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 $\boxed{\textbf{(B)}\ 4}$ .

Video Solution

https://youtu.be/CpOhH3ZYRfE ~savannahsolver

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2012 AMC 8 Problems/Problem 17

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Video Solution

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