Have given one more question from Factorial. For the previous two questions have a look here
The previous question includes a detailed discussion on how to find the number of trailing zeroes of n!, for any natural number n.We see that 24! has [24/5] = 4 zeroes
25! ends with [25/5] + [25/25] = 6 zeroes. There is no natural number m such that m! has exactly 5 zeroes.Similarly, we see that 49! ends with [49/5] + [49/25] = 10 zeroes, whereas 50! ends with [50/5] + [50/25] = 12 zeroes. No factorial ends with 11 zeroes.
So, any time we have a multiple of 25, we ‘skip’ a zero. This is because a multiple of 25 adds two zeroes to the factorial.
In order to jump three zeros, think about what we need to look at. Every multiple of 25 gives us one ‘skipped’ zero. Every multiple of 125 gives us two ‘skipped’ zeroes.
In order to have three skipped zeroes, we need to look at 624! and 625!