2002 AMC 8 Problem 11

Attempt Problem 11 of the 2002 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
1111
1212
1313
1414
1515
Answer: C
Concepts:perfect squarepattern recognition
Small hint:
The sixth and seventh squares have side lengths 66 and 7.7.
Big hint:
Compare 727^2 and 62.6^2.
Solution:
The sixth and seventh squares have side lengths 66 and 77 tile lengths.
They use 62=366^2=36 and 72=497^2=49 tiles, respectively, so the seventh square requires 49−36=1349-36=13 more tiles.
Thus, C is the correct answer.
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