 (odd); E₇ = 2 (even); E₁1 = 1 (odd); E₁3 = 1 (odd)
5STEP 5

Combine K and 16! parities: 11 and 13 pair to even, leaving odd exponents on 2, 5, 7 only.

K · 16! = 2^odd · 3^even · 5^odd · 7^odd · 11^even · 13^even · (square)
6STEP 6

The odd-exponent primes are 2, 5, 7, so the smallest m supplies one of each: m = 2 · 5 · 7 = 70.

m = 2 · 5 · 7 = 70
7STEP 7

70 is choice (C); the others miss primes — 30 drops 7, 1001 = 7 · 11 · 13 forgets 16!, 1430 and 30030 mis-include.

m = 70 → (C)
Answer
70
Cross-check the parity of 7 in N. From K: 7 appears once. From 16!: E₇(16!) = ⌊ 167\frac{16}{7} ⌋ + ⌊ 1649\frac{16}{49} ⌋ = 2 + 0 = 2. Total exponent of 7 in N: 1 + 2 = 3 — odd, so m needs a factor of 7. Consistent. Cross-check 11: K contributes 11¹, 16! contributes 11¹, total 11² — even, m does not need 11. Consistent. Cross-check 2: K contributes 2⁰, 16! contributes 2¹⁵, total odd — m needs 2. Consistent. So m = 2 · 5 · 7 = 70 is square-free and minimal. Magnitude check: 70 is much smaller than 30030 = 70 · 11 · 13 · 3, so 70 being the smallest is reasonable.
💡Key takeaway

This AMC 10 problem only needs Grade 8 "properties of integer exponents" you already know — pairing consecutive factorials peels off perfect squares, and the only primes left with odd exponents in the leftover are 2, 5, 7, so the smallest m is 2 · 5 · 7 = 70.