(Q). Find the number of trailing zeros in 1!.2!.3!.4!.5!………148!.149!.150!.
Solution:-
1! till 4! will have same number of trailing zeros.
5! till 9! will have same number of trailing zeros.
10! till 14! will have same number of trailing zeros.
15! till 19! will have same number of trailing zeros.
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95! till 99! will have same number of trailing zeros.
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140! till 144! will have same number of trailing zeros.
& 145! till 149! will have same number of trailing zeros.
So divide above number in group of five-five factorial numberes which have same number of trailing zeros.
➟ (1!.2!.3!.4!)(5!.6!.7!.8!.9!)(10!.11!.12!.13!.14!)(15!.16!.17!.18!.19!)(20!.21!.22!.23!.24!)………..(95!.96!.97!.98!.99!)……..(145!.146!.147!.148!.149!)(150!).
⇨ now number of trailing zeros in first group i.e. (1!.2!.3!.4!) = 0
Hence number of trailing zeros or highest power of 5 in given question will be the sum of all theser total values = 2612 Answer