AMC 8 · 2012 · #17
Grade 4 geometry-2dProblem
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.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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 $S^2$.
💡 Area of a square is side $\times$ side, and areas of non-overlapping pieces add — both are Grade 3 area ideas.
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 $8 + 1 + 1 = 10$.
💡 Adding up the smallest possible areas gives the smallest possible total — a Grade 4 multi-step reasoning move.
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).
💡 Comparing the area of a square with side $3$ to the required minimum $10$ is exactly the Grade 4 area-formula skill.
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.
💡 Looking through small squares ($1, 4, 9, \dots$) to find a pair summing to $8$ is the Grade 4 factor/multiple search habit.
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.
💡 Splitting a rectangle into equal unit squares is the Grade 3 "partition shapes into equal-area pieces" idea.
3.MD.C.7 Use the area equation. The $10$ small squares together cover the original $S \ti 4.OA.A.3 Find a lower bound for $S^2$. At least $8$ pieces have area $1$, and the other $ 4.MD.A.3 Test the smallest choice $S = 3$. Then $S^2 = 9$, which is less than $10$, so $S 4.OA.B.4 Test the next choice $S = 4$. Then $S^2 = 16$. The $8$ unit squares contribute $ 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.7Relate 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.2Partition 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.3Solve 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.4Find 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.3Apply 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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