Find the last two digits of the following sum: \(5! + 10! + 15! + \cdots + 100!\)
All you have to do is to sum first 7-8 terms and you will end up with 80 for the last 2 digits.
All the terms from 10! have 2*5*10 = 100 as a factor so none of these will affect the last 2 places in the sum
So I only need the last 2 places of 5!
The last place will be 0 (2*5)
1*2*3*4*5 = 120
So the last 2 digits are 20
