Find all positive integers
such that the decimal expansion of
n
! ends with exactly 2020 zeros.
Add all values of together and input that number as your answer.
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Every zero at the end of n ! corresponds to a factor 5. (It is easy to see that there are always more factors 2 than 5 in n ! )
In the product n ! = 1 ⋅ 2 ⋅ 3 ⋯ n , every fifth number contributes a factor 5; every twenty-fifth number contributes an additional factor 5; and so on. Therefore the number of factors 5 is ⌊ 5 n ⌋ + ⌊ 2 5 n ⌋ + ⌊ 1 2 5 n ⌋ + ⋯ Ignoring the ⌊ ⋅ ⌋ , we find that the number of factors 5 is approximately (and no greater than) n ⋅ ( 5 1 + 2 5 1 + 1 2 5 1 + ⋯ ) = 4 n . If there are 2020 zeroes, and therefore 2020 factors 5, we expect n ≈ 4 ⋅ 2 0 2 0 = 8 0 8 0 . In fact, for 8080 we find that there are ⌊ 5 8 0 8 0 ⌋ + ⌊ 2 5 8 0 8 0 ⌋ + ⌊ 1 2 5 8 0 8 0 ⌋ + ⌊ 6 2 5 8 0 8 0 ⌋ + ⌊ 3 1 2 5 8 0 8 0 ⌋ = 1 6 1 6 + 3 2 3 + 6 4 + 1 2 + 2 = 2 0 1 7 factors 5. The remaining three factors 5 come from 8085, 8090, and 8095.
Since the next factor 5 is found at 8100, the required values of n are n = 8 0 9 5 , 8 0 9 6 , 8 0 9 7 , 8 0 9 8 , 8 0 9 9 . The sum is easily found as five times the middle value, ∑ n = 5 ⋅ 8 0 9 7 = 4 0 4 8 5 .