AMC 10 · 2024 · #2

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

factorialmulti-digit-arithmeticexponents identify-subproblemsconvert-to-algebra ↑ Prerequisites: factorialmulti-digit-arithmeticorder-of-operations

📏 Short solution 💡 2 insights

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Problem

Compute the value of 10! - 7! · 6!, where n! means the product of all positive integers from 1 up to n.

Pick an answer.

(A)

-120

(B)

(C)

120

(D)

600

(E)

720

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Computing 10! and 7! · 6! from scratch is many digits of arithmetic — exactly when Tool #7 (Identify Subproblems) helps: notice 10! already contains 7! inside it, so factor 7! out and let the small parenthesis decide the whole answer. Tool #5 (Look for a Pattern) is the supporting insight — spotting that 10 · 9 · 8 = 720 = 6! collapses the parenthesis to zero, which is the whole punchline.

1STEP 1

Because 10! = 10 · 9 · 8 · 7!, both terms share the factor 7!, so factor it out.

10! - 7! · 6! = (10 · 9 · 8) · 7! - 7! · 6! = 7! · (10 · 9 · 8 - 6!)

2STEP 2

Multiply the product in pairs: 10 · 9 = 90, then 90 · 8 = 720.

10 · 9 · 8 = 90 · 8 = 720

3STEP 3

Build 6! the same way: 6 · 5 = 30, 30 · 4 = 120, 120 · 3 = 360, 360 · 2 = 720.

6! = 6 · 5 · 4 · 3 · 2 · 1 = 720

4STEP 4

Both products are 720, so the parenthesis is 0, and 7! · 0 = 0.

7! · (720 - 720) = 7! · 0 = 0 → (B)

Answer

Double-check by brute force: 10! = 3,628,800. And 7! · 6! = 5040 · 720 = 3,628,800. Subtracting gives exactly 0. So 10! really does equal 7! · 6! — the identity is real, not a coincidence — and the answer matches (B).

💡Key takeaway

This AMC 10 problem only needs Grade 5 multi-digit multiplication plus the Grade 3 distributive property to spot that 10 · 9 · 8 = 6!, making the whole thing zero!

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