Q. $85!$ ends with exactly $20$ trailing zeros. When $85!$ is converted to base $N$, $N$ being any natural number, it so happens that it has the same number of zeros at the end. What could be the largest possible value of $N$?
5. None of these.
I think the answer is base $80$. $80=5\cdot 2^4$. Now, as the $80$th number in base $80$ is $10$, we can see clearly that one $5$ and a group of four $2$s will yield one zero. $85!$ in base $10$ has $20$, $5$s and $81$, $2$s. A group of $20$, $2^4$s can be selected. All other cases will yield less zeros. So, I think this is the answer.
But the given answer is $5$. None of these. What is wrong in what I am doing?