If I understood your question correctly I think you are trying to find the number of 0's at the end of the result of the factorial, hence how many 10's you can take out of the result.
Think of it as counting the number of 10's you can divide out of the factorial, 10 = 5*2 so we need a pair of 5 and a 2 for each ten. Since factorials will always have more 2's than 5's, the question can be furthered simplified to how many 5's are there in the prime factorization of the factorial? Now typically you will have one every five numbers with the exception of numbers like 25 and 125, which are 5*5 and 5*5*5 respectively.
For 100! there are 20 multiples of 5 from 5, 10, 15 ... 95, 100. Then we need to identify the multiples which contain more than one 5. The numbers are: 25 (5*5), 50 (5*5*2), 75 (5*5*3), and 100 (5*5*2*2). There are 4 numbers with an additional five in its prime factorization, so in total we have 20 + 4 = 24 copies of fives. This means that there are 24 trailing zero's for 100! Hopefully, this answers what you were asking for!