What are the last two digits in the sum of factorials of the first 100 positive integers?
I need to find the sum of 1! + 2! + 3! + ... + 100! and then look at the last two digits. I notice that for any n >= 10, n! ends with two zeros, since it is a multiple of 10 and 100. This means that the last two digits of the sum are only affected by the terms from 1! to 9!. So I can ignore the terms from 10! onwards and just focus on adding 1! + 2! + ... + 9!. I can use a calculator or do this by hand: 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320, and 9! = 362880. Adding these up, I get 409113. The last two digits are 13. 