Question 1115766
The number of trailing zeros in the decimal representation of {{{n!}}}, the factorial of a non-negative integer {{{n}}}, can be determined with this formula:

{{{n/5+n/5^2+n/5^3}}}+...+{{{n/5^k}}} where {{{k}}} must be chosen such that {{{5^(k+1)>n}}}


so, {{{100!}}} has {{{100/5+100/5^2=100/5+100/25=20+4=24}}} trailing zeros