0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ cut-the-knot.org/blue/LegendresTheorem.shtml $\endgroup$ – Will Jagy Jan 1 '15 at 5:21
  • $\begingroup$ I know the prime factorization rule. what is your answer @ Will Jagy $\endgroup$ – archangel89 Jan 1 '15 at 5:24
  • $\begingroup$ Have you tried base $720$? $\endgroup$ – JimmyK4542 Jan 1 '15 at 5:29
  • 1
    $\begingroup$ 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 $N$ $\endgroup$ – Will Jagy Jan 1 '15 at 5:30
  • 1
    $\begingroup$ I gave you a counterexample, but you didn't verify it correctly. Note that $3^{41} \mid 85!$. $\endgroup$ – JimmyK4542 Jan 1 '15 at 5:42
2
$\begingroup$

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 $85!$.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for answering. Actually I calculated the power of 3 wrong. Tha's why I was perplexed. Thank you again $\endgroup$ – archangel89 Jan 1 '15 at 5:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.