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2002 AMC 12B Problems/Problem 13

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Problem

The sum of 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 be the consecutive positive integers. Their sum, 18a + \frac{17(18)}{2} = 9(2a+17), is a perfect square. Since is a perfect square, it follows that is a perfect square. The smallest possible such perfect square is when , and the sum is .

See also

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
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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