1989 AMC 12 Problem 17

Attempt Problem 17 of the 1989 AMC 12 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 1989 AMC 12 solutions, or check the answer key.

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

17.

The perimeter of an equilateral triangle exceeds the perimeter of a square by 19891989 cm. The length of each side of the triangle exceeds the length of each side of the square by dd cm. The square has perimeter greater than 0.0. How many positive integers are not possible values for d?d?
00
99
221221
663663
infinitely many
Answer: D
Concepts:perimeterlinear equationcounting integers in a range
Small hint:
Let the square’s side length be ss
Big hint:
Translate the positive-perimeter condition into a strict inequality for dd
Solution:
If the square has side s,s, the triangle has side s+d.s+d. The perimeter condition gives 3(s+d)−4s=1989, 3(s+d)-4s=1989, so s=3d−1989.s=3d-1989. The condition s>0s\gt0 is equivalent to d>663.d\gt663. Thus the positive integers 11 through 663663 are precisely the impossible values.
Thus the correct answer is D.
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