TITLE: About the sum of the digits of $k^{105}$.
QUESTION [10 upvotes]: I read here that 

We cannot find an integer $k>2$ such that the sum of the digits of $k^{105}$ is $k$. 

Does anyone know a proof of this?

REPLY [8 votes]: A dumb-but-working approach could look like this:

$k^{105}$ has $1+\lfloor\log_{10} k^{105}\rfloor=1+\lfloor 105\log_{10} k\rfloor$ digits.
A number with $n$ digits has digit sum between $1$ and $9n$.
Thus, the maximum possible digit sum of $k^{105}$ is $9(1+\lfloor 105\log_{10}k\rfloor)$.
This quantity is smaller than $k$ for $k\geq 3330$.
This reduced the problem to a finite one, which can be easily finished by a computer.