AMC 10 · 2020 · #7

Grade 8 geometry-3d
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, and 343 — biggest on the bottom, smallest on top — each resting fully 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
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

Find each side length

The cube root gives each side.

s_k = ∛(k³) = k for k = 1, 2, …, 7
2STEP 2

Add the side faces

Every side face is fully exposed.

L = 4(1² + 2² + 3² + 4² + 5² + 6² + 7²) = 4 · 140 = 560
3STEP 3

Count the top faces

From above it is one big square.

T = 1² + (2² - 1²) + (3² - 2²) + … + (7² - 6²) = 7² = 49
4STEP 4

Add the bottom face

The bottom face is the same size.

B = 7² = 49
5STEP 5

Add it all up

Add the three parts.

Total = L + T + B = 560 + 49 + 49 = 658
6STEP 6

Match the choice

The surface area is 658.

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 12 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.