2002 AMC 12B Problems/Problem 13

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Problem

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

$\mathrm{(A)}\ 169\qquad\mathrm{(B)}\ 225\qquad\mathrm{(C)}\ 289\qquad\mathrm{(D)}\ 361\qquad\mathrm{(E)}\ 441$

Solution

Let $a, a+1, \ldots, a + 17$ be the consecutive positive integers. Their sum, $18a + \frac{17(18)}{2} = 9(2a+17)$ , is a perfect square. Since $9$ is a perfect square, it follows that $2a + 17$ is a perfect square. The smallest possible such perfect square is $25$ when $a = 4$ , and the sum is $225 \Rightarrow \mathrm{(B)}$ .

2002 AMC 12B (Problems)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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2002 AMC 12B Problems/Problem 13

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Problem

Solution

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