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This is my first time posting here so sorry if I done something wrong, and also my first time encountering a problem like this.

Besides trivial $0$, the only solution I found by simply writing down the powers of $2$ and $5$ parallely, is $5$ ($5^5=3125$, $2^5=32$). I couldn't find any kind of period. I've done problems which are about last digit, but not about the first one.

Hopefully I could get a hint. Thanks.

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Hint: $5^n\times2^n=10^n$. ${}$

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  • 4
    $\begingroup$ +1 This is just so... I'm gonna use this as an exercise given half a chance. $\endgroup$ – Jyrki Lahtonen Feb 6 '12 at 12:04

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