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Write a program or function that finds the number of zeroes at the end of n! in base 10, where n is an input number (in any desired format).

It can be assumed that n is a positive integer, meaning that n! is also an integer. There are no zeroes after a decimal point in n!. Also, it can be assumed that your programming language can handle the value of n and n!.

Test cases

1
==> 0

5
==> 1

100
==> 24

666
==> 165

2016
==> 502

1234567891011121314151617181920
==> 308641972752780328537904295461

This is code golf. Standard rules apply. The shortest code in bytes wins.

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  • \$\begingroup\$ Related. \$\endgroup\$
    – xnor
    May 12, 2016 at 2:39
  • \$\begingroup\$ Can we assume that n! will fit within our languages' native integer type? \$\endgroup\$
    – Alex A.
    May 12, 2016 at 2:45
  • \$\begingroup\$ @AlexA. Yes you can. \$\endgroup\$
    – Arcturus
    May 12, 2016 at 3:01
  • 17
    \$\begingroup\$ I think this would be a better question if you were not allowed to assume n! would fit into your integer type! Well, maybe another time. \$\endgroup\$
    – A Simmons
    May 12, 2016 at 10:45
  • 1
    \$\begingroup\$ @ASimmons Most of the answers so far, or at least the ones that use the floor division trick, don't rely on that assumption anyway. \$\endgroup\$
    – Alex A.
    May 12, 2016 at 16:53

61 Answers 61

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0
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Risky, 6 bytes

!!?+_0__/\\?

Try it online!

Explanation

! length
!       factorial
?         input
+     +
_
0         0
_   apply function
_
/         5
\     find indices in
\       prime factors
?         argument
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