AMC 10 · 2017 · #6

Grade 5 arithmetic
Archetype Grid Tiling / Packing with Divisibility └ Sub-archetype Rectangular Tiling — Minimize Filler Count
volume-rectangular-prismspatial-visualization bound-inequality-then-enumeratespatial-visualization ↑ Prerequisites: volume-rectangular-prism
📏 Medium solution 💡 2 insights
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Problem
Solid blocks measure 2 in×2 in×1 in. They are packed, without overlapping and without sticking out, into a 3 in×2 in×3 in box. The question asks for the greatest number of blocks that can fit.

Pick an answer.

(A)
3
(B)
4
(C)
5
(D)
6
(E)
7

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

The word "largest" makes this a maximum question, so Tool #14 (Extreme Principle) sets the strategy: find a ceiling no packing can beat, then show that ceiling is actually reachable. Tool #8 (Analyze the Units) supplies the ceiling — total block volume can never exceed the box's volume, and dividing the two volumes caps the count. A volume cap alone is not proof you can reach it, because solid blocks might not tile perfectly, so Tool #17 (Visualize Spatial Relationships) and Tool #1 (Draw a Diagram) build an actual arrangement that hits the cap. Ceiling plus matching construction pins the answer exactly.

1STEP 1

Compute both volumes

Volume is length×width×height: the box holds 3×2×3=18 cubic inches and each block uses 2×2×1=4 cubic inches.

V_box=3×2×3=18, V_block=2×2×1=4
2STEP 2

Cap the count with volume

Non-overlapping blocks total at most the box's volume, so the count is at most 18÷4=4.5; a partial block is impossible, so at most four fit.

18/4=4.5 → at most 4 blocks
3STEP 3

Build a packing that reaches 4

Stand three blocks upright to fill the front 3×2×2 chunk, then lay one flat in the 3×2×1 slab — a real packing of 4 blocks.

3₁-wide blocks+1_flat block=4
4STEP 4

State the maximum

Volume forbids more than four and the packing fits four, so the two meet: the largest number of blocks is 4, choice (B).

max=4 → (B)
Answer
4
Check the bookkeeping: 4 blocks use 4×4=16 cubic inches inside an 18-cubic-inch box, leaving 2 cubic inches empty — a small 1×2×1 gap in one corner, which is fine since no block can squeeze into a 1-inch-thick leftover. A fifth block would need 20 cubic inches of contents in an 18-cubic-inch box, which is impossible. So (B) 4 is both achievable and the most that fits.
💡Key takeaway

Box volume 18 divided by block volume 4 is 4.5, so at most 4 blocks fit — and stacking them in flat layers really does fit 4, choice (B).

  • Compute both volumes
  • Cap the count with volume
  • Build a packing that reaches 4
  • State the maximum

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