Factorial Rocks!

The number of trailing zero's of $500!$ will be $\displaystyle \sum_{k=1}^3 \left\lfloor{\frac{500}{5^k}}\right\rfloor= \left\lfloor{\frac{500}{5}}\right\rfloor+ \left\lfloor{\frac{500}{5^2}}\right\rfloor+ \left\lfloor{\frac{500}{5^3}}\right\rfloor\\=100+20+4=\boxed{124}$

As Sravanth pointed out, we just need to count the power of 5's.

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n = int(raw_input())
c = 5

r = n // c

count = 0

while r > 0:
    count += r
    c *= 5
    r = n // c

print count