AMC 10 · 2020 · #10

Grade 8 geometry-3d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Rectilinear Frame / Box Border

surface-areaspatial-visualizationperfect-squaressequences-arithmetic identify-subproblemspattern-recognition ↑ Prerequisites: surface-areaspatial-visualization

📏 Medium solution 💡 3 insights

Problem

Stack seven cubes with volumes 1, 8, 27, 64, 125, 216, 343 — biggest on the bottom, smallest on top — each centered (or at least fully resting) on the cube below. Find the total surface area of the tower, including the bottom face of the biggest cube.

Pick an answer.

(A)

644

(B)

658

(C)

664

(D)

720

(E)

749

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The tower's surface naturally splits into three pieces: (a) the four vertical sides of every cube, (b) all exposed top surface looking down from above, (c) the single bottom face. Tool #7 isolates these three subproblems. Tool #17 (Visualize) handles the key insight for (b) — looking straight down from above, the visible top area is just the top face of the biggest cube (7 × 7), because each smaller cube only blocks part of the cube below. Tool #5 (Pattern) lets us compute the side-area sum 4(1² + 2² + … + 7²) as a known square-sum. Tool #3 matches to a choice.

1STEP 1

Cube-root each volume: 1, 8, 27, …, 343 are 1³ through 7³, so the side lengths are 1, 2, 3, 4, 5, 6, 7.

s_k = sqrt[3]{k³} = k for k = 1, 2, …, 7

2STEP 2

Each cube has 4 vertical faces of area s², so the side area totals 4(1² + … + 7²) = 4 · 140 = 560.

L = 4(1² + 2² + 3² + 4² + 5² + 6² + 7²) = 4 · 140 = 560

3STEP 3

Look straight down: the tower's silhouette is just the 7 × 7 top of the biggest cube, so the exposed top area is 49.

T = 1² + (2² - 1²) + (3² - 2²) + … + (7² - 6²) = 7² = 49

4STEP 4

The bottom face is just the biggest cube's base: 7 × 7 = 49.

B = 7² = 49

5STEP 5

Add the three parts: 560 + 49 + 49 = 658.

Total = L + T + B = 560 + 49 + 49 = 658

6STEP 6

658 matches choice (B).

658 → (B)

Answer

658

Quick sanity check on the top-view trick. If we'd wrongly counted each cube's full top face we'd add 1² + 2² + … + 7² = 140, making the total 560 + 140 + 49 = 749 — exactly choice (E), a classic trap. But each cube above hides part of the one below, so the exposed top is just the silhouette 7 × 7 = 49. Adding 560 + 49 + 49 = 658 falls cleanly between 644 (bottom omitted) and 664 (bottom double-counted), confirming (B).

💡Key takeaway

This AMC 10 problem only needs Grade 8 cube-root and Grade 6 surface-area ideas you already know — side lengths 1 through 7, side strips total 4 · 140 = 560, and the top view is a 7 × 7 silhouette (49), so total = 560 + 49 + 49 = 658.

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