AMC 8 · 2020 · #12

Grade 4 algebranumber-theory

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

factorialprime-factorization identify-subproblems ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

The factorial n! means the product of all positive integers from n down to 1 (so 4! = 4 × 3 × 2 × 1 = 24). The equation 5! · 9! = 12 · N! asks: which positive integer N makes the two sides equal?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Identify Subproblems) splits the equation 5! · 9! = 12 · N! into three small, friendly pieces: (a) compute 5! since it is the only "small" factorial in sight, (b) divide both sides by 12 to clear the loose coefficient, and (c) read off the surviving expression as a single factorial. Each subproblem is a Grade 3-4 arithmetic move; the big-looking factorial expression never has to be expanded. Tool #3 (Eliminate Possibilities) is a natural backup for any AMC multiple-choice question — once we have 10 · 9! = N!, the five choices 10, 11, 12, 13, 14 can each be tested against the recursive rule n! = n · (n-1)! until only one fits.

1STEP 1

Only 5! is small enough to expand — 9! we leave untouched — and its value is 120.

5! = 5 × 4 × 3 × 2 × 1 = 120

2STEP 2

Substitute 5! = 120 into the equation to get 120 · 9! = 12 · N!.

120 · 9! = 12 · N!

3STEP 3

Divide both sides by 12 (since 120 ÷ 12 = 10) to get 10 · 9! = N!.

(120 · 9!)/12 = (12 · N!)/12 ⟹ 10 · 9! = N!

4STEP 4

By the rule n! = n · (n-1)!, the left side folds into one factorial: 10 · 9! = 10!, so 10! = N!.

10 · 9! = 10 · (9 · 8 … 1) = 10! ⟹ 10! = N!

5STEP 5

Equal factorials of positive integers force equal integers, so N = 10 — choice (A).

10! = N! ⟹ N = 10 ⟹ (A)

Answer

The left side 5! · 9! = 120 · 9! already contains 9! as a factor, and 120 = 12 · 10 contributes one extra factor of 10 after we cancel the 12. So the left side is exactly 10 · 9! = 10!, a single factorial just one step beyond 9!. An answer of N = 10 — the next integer up from 9 — is therefore the most natural outcome; jumping all the way to N = 14 would require an extra factor of 14 · 13 · 12 · 11 = 24024 on the left, which is not there.

💡Key takeaway

This AMC 8 problem only needs Grade 4 multiplication patterns — like 10 × 9! = 10! — that you already know!

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