2007 Cyprus MO/Lyceum/Problem 29

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Problem

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

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

Thus, $k+5$ must be divisible by $11$, and the minimum for $k$ is $6\Longrightarrow\mathrm{ B}$.

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 28
Followed by
Problem 30
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