# Difference between revisions of "2012 AMC 8 Problems/Problem 17"

## 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?

$\text{(A)}\hspace{.05in}3\qquad\text{(B)}\hspace{.05in}4\qquad\text{(C)}\hspace{.05in}5\qquad\text{(D)}\hspace{.05in}6\qquad\text{(E)}\hspace{.05in}7$

## Solution

The first answer choice ${\textbf{(A)}\ 3}$, can be eliminated since there must be $10$ squares with integer side lengths. We then test the next smallest sidelength which is $4$. The square with area $16$ can be partitioned into $8$ squares with area $1$ and two squares with area $4$, which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is $\boxed{\textbf{(B)}\ 4}$.

## See Also

 2012 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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