Question

What are the last three digits of
N=(2^(1!)!) + (2^(2!)!) + ... + (2^(1000!)!)?

please help..

Answer 1

N = 2^1 + 2^2! + 2^(1000!)!)
The last number has no ending digits as small as 2 or 4

006 would be the last three digits of the complete number.

Answer 2

Since any number n! where n>=5 ends in zero,

(2^1)! + (2^2)! + ...
= 2 + 24 + ...
ends in 6