Problem 44071. Smallest n, for n! to have m trailing zero digits

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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69.81% Correct | 30.19% Incorrect
Last Solution submitted on Jul 10, 2020

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