AMC 8 · 2017 · #5

Grade 4 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

factorsfraction-arithmeticsequences-arithmeticfactorial identify-subproblems ↑ Prerequisites: multi-digit-arithmeticfraction-arithmetic

📏 Medium solution 💡 3 insights

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Problem

Evaluate the fraction whose numerator is the product 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 and whose denominator is the sum 1+2+3+4+5+6+7+8, then pick the matching answer choice.

Pick an answer.

(A)

1020

(B)

1120

(C)

1220

(D)

2240

(E)

3360

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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 = $\frac{n(n+1)}{2}$ gives the denominator instantly as $\frac{8 · 9}{2}$ = 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.

1STEP 1

Add the bottom: pair the ends (1+8=9, four nines) or use $\frac{n(n+1)}{2}$ , so 1+…+8 = 36.

1+2+3+4+5+6+7+8 =

\frac{8 · 9}{2}

= 36

2STEP 2

Substitute the denominator back in, turning the problem into the product 1·2·…·8 over 36.

\frac{1 · 2 · 3 · 4 · 5 · 6 · 7 · 8}{1+2+3+4+5+6+7+8}

\frac{1 · 2 · 3 · 4 · 5 · 6 · 7 · 8}{36}

3STEP 3

Factor the denominator: 36 = 4 × 3 × 3 — and a 4, a 3, and a 6 = 2 × 3 already sit in the numerator.

36 = 4 × 9 = 4 × 3 × 3

4STEP 4

Cancel the matched factors (the 4, and both 3s from the 3 and the 6 = 2 · 3), leaving 1 · 2 · 5 · 2 · 7 · 8.

\frac{1 · 2 · 3 · 4 · 5 · 6 · 7 · 8}{36}

\frac{1 · 2 · 3 · 4 · 5 · (2 · 3) · 7 · 8}{4 · 3 · 3}

= 1 · 2 · 5 · 2 · 7 · 8

5STEP 5

Multiply what remains, grouping 2 × 5 = 10 first: 10 · 14 · 8 = 1120, which is choice (B).

(2 · 5) · (2 · 7) · 8 = 10 · 14 · 8 = 1120 → (B)

Answer

1120

Sanity-check by computing 8! the long way: 8! = 40320, and 40320 ÷ 36 = 1120, matching choice (B). Magnitude makes sense too —

\frac{40320}{36}

should land near

\frac{40000}{40}

= 1000, so a four-digit answer just above 1000 is exactly the right size, ruling out (D) 2240 and (E) 3360 on size alone.

💡Key takeaway

This AMC 8 problem only needs Grade 4 factor pairs and equivalent fractions you already know — no calculator, just smart cancellation!

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