AMC 10 · 2024 · #2
Grade 5 arithmeticPick an answer.
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.
Factor out the shared factorial
Pull 7 factorial out of both.
Splitting 10! into (10 · 9 · 8) · 7! and pulling 7! out front uses the distributive property — the same Grade 3 "properties of multiplication" idea that lets us turn a · c - b · c into (a - b) · c.
Pulling the shared factorial out front leaves a small bracket behind.
▸ Why?
A factor shared by every term can be lifted out of the sum, leaving something small inside.
▸ Why?
A bigger factorial is the smaller one with a few extra factors stacked on, so the shared piece is easy to name.
Compute the first piece
Ten times nine times eight is 720.
Multiplying 90 by a one-digit 8 is a Grade 4 multi-digit-by-one-digit calculation, easy to do without a pencil.
4.NBT.B.5Identify SubproblemsCompute the second piece
Six factorial is also 720.
The familiar pattern that 6! = 720 matches the 10 · 9 · 8 result — this matching is the heart of the problem and is exactly Grade 5 multi-digit multiplication fluency.
5.NBT.B.5Look For A PatternThe bracket vanishes
The bracket is zero, so the answer is 0.
Anything times zero is zero — a Grade 3 multiplication property. No need to compute 7! = 5040.
3.OA.B.5Identify SubproblemsThis AMC 12 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!