# 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?

• cut-the-knot.org/blue/LegendresTheorem.shtml – Will Jagy Jan 1 '15 at 5:21
• I know the prime factorization rule. what is your answer @ Will Jagy – archangel89 Jan 1 '15 at 5:24
• Have you tried base $720$? – JimmyK4542 Jan 1 '15 at 5:29
• my answer is that a person who knows that rule ought to have little difficulty finding the number of zeros in bases 2,3,4,5,6,7,8,9,11,12,13,14,15, thus giving good grounds for conjecturing the largest possible value of their $N$ – Will Jagy Jan 1 '15 at 5:30
• I gave you a counterexample, but you didn't verify it correctly. Note that $3^{41} \mid 85!$. – JimmyK4542 Jan 1 '15 at 5:42

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 $85!$.)