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.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
17.
A square with an integer side length is cut into squares, all of which have integer side length and at least of which have area What is the smallest possible value of the length of the side of the original square?
Answer: B
Small hint:
A side length of gives area too small for ten integer squares.
Big hint:
A by square can be cut as shown.
Video solution:
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Written solution:
Since all squares have positive integer side lengths, each has area at least . Their total area is therefore at least . A square with integer side length at most has area at most , so its side length cannot be less than .
The following configuration cuts a square into ten integer-sided squares, eight of which are unit squares. Therefore side length is attainable and is the minimum.
Thus, the answer is B .
Problem 17 in other years
- 1985 AMC 8
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