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.

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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

The sum of

n,

n+1,\ldots,

n+17

equals

9(2n+17)

Since

9

is already a square,

2n+17

must be an odd perfect square; take the smallest valid one

The sum of

n,

n+1,\ldots,

n+17

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.

Problem 13 in other years