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AMC 10 · 2020 · #10

Grade 8 geometry-3d
Archetype Compound Area by Decomposition / Difference
surface-areaspatial-visualizationperfect-squaressequences-arithmetic identify-subproblemspattern-recognition ↑ Prerequisites: surface-areaspatial-visualization
📏 Medium solution 💡 3 insights

Problem

Seven cubes, whose volumes are 11, 88, 2727, 6464, 125125, 216216, and 343343 cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

Pick an answer.

(A)
644
(B)
658
(C)
664
(D)
720
(E)
749
View mode:

Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Seven cubes with volumes $1, 8, 27, 64, 125, 216, 343$ cubic units; Side lengths (cube roots): $1, 2, 3, 4, 5, 6, 7$ — these are the cubes of $1$ through $7$; Stacked vertically: bottom cube has side $7$, top cube has side $1$; Each upper cube's bottom face lies entirely on the cube below it; Surface area includes the bottom face (touching the ground); Choices: (A) $644$, (B) $658$, (C) $664$, (D) $720$, (E) $749$

Unknowns: Total exposed surface area in square units

Understand

Restated: 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.

Givens: Seven cubes with volumes $1, 8, 27, 64, 125, 216, 343$ cubic units; Side lengths (cube roots): $1, 2, 3, 4, 5, 6, 7$ — these are the cubes of $1$ through $7$; Stacked vertically: bottom cube has side $7$, top cube has side $1$; Each upper cube's bottom face lies entirely on the cube below it; Surface area includes the bottom face (touching the ground); Choices: (A) $644$, (B) $658$, (C) $664$, (D) $720$, (E) $749$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #17 Visualize Spatial Relationships, #5 Look for a Pattern, #3 Eliminate Possibilities

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 ($7 \times 7$), because each smaller cube only blocks part of the cube below. Tool #5 (Pattern) lets us compute the side-area sum $4(1^2 + 2^2 + \cdots + 7^2)$ as a known square-sum. Tool #3 matches to a choice.

Execute — Answer: B

#7 Identify Subproblems 8.EE.A.2 Step 1
  • Find the side length of each cube.
  • Volume $V = s^3$ gives side $s = \sqrt[3]{V}$.
  • The seven volumes $1, 8, 27, 64, 125, 216, 343$ are the perfect cubes $1^3$ through $7^3$, so the side lengths are $1, 2, 3, 4, 5, 6, 7$.
$$s_k = \sqrt[3]{k^3} = k\;\text{for }k = 1, 2, \ldots, 7$$

💡 Cube roots of perfect cubes are whole — read $s_k = k$ off the volume.

#5 Look for a Pattern 6.G.A.4 Step 2
  • Compute the total lateral (vertical-face) area.
  • A cube of side $s$ has $4$ vertical faces of area $s^2$ each, so $4s^2$ per cube.
  • Sum over all seven cubes.
$$L = 4(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2) = 4 \cdot 140 = 560$$

💡 Side area is $4s^2$ per cube; sum-of-squares $1^2 + \cdots + 7^2 = 140$.

#17 Visualize Spatial Relationships 6.G.A.4 Step 3
  • Compute the total exposed top area.
  • Look straight down at the tower from above — the projection is the top of the biggest cube ($7 \times 7 = 49$), because the smaller cubes only cover bits of the cubes beneath them, while every unshaded patch on a lower cube's top face is still visible from above.
  • Telescoping: the exposed top of cube $k$ (under cube $k-1$ above it) is $k^2 - (k-1)^2$, and these pieces add to $7^2 = 49$.
$$T = 1^2 + (2^2 - 1^2) + (3^2 - 2^2) + \cdots + (7^2 - 6^2) = 7^2 = 49$$

💡 Looking down from above the tower fills a $7 \times 7$ silhouette — no holes.

#7 Identify Subproblems 3.MD.C.7 Step 4

The bottom face is just the bottom of the biggest cube.

$$B = 7^2 = 49$$

💡 Only the biggest cube touches the floor.

#7 Identify Subproblems 4.MD.A.3 Step 5

Add the three subtotals for the total surface area.

$$\text{Total} = L + T + B = 560 + 49 + 49 = 658$$

💡 Three disjoint surface regions — just add their areas.

#3 Eliminate Possibilities 4.NBT.A.2 Step 6

$658$ matches choice (B).

$$658\;\Rightarrow\;\textbf{(B)}$$

💡 Read the matching answer choice.

[1] #7 8.EE.A.2 Find the side length of each cube. Volume $V = s^3$ gives side $s = \sqrt[3]{V}$
[2] #5 6.G.A.4 Compute the total lateral (vertical-face) area. A cube of side $s$ has $4$ verti
[3] #17 6.G.A.4 Compute the total exposed top area. Look straight down at the tower from above —
[4] #7 3.MD.C.7 The bottom face is just the bottom of the biggest cube.
[5] #7 4.MD.A.3 Add the three subtotals for the total surface area.
[6] #3 4.NBT.A.2 $658$ matches choice (B).

Review

Reasonableness: Quick sanity check on the top-view trick. If we'd wrongly counted each cube's full top face we'd add $1^2 + 2^2 + \cdots + 7^2 = 140$, 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 $7 \times 7 = 49$. Adding $560 + 49 + 49 = 658$ falls cleanly between $644$ (bottom omitted) and $664$ (bottom double-counted), confirming (B).

Alternative: Tool #10 (Physical Representation): physically stack seven nested boxes (largest to smallest). Wrap the outside with paper to feel the three regions — lateral side strips ($4 \cdot 1 + 4 \cdot 4 + \cdots + 4 \cdot 49$), the flat-top silhouette square, and the floor square. Reading off the wrapping confirms $560 + 49 + 49 = 658$.

CCSS standards used (min grade 8)

  • 8.EE.A.2 Use square root and cube root symbols to represent solutions (Recovering each cube's side length from its volume: $s = \sqrt[3]{V}$.)
  • 6.G.A.4 Represent three-dimensional figures using nets and find surface area (Summing $4s^2$ over the seven cubes for total lateral area and recognizing the $7 \times 7$ top silhouette.)
  • 3.MD.C.7 Relate area to multiplication and addition operations (Computing each face area as $s^2$.)
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems (Adding the lateral, top, and bottom subtotals to a single grand total.)
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Matching the computed value $658$ to choice (B).)

⭐ 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 $4 \cdot 140 = 560$, and the top view is a $7 \times 7$ silhouette ($49$), so total $= 560 + 49 + 49 = 658$.

⭐ 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 $4 \cdot 140 = 560$, and the top view is a $7 \times 7$ silhouette ($49$), so total $= 560 + 49 + 49 = 658$.

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