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Fun is, the prime multipliers in 100! that contribute trailing zeros are 2 and 5. And the exponent of 2 is more than exponent of 5 as we have 10 = 2 x 5. So, number of trailing zeros determined by the exponent of 5.There will be  fives contributed by 5, 10, 15, 20, 25, 30 ... 100. Further there are few more fives due to the number 25 which are less than 100 and multiples of 5 (higher powers). We have  fives contributed by 25. Overall we have 20+4 = 24 fives in 100!. Thus the exponent of 5 in expressing 100! as powers of prime numbers is 24.We will have 24 trailing zeros in 100!.Mathematically, we can say that there will be X/Y integers that are <= X and divisible by Y. We have 100/5 = 20 numbers divisible by 5, and 100/25 = 4 numbers divisible by 25. As 5 and its powers <= 100 contributes exponent of 5, we have 24 fives.
I like this.The answer is 24 as it is the number of 5s considering the factorization of all the number in the factorial. Each 5 gets multiplied by a factor 2, generating a 0 in the final number. There are a lot of "2"s in the factorization so we can just consider the number of 5s in it. The only tricky part is to remember that 25, 50, 75 and 100 bring in respectively 2 fives instead of just one (like for instance 10 or 20). Hope this is clear. Best!