How many zeros are there in $25!$?

Question

I don't know how to really calculate it the number of zeros in the right hand can easily find by Legendre's formula.

That gives us:

$\lfloor{\frac{25}{5}}\rfloor+\lfloor{\frac{25}{25}}\rfloor=5+1=6$

But I got stuck with finding the zeros between.Any hints?

Not following...you have correctly observed that $v_5(25!)=6$. As $v_2(25!)$ is certainly bigger than that, the answer must be $6$. Or have I misunderstood? — lulu, Jul 28 '16 at 16:37
You need to count all the zeros, not just the ones at the end? — , Jul 28 '16 at 16:38
I think the mean counting the number of 0's in the string $25!= 15511210043330985984000000$. — , Jul 28 '16 at 16:38
@T.Bongers Yikes! You think? I doubt there is a good general way of doing that...though, as you point out, this particular case is fairly easy to handle directly. — lulu, Jul 28 '16 at 16:39
@TahaAkbari Got it. That sounds hard. Wolfram alpha (or other sources) can certainly handle the raw computation....don't see a more elegant way to proceed. — lulu, Jul 28 '16 at 16:40
@TahaAkbari This is easy to do numerically, but I don't know of any nice patterns. For example, $1000!$ is a 2568 digit number in which zero occurs 472 times. (Computed in a few milliseconds in python) — , Jul 28 '16 at 16:43
This depends on base-10 arithmetic coincidences, so I doubt that there is any reasonable formula. I could not find this in OEIS. — marty cohen, Jul 28 '16 at 16:52
@TahaAkbari Your question was how many. No one has come up with a better way than evaluating 25! explicitly and counting the zeros, so please accept the CW answer below if you are content, or explain why you are not. — almagest, Jul 28 '16 at 17:17

score 3 · Accepted Answer · edited Jul 28 '16 at 17:24

3

25!=15511210043330985984000000, so the answer is 9 zeros. [Thanks to @T.Bongers]

Or:

def fact(n):
  if n < 2:
    return 1
  else:
    return n*fact(n-1)

print str(fact(25)).count('0')

Thanks to Python.

edited Jul 28 '16 at 17:24

David C. Ullrich

89,985

answered Jul 28 '16 at 17:15

almagest

18,380

Depending on the system, this will break once $n$ is large because the recursion limit will be hit. I'd suggest a loop instead. – Jul 28 '16 at 18:29

score 0 · Answer 2 · answered Jul 28 '16 at 18:19

Your calculation gives the ended zeroes but for the other appearing in the decimal expression of $25!$ I think you have no other way that to get a such expression. The calculator of Windows gives $$15\space 511\space 210\space 043\space 330\space 985\space 984\space 000\space 000$$ hence you have $6+3=\color{red}{9}$

How many zeros are there in $25!$?

2 Answers2