Q:

Number of trailing zeros at other bases

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 $N$?
1. $80$  2. $160$ 3. $240$ 4. $200$ 5. None of these.
My attempt
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?

A:

If $85!$ has exactly $20$ zeros in base $N$, then $N^{20}$ divides $85!$ and $N^{21}$ does not divide $85!$. 
Using the rule Will Jagy gave you, $85! = 2^{81} \cdot 3^{41} \cdot 5^{21} \cdot 7^{13} \cdots 83^1$. (It is easy to see that the exponents on each prime decrease, so the exponents from $7$ until $83$ are all less than $20$.)
Let $N = 2^{a_1} \cdot 3^{a_2} \cdot 5^{a_3} \cdot 7^{a_4} \cdots 83^{a_n}$. How big can you make each of the $a_k$'s such that $N^{20}$ divides $85!$? (For this largest value of $N$, you should see that $N^{21}$ does not divide $85!$.)

