2002 AMC 12B Problem 13 — Solution | LIVE by Po-Shen Loh

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2002 AMC 12B Problem 13

Attempt Problem 13 of the 2002 AMC 12B 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 12B solutions, or check the answer key.

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

13.

The sum of $18$ consecutive positive integers is a perfect square. The smallest possible value of this sum is

$169$

$225$

$289$

$361$

$441$

Answer: B

Concepts:perfect square arithmetic sequence

Difficulty rating: 1430

Solution:

The sum of $n,$ $n+1,\ldots,$ $n+17$ is $18n+\dfrac{17\cdot18}{2}=9(2n+17).$ Since $9$ is a perfect square, $2n+17$ must be one too.

The smallest positive integer $n$ making $2n+17$ a perfect square is $n=4,$ giving $2n+17=25$ and a sum of $9\cdot25=225.$

Thus, the correct answer is B.

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