n! means n × (n − 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!
Looking back at my solution, it's disgusting. I calculated 100! and find the digits as previous problems. What should have been done is divide by 10 whenever the number becomes divisible by 10 which first occurs with 5! This works because trailing zeros do not affect the summation of digits... However, the code seems to run quickly. This definitely does not work for bigger factorials...
Answer: 648
Runtime: 0.53 ms