Number of zero digits of $\frac{41!}{100}$

Question

Let's consider a factorial number $\frac{41!}{100}$. How can we find how many digits are zero starting from the last digit of the number?

If we multiply the factorial by $100$, then the number of $0$ digits increases by $2$ and we can work with $41!$ directly. We're looking for the largest $k$ such that

$$41!\equiv 0\pmod{10^k}$$

Furthermore, since $41!$ contains $2$ and $5$, a trailling zero exists. But I am not certain as to what can be done. Could you provide a feedback?

Find the number of trailing zeros of $41!$ (by counting all the times $5$ contributes to the product), and subtract $2$. — Ekene E., Jan 13 '24 at 22:36
Welcome to Math SE. FYI, I suspect you'll find Legendre's formula to be useful here, but especially so when dealing with larger factorial numbers. — John Omielan, Jan 13 '24 at 22:38
@EkeneE. Why subtract $2$? Don't you just count the number of times $10$ divides the $41!$ and since $2$ occurs much more as a factor than $5$, you just count the number of times $5$ divides the product right? — Vincent Batens, Jan 13 '24 at 23:00
@VincentBatens Subtracting $2$ accounts for dividing by $100$. There are two related questions appearing in the question — Brian Moehring, Jan 13 '24 at 23:10
@BrianMoehring Ah thank you, I apparently didn't read it carefully enough and didn't consider the dividing by $100$. — Vincent Batens, Jan 14 '24 at 00:05
Does this answer your question? Find the number of trailing zeros in $n!$. Please look for related posts on this site before (or instead of) posting a duplicate question. — Anne Bauval, Jan 15 '24 at 15:50

score 5 · Answer 1 · answered Jan 13 '24 at 23:04

5

By Legendre's formula the number of trailing zeros of $41!$ is $$\nu_5(41!) = \left\lfloor \frac{41}{5} \right\rfloor + \left\lfloor \frac{41}{25} \right\rfloor = 9,$$ and dividing $41!$ by $100$ takes off two of those zeros, so there are $7$ trailing zeros.

answered Jan 13 '24 at 23:04

Ekene E.

537

4

you should mention trailing zeros require factors of 2 and 5, just that there's more factors of 2 in this case. – qwr Jan 13 '24 at 23:40
Please don't post an answer to such a FAQ. Instead, look for a good duplicate target. – Anne Bauval Jan 15 '24 at 15:51

Number of zero digits of $\frac{41!}{100}$

1 Answers1