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Problem of the Day #116: Binomial coefficient sum July 13, 2011

Posted by Mitchell in : potd , trackback

Given that \[\sum_{n = 1003}^{2006} \frac{1}{n} \binom{n}{2006 - n} = \frac{p}{q},\] where $p$ and $q$ are relatively prime integers with $q > 0$, find the smallest positive integer $m$ such that $qm – p$ is divisible by $2011$.

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