Number of zeros in 100!

no. of zeroes in the end = no. of fives (since it will always be less than the no. of 2s)= $[\frac{100}{5}] + [\frac{100}{25}]$ = 20 + 4 = 24

In 100! we have 3 possibilities of getting zero at the end . They are: *When the number is a multiple of 10. From this we get 11 0s including 100. *When a number ending with 5 is multiplied with an even number. Here 25, 50, 75 are exceptions because these numbers will yield 2 zeros when multiplied with single digit even numbers like 4, 2, 8,6. therefore we get 13 0s from this. There fore 11+13=24

to find number of zeroes at the end in 100! you have to count number of 5's and 10's in the series.

consider 10! in this number of 5' = 1 and number of 10's = 1 so total number of zeroes in 10! = 2

consider 30! number of 5's are 5, 15, 25 (in 25 there are are 2 5's as 25 = 5× 5) total 5's = 4

number of 10's are 10, 20, total =

so total number of zeroes = 4 + 2 = 6

but in 100! or still greater number factorial it is not so easy to count number of 5 in the series

easy way find quotient by dividing by 5 divide the quotient again till the quotient is less than 5 then add all quotients will give number of zeroes. 100/5 = 20, 20/5 = 4 (4 is less than 5) so total zeroes = 20 + 4 = 24

By the way, I know the math behind it, But i made a program in C++ which calculates this. Wanna see?