So many zeros!

It simple sir. Use a floor minimum integer divided by 5^n. | 100/5 | + | 100/25 | + | 100/125 | + ... = 20 + 4 + 0 + ... + 0 = 24. Answer : 24

Just count how many multiple of 5 in 100, or to make it more easy, just divide 100 with 5 which the answer is 20, but don't forget that 5 x 5 = 25 has two 5 number in it, so divide 100 with 25 and you will get 4 then 20+4 = 24

When an integer which doesn't end with 0 is multiplied by 10 then the number ends with a 0. 10=2 x 5

So to find out how many zeroes are there at the end of 100! we have to see how many 2s and 5s are there in its prime factorization. The number of multiples of 2 exceeds the number of multiples of 5. So logically we can say that the number of zeroes at the end 100! will be the number of prime factors of 100! which are 5. There are 20 multiples of 5 below 100 but multiples of 25 will have two 5s as prime factors. There are 4 multiples of 25 which are less than or equal to 100. So the number of fives will be (4 x 2) + 20-4 = $\boxed {24 }$

So the product of what numbers when multiplied ends in a zero: When one of the things being multiplied ends in zero itself. A number ending in 5 multiplied by an even number. 25, 50 and 75 when multiplied by some of the small numbers available eg (4, 2 and 6) generate an extra zero. using this logic we can calculate how many zeros each no.in 100! contributes to and it comes out to be 24!!!