However, if your question is modified a little to "How many trailing zeros are there in 100!?" then we can definitely work it out.

The only trick is to remember that a zero is created by the factor pair (2,5) as 2*5 = 5*2 =10. Therefore, it is necessary to count the number of 5s and 2s. One could possibly argue that the count of 5s alone or 2s alone might give us the result. But that would actually cause a problem by giving us an exaggerated or underestimated count of 0s. As it is possible that the count of 2s is greater or lesser than the count of 5s. So, the count of 0s in the end result would rather depends on the minimum count between the count of 2s and the count of 5s.

100! represents the product of all the natural numbers up to and including 100.

Clearly, the count of 2s is more than the count of 5s. (Let me know if you could visualize it!)

To count the number of 5s, we must note that there are 20 multiples of 5 among the natural numbers up to and including 100. Only these multiples can contribute 5!!

It would be worth to note that 16 multiples of 5 would contribute one 5 each except 25, 50, 75 and 100. The reason to expel out these 4 multiples of 5 from the first group is that each of these multiples is a multiple of 25 too. That means each of these multiples contribute two 5s each.

So, the total count of 5s = 16*1 + 4*2 = 16 +8 =24

Therefore, the total number of trailing 0s in 100! is =24

Otherwise the total count of 0s in 100! is30if all the 0s , not just the trailing 0s, are to be counted.