AMC 8 · 2012 · #17

Grade 4 geometry-2d

Archetype Grid Tiling / Packing with Divisibility └ Sub-archetype Divisibility-Selected Items — Minimize Bounding Footprint

area-rectanglesperfect-squaresspatial-visualizationsystematic-enumeration bound-inequality-then-enumeratephysical-representation ↑ Prerequisites: area-rectanglesperfect-squares

📏 Medium solution 💡 3 insights

Problem

A square has an integer side length S. It is cut into exactly 10 smaller squares, each with an integer side length, and at least 8 of those 10 have area 1 (so side length 1). What is the smallest possible value of S?

Pick an answer.

(A)

$hspace{.05in}3$

(B)

$hspace{.05in}4$

(C)

$hspace{.05in}5$

(D)

$hspace{.05in}6$

(E)

$hspace{.05in}7$

AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Instead of jumping to algebra, Tool #9 (Easier Related Problem) tells us to test the answer choices from smallest to largest: just ask "can S = 3 work? can S = 4 work?" The total-area equation S² = (sum of 10 small areas) gives a quick lower bound, so most choices die immediately. For the surviving choice, Tool #2 (Systematic List) enumerates the integer pairs (s₉, s₁0) whose squares fill the leftover area, and Tool #1 (Draw a Diagram) confirms the pieces actually fit together as a tiling, not just an area match.

1STEP 1

The 10 pieces cover the S × S square exactly once, so their areas add up to S².

S² = s₁² + s₂² + … + s₁0²

2STEP 2

8 pieces have area 1 and the other 2 have integer sides (area ≥ 1 each), so the total area S² is at least 10.

S² ≥ 8 · 1 + 1 + 1 = 10

3STEP 3

Try the smallest choice S = 3: then S² = 9, under 10, too small to hold even 10 unit squares. Eliminate (A).

3² = 9 ＜ 10 → S = 3 fails

4STEP 4

Try S = 4: S² = 16. The 8 unit squares use 8, leaving 8 for two squares; s₉² + s₁0² = 8 forces (2, 2) — two 2 × 2 squares.

s₉² + s₁0² = 16 - 8 = 8 → (s₉, s₁0) = (2, 2)

5STEP 5

The two 2 × 2 squares form a 4 × 2 top strip; the eight 1 × 1 squares fill the bottom 4 × 2 strip. So S = 4 truly tiles — (B) 4.

2 × 2_top-left 2 × 2_top-right / 8 × (1 × 1)_bottom 4 × 2 strip → (B) 4

Answer

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The pieces really do account for the area: 8 · 1 + 2 · 4 = 8 + 8 = 16 = 4². The count also checks: 8 + 2 = 10 small squares, with 8 of them area 1 as required. And we ruled out S = 3 by the area lower bound, so 4 is the true minimum — choice (B) is the smallest legal side.

💡Key takeaway

This AMC 8 problem only needs Grade 4 area and arithmetic you already know — try the smallest sides first and check that the pieces really fit!

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