Parity of number of digits

Question

Let $s(n)$ be the number of digits of $n$. Prove that $s(4^{2017}) + s(25^{2017})$ is odd. A solution without using logarithms would be nice.

Answer 1

The product of $4^{2017}$ and $25^{2017}$ is $100^{2017}$. We can write $4^{2017}=10^ra$ and $25^{2017}=10^sb$ where $1<a,b<10$. Then $4^{2017}$ has $r+1$ digits and $25^{2017}$ has $s+1$ digits. But $ab=10$ and $r+s=4033$, etc.

Answer 2

Just calculate $[2017\log25]+1=2820$ and $[2017\log4]+1=1215$.