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P: 1,422
 Quote by SteveRives Your question is the same as finding out how often 5 is a prime factor on the natural numbers. So I am guessing that 100! has 20 zeros, because in counting by fives, we get to 100 after 20 steps.
You are WRONG, SteveRives.
$$C = \left[ \frac{n}{5} \right] + \left[ \frac{n}{5 ^ 2} \right] + \left[ \frac{n}{5 ^ 3} \right] + ... + \left[ \frac{n}{5 ^ q} \right]$$ zeros at the end.
$$\left[ \frac{n}{5 ^ q} \right] = 0$$