Competition · AMC preparation · step 4 of 4
AMC 8 · 2017 · #5
Grade 4 arithmeticPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The expression splits cleanly into two independent subproblems: the denominator (a sum) and the numerator (a product), so Tool #7 says 'solve each piece, then combine'. Tool #5 helps with the sum step: the familiar pattern 1+2+…+n = (n(n+1))/2 gives the denominator instantly as (8 · 9)/2 = 36. Once the denominator is in hand, we treat the big fraction as a third subproblem — simplify by spotting 36 = 4 × 9 inside the numerator and canceling — instead of multiplying out 8! = 40320 and dividing.
Add 1 through 8
Add the bottom: pair the ends (1+8=9, four nines) or use , so 1+…+8 = 36.
Adding small whole numbers up to 100 is a Grade 2 fluency skill; the pairing pattern is just a faster way to do that same addition.
2.NBT.B.5Look For A PatternRewrite with the new denominator
Substitute the denominator back in, turning the problem into the product 1·2·…·8 over 36.
Substituting the computed sum back into the fraction is a multi-step Grade 4 word-problem move — finish one piece, then use it.
4.OA.A.3Identify SubproblemsFactor the denominator
Factor the denominator: 36 = 4 × 3 × 3 — and a 4, a 3, and a 6 = 2 × 3 already sit in the numerator.
Listing factor pairs of 36 and checking which factors are 'already there' is exactly the Grade 4 factor-pair / divisibility skill.
4.OA.B.4Identify SubproblemsCancel the matching factors
Cancel the matched factors (the 4, and both 3s from the 3 and the 6 = 2 · 3), leaving 1 · 2 · 5 · 2 · 7 · 8.
Dividing the top and the bottom by the same number to get an equivalent fraction is the Grade 4 equivalent-fractions rule applied piece by piece.
Canceling the factors that build up the denominator 36 out of the numerator of (1 · 2 · 3 · 4 · 5 · 6 · 7 · 8)/36 leaves the smaller product 1 · 2 · 5 · 2 · 7 · 8, and that leftover has the same value as the whole fraction.
▸ Why?
The numerator can be rearranged and regrouped into 36 times the leftover product 1 · 2 · 5 · 2 · 7 · 8, because it holds a 4, a 3, and a 6 = 2 · 3, and 4 · 3 · 3 = 36.
▸ Why?
The numbers in a product may be multiplied in any order, so the 4, the 3, and the 6 can be slid next to each other and apart from the rest.
▸ Why?
Once those factors sit together they can be grouped as (4 · 3 · 3) on one side and the leftover on the other, and regrouping does not change a product, so the numerator equals 36 times the leftover.
▸ Why?
With the fraction written as (36 × leftover) ÷ 36, dividing by 36 exactly undoes the multiplying by 36 and hands back just the leftover.
Multiply what is left
Multiply what remains, grouping 2 × 5 = 10 first: 10 · 14 · 8 = 1120, which is choice (B).
Each remaining multiplication (2 · 5, 2 · 7, 14 · 8, × 10) is a within-100 basic fact — Grade 3 multiplication fluency.
3.OA.C.7Identify SubproblemsThis AMC 8 problem only needs Grade 4 factor pairs and equivalent fractions you already know — no calculator, just smart cancellation!
- Add 1 through 8
- Rewrite with the new denominator
- Factor the denominator
- Cancel the matching factors
- Multiply what is left
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