Posted by anonymous on Sunday, August 4, 2013 at 12:07am.
What are the last three digits of
N=(2^(1!)!) + (2^(2!)!) + ... + (2^(1000!)!)?
please help..

math  drwls, Sunday, August 4, 2013 at 12:27am
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.

math  Steve, Sunday, August 4, 2013 at 5:44am
Since any number n! where n>=5 ends in zero,
(2^1)! + (2^2)! + ...
= 2 + 24 + ...
ends in 6
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