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2007 Cyprus MO/Lyceum/Problem 29

Revision as of 14:37, 6 May 2007 by I_like_pie (talk | contribs)
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Problem

The minimum value of a positive integer $k$, for which the sum $S=k+(k+1)+(k+2)+\ldots+(k+10)$ is a perfect square, is

$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } \mathrm{None\;of\;these}$

Solution

$k+(k+1)+(k+2)+\ldots+(k+10)=11k+55=11(k+5)$

The smallest value of $k$ is $6\Rightarrow\mathrm{ B}$.

See also

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