Difference between revisions of "2002 AIME II Problems/Problem 7"

Revision as of 13:07, 19 April 2008

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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 == Problem ==
+It is known that, for all positive integers <math>k</math>,
+<center><math>1^2+2^2+3^2+\ldots+k^{2}=\frac{k(k+1)(2k+1)}6</math>.</center>
+Find the smallest positive integer <math>k</math> such that <math>1^2+2^2+3^2+\ldots+k^2</math> is a multiple of <math>200</math>.
 == Solution ==

2002 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AIME Problems and Solutions