2002 AIME II Problems/Problem 7

Problem

It is known that, for all positive integers $k$ ,

$1^2+2^2+3^2+\ldots+k^{2}=\frac{k(k+1)(2k+1)}6$ .

Find the smallest positive integer $k$ such that $1^2+2^2+3^2+\ldots+k^2$ is a multiple of $200$ .

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2002 AIME II (Problems • Answer Key • Resources)
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