Never mind. I figured it out.

1. in 100!, we have 100 / 5 numbers that have 5 as the prime factor. So we get a count of 20.

2. Now 25 integer have an additoinal 5 as a factor. So 100 / 25 gives an additional 5 as the prime factor. So we get a count of 4.

Now we have the power of 5 to be 20 + 4 which is 24.

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Simple Rule worth noting to find the number of terminating zeroes of n!

1. Divide the number n by 5^m where m ranges from 1 thru a max value x such that 5^x is less than the number.

2. Sum of all the quotients obtained is the Ans. For in our case n = 100.

Divide 100 by 5, 5^2. Quotients are 20 and 4. So the final ans is 24.

another example for 200!.

200 / 5 = 40

200 / 25 = 8

200 / 125 = 1. (Stop at this point. 5^4 = 625 < 200 and division results zero from here onwards. )

So ans is 49.

Some good alternatives are provided in the below link.

**Edit: **URL was broken