Question 388588
It suffices to count the number of 5's in 95! since there will be enough 2's to make multiples of 10.


The multiples of 5 less than or equal to 95 are 5, 10, 15, ..., 95, so there are 19 multiples of 5. The multiples of 25 are 25, 50, 75, so 3 "additional" 5's, so there are 22 5's in the factorization of 95!. Therefore there are 22 zeroes at the end of 95!.


(The faster algorithm is to repeatedly divide by 5, i.e. 95/5 = 19, 19/5 = 3, 19+3 = 22)