math
posted by harsh on .
What are the last three digits of the number N=2(1!)!+2(2!)!+...+2(1000!)!?

let's look at the first few terms
2(1!) = 2
2(2!) = 4
2(3!) = 12
2(4!) = 48
2(5!) = 240
2(6!) = 1440
2(7!) = 10080
2(8!) = 80640
2(9!) = 725760
2(10!) = 7257600
2(11!) = .....3600 , since we are only concerned about the last 3 digits
2(12!) = ..... 3200
2(13!) = ... 1600
2(14!) = ...2400
2(15!) = ...36000
after that no result will contribute to the last 3 digits.
so add them up , to the end of the first 3 digits.
Unless I made a silly arithmetic error I got 626 
N=2^(1!)!+2^(2!)!+...+2^(1000!)!?
this is the actual series.. 
stop this.