AMC 8 · 2014 · #6

Grade 4 arithmetic
area-rectanglesperfect-squaresmulti-digit-arithmetic identify-subproblems ↑ Prerequisites: area-rectanglesmulti-digit-arithmetic
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Problem
Six rectangles each have width 2. Their lengths are the first six perfect squares: 1, 4, 9, 16, 25, 36. Find the sum of their areas.

Pick an answer.

(A)
91
(B)
93
(C)
162
(D)
182
(E)
202

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Find a Pattern

Each rectangle's area is 2 × length, so the total area is 2 × 1 + 2 × 4 + … + 2 × 36. The shared factor 2 is a pattern (Tool #4) — every term has it — so we can factor it out and add the lengths just once instead of six times: total = 2 × (1 + 4 + 9 + 16 + 25 + 36). Tool #7 (Identify Subproblems) then splits the work into two clean pieces — first add the six numbers, then multiply by 2.

1STEP 1

Write the total as a sum of six rectangle areas, each equal to width × length = 2 × its length.

Total = 2·1 + 2·4 + 2·9 + 2·16 + 2·25 + 2·36
2STEP 2

Every term shares a factor 2 — pull it out by the distributive property, so the lengths get added just once.

Total = 2 × (1 + 4 + 9 + 16 + 25 + 36)
3STEP 3

Add the six squares by easy pairs: (1 + 9) + (4 + 16) + (25 + 36) = 10 + 20 + 61 = 91.

1 + 4 + 9 + 16 + 25 + 36 = 10 + 20 + 61 = 91
4STEP 4

Multiply the sum by the common width: 2 × 91 = 182 → (D).

Total = 2 × 91 = 182 → (D)
Answer
182
Choice (A) 91 is the sum of the lengths alone, before multiplying by the width 2 — a tempting trap if you forget the last step. Doubling 91 gives 182, which is choice (D). Quick sanity check: the largest single rectangle is 2 × 36 = 72, and the six areas 2, 8, 18, 32, 50, 72 are all well under 200, so a total near 182 is reasonable.
💡Key takeaway

This AMC 8 problem only needs Grade 4 arithmetic — once you factor out the shared width, it's just one addition and one multiplication!