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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The Prime Directive
- 14 Problems
- 36 Finishers
- Find the nearest prime number
- Extra safe primes
- Prime factor digits
- Twin Primes
- Twins in a Window
- The Goldbach Conjecture
- The Goldbach Conjecture, Part 2
- Goldbach's marginal conjecture - Write integer as sum of three primes
- Sophie Germain prime
- Mersenne Primes vs. All Primes
- Circular Primes (based on Project Euler, problem 35)
- Numbers spiral diagonals (Part 2)
- Pernicious Anniversary Problem
- Prime Ladders
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