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

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2002 AMC 12B Problem 13

Below is the professionally curated solution for Problem 13 of the 2002 AMC 12B, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:perfect square arithmetic sequence

Difficulty rating: 1430

13.

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

$169$

$225$

$289$

$361$

$441$

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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