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it can be found as follows
- abybaby January 06, 2008num zeros = floor(n/5) + 2*floor(n/5^2) + 3*floor(n/5^3)...so on
Th logic is straight forward. There will be as many zeros as there are no of 5s in the numbers, as 5 * 2 = 10 and num(5) < num (2)
Note that these 5s also include the multiples of 10. So we don't need extra
logic for them