## Problem 699

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pjt33
Posts: 27
Joined: Mon Oct 06, 2008 5:14 pm

### Problem 699

There's an error in the problem statement. It currently says
the denominator is a power of 3 i.e. $b=3^k, k>0$.
It should say
the denominator is a power of 3 other than 1, i.e. $b=3^k, k>0$.
(The other possible correction, $>$ to $\ge$, doesn't match up with the test cases).