TITLE: Find all $n$ so that $s(2^n)=s(5^n)$
QUESTION [7 upvotes]: Let $s(k)$ be the sum of digits of the number $k$. Find all $n$ so that $s(2^n)=s(5^n)$
I have got that $n=3$ is one solution.
And looking modulo 9 it's easy to get that 3 has to divide $n$.

REPLY [3 votes]: As was noted during the discussion of this question, the digits of $k^n$ are created quite similar to a pseudo random number generator, so the heuristic is that $s(k^n)$ is about the average digit value times the length of the decimal number string:
$$
s(k^n) \approx h(k, n) = 4.5 \frac{\ln(k)}{\ln(10)} n
$$
Like Ross Millikan noted, this will make it hard for $s(2^n)$ to catch up with $s(5^n)$.
$s(2^n)$ would need a $n$ with a large deviation up and/or a large deviation down from $s(5^n)$.
Example: 
$$
\begin{array}{c|c|c|c|c}
n & s(2^n) & h(2, n) & s(5^n) & h(5, n) \\ 
\hline
10     & 7      & 13     & 40     & 31 \\ 
100    & 115    & 135    & 283    & 314 \\ 
1000   & 1366   & 1354   & 3172   & 3145 \\ 
10000  & 13561  & 13546  & 31117  & 31453 \\ 
100000 & 135178 & 135463 & 313339 & 314536 \\ 
\end{array}
$$