In how many zeros does 40!/(4 (5!)) end?

How many multiples of 5 are between 1 and 40?

5, 10, 15, 20, 25, 30, 35, 40

Plus one more because we have 5x5=25.

So, there are 9 trailing 0's on 40!.

But, \(\displaystyle 4\cdot 5!=480\)

This 0 must be subtracted off.

Thus, there are 8 trailing 0's on \(\displaystyle \frac{40!}{4\cdot 5!}\)
 
Hi galactus,
Can you please explain me. I couldn't catch up with the solution of 9 zeros.
Please guide me.
 
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