2012 AMC 8 Problem 17

Attempt Problem 17 of the 2012 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 8 solutions, or check the answer key.

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17.

A square with an 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?

3 3

4 4

5 5

6 6

7 7

Answer: B
Concepts:tilingextremal argument
Difficulty rating: 1540
Video solution:
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Written solution:

Since all 1010 squares have positive integer side lengths, each has area at least 11. Their total area is therefore at least 1010. A square with integer side length at most 33 has area at most 32=93^2=9, so its side length cannot be less than 44.

The following configuration cuts a 4×44\times4 square into ten integer-sided squares, eight of which are unit squares. Therefore side length 44 is attainable and is the minimum.

Thus, the answer is B .

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