For given positive integer n, its factorial often has many trailing zeros, in other words many factors of 10s. In order for n! to have at least "m" trailing zeros, what is the smallest "n" ?
Example: factorial(10) = 3628800 factorial(9) = 362880 In order to have 2 trailing zeros on factorial, the smallest n is 10.
Optional: Can you make an efficient algorithm for a very large m?
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Solution 1135219
I think this is (so far) the most efficient solution for large m. It requires at most log(4*m)/log(5) iterations; while the smaller sized solutions that have been suggested require at least 0.8*m iterations (or a very large array). For example, for m = 10^5, this solution takes at most 8 iterations while the other solutions of smaller size require at least 80,000 iterations.