About the sum of the digits of $k^{105}$.
Solution 1:
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.