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AMC 8 · 2012 · #17

Grade 4 geometry-2d
Archetype Grid Tiling / Packing with Divisibility
area-rectanglesperfect-squaresspatial-visualizationsystematic-enumeration bound-inequality-then-enumeratephysical-representation ↑ Prerequisites: area-rectanglesperfect-squares
📏 Medium solution 💡 3 insights

Problem

A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?

Pick an answer.

(A)
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(B)
$hspace{.05in}4$
(C)
$hspace{.05in}5$
(D)
$hspace{.05in}6$
(E)
$hspace{.05in}7$
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Toolkit + CCSS Solution

Understand

Restated: 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 $S$?

Givens: Original side length $S$ is a positive integer; The original square is cut into exactly $10$ smaller squares; Every small square has integer side length; At least $8$ of the $10$ small squares have area $1$ (side length $1$); Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$

Unknowns: The smallest integer side length $S$ of the original square

Understand

Restated: 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 $S$?

Givens: Original side length $S$ is a positive integer; The original square is cut into exactly $10$ smaller squares; Every small square has integer side length; At least $8$ of the $10$ small squares have area $1$ (side length $1$); Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #2 Make a Systematic List, #1 Draw a Diagram

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 $S^2 = $ (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 $(s_9, s_{10})$ 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.

Execute — Answer: B

#9 Solve an Easier Related Problem 3.MD.C.7 Step 1
  • Use the area equation.
  • The $10$ small squares together cover the original $S \times S$ square exactly once, so their areas add up to $S^2$.
$$S^2 = s_1^2 + s_2^2 + \dots + s_{10}^2$$

💡 Area of a square is side $\times$ side, and areas of non-overlapping pieces add — both are Grade 3 area ideas.

#9 Solve an Easier Related Problem 4.OA.A.3 Step 2
  • Find a lower bound for $S^2$.
  • At least $8$ pieces have area $1$, and the other $2$ pieces have integer sides so their areas are at least $1$ each.
  • So the total area is at least $8 + 1 + 1 = 10$.
$$S^2 \ge 8 \cdot 1 + 1 + 1 = 10$$

💡 Adding up the smallest possible areas gives the smallest possible total — a Grade 4 multi-step reasoning move.

#9 Solve an Easier Related Problem 4.MD.A.3 Step 3
  • Test the smallest choice $S = 3$.
  • Then $S^2 = 9$, which is less than $10$, so $S = 3$ cannot hold even $10$ unit squares.
  • Eliminate (A).
$$3^2 = 9 < 10 \;\Rightarrow\; S = 3 \text{ fails}$$

💡 Comparing the area of a square with side $3$ to the required minimum $10$ is exactly the Grade 4 area-formula skill.

#2 Make a Systematic List 4.OA.B.4 Step 4
  • Test the next choice $S = 4$.
  • Then $S^2 = 16$.
  • The $8$ unit squares contribute $8$, so the remaining two squares must contribute $16 - 8 = 8$.
  • List integer pairs $(s_9, s_{10})$ with $s_9^2 + s_{10}^2 = 8$ and $s_9, s_{10} \ge 1$: $(1, ?)$ needs $? ^2 = 7$, not a perfect square; $(2, 2)$ gives $4 + 4 = 8$.
  • So the only integer fit is two $2 \times 2$ squares.
$$s_9^2 + s_{10}^2 = 16 - 8 = 8 \;\Rightarrow\; (s_9, s_{10}) = (2, 2)$$

💡 Looking through small squares ($1, 4, 9, \dots$) to find a pair summing to $8$ is the Grade 4 factor/multiple search habit.

#1 Draw a Diagram 3.G.A.2 Step 5
  • Confirm a tiling exists, not just an area match.
  • Place the two $2 \times 2$ squares side by side at the top: they form a $4 \times 2$ strip.
  • The bottom of the $4 \times 4$ is another $4 \times 2$ strip, which the eight $1 \times 1$ squares fill perfectly in a $4 \times 2$ grid.
  • So $S = 4$ really works.
$$\underbrace{2 \times 2}_{\text{top-left}} \;\;\underbrace{2 \times 2}_{\text{top-right}} \;\Big/\; \underbrace{8 \times (1 \times 1)}_{\text{bottom } 4 \times 2 \text{ strip}} \;\Rightarrow\; \textbf{(B) } 4$$

💡 Splitting a rectangle into equal unit squares is the Grade 3 "partition shapes into equal-area pieces" idea.

[1] #9 3.MD.C.7 Use the area equation. The $10$ small squares together cover the original $S \ti
[2] #9 4.OA.A.3 Find a lower bound for $S^2$. At least $8$ pieces have area $1$, and the other $
[3] #9 4.MD.A.3 Test the smallest choice $S = 3$. Then $S^2 = 9$, which is less than $10$, so $S
[4] #2 4.OA.B.4 Test the next choice $S = 4$. Then $S^2 = 16$. The $8$ unit squares contribute $
[5] #1 3.G.A.2 Confirm a tiling exists, not just an area match. Place the two $2 \times 2$ squa

Review

Reasonableness: The pieces really do account for the area: $8 \cdot 1 + 2 \cdot 4 = 8 + 8 = 16 = 4^2$. 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.

Alternative: Tool #3 (Eliminate Possibilities) on the choices: the lower bound $S^2 \ge 10$ immediately kills (A) $3$. For each remaining choice, the leftover area for the $2$ non-unit squares is $S^2 - 8$: that gives $8, 17, 28, 41$ for $S = 4, 5, 6, 7$. Of those, only $8 = 2^2 + 2^2$ splits as a sum of two positive perfect squares ($17, 28, 41$ each fail the two-squares check), and $S = 4$ is the smallest survivor, so (B) wins.

CCSS standards used (min grade 4)

  • 3.MD.C.7 Relate area to multiplication and addition operations (Writing the area equation $S^2 = s_1^2 + s_2^2 + \dots + s_{10}^2$ — the area of each square is side $\times$ side, and non-overlapping pieces' areas add.)
  • 3.G.A.2 Partition shapes into equal parts with equal areas (Verifying that a $4 \times 4$ square can actually be tiled by two $2 \times 2$ squares and eight $1 \times 1$ squares — a partition-of-shape question, not just an area check.)
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Combining "$\ge 8$ unit squares" plus "$\ge 1$ each for the other two" to derive the lower bound $S^2 \ge 10$.)
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Searching small perfect squares $1, 4, 9, \dots$ for a pair that sums to $8$, and confirming $(2, 2)$ is the only integer solution.)
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems (Computing $S^2$ for the candidate side lengths $S = 3$ ($9$) and $S = 4$ ($16$) and comparing to the required minimum area $10$.)

⭐ 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!

⭐ 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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