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AoPS Wiki talk:Problem of the Day/June 20, 2011

Revision as of 21:10, 19 June 2011 by Djmathman (talk | contribs) (Added Solution)
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Problem

AoPSWiki:Problem of the Day/June 20, 2011

Solution

Since it is both a perfect cube and a perfect square, it must be a perfect sixth power. This is because the GCD of $2$ and $3$ is $6$, meaning it will leave an integer power when either the square root or the cube root is taken. The fourth perfect sixth power is $4^6=2^{12}=\boxed{4096}$.

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