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December 21, 2014

December 21, 2014

Posted by **anonymous** on Sunday, August 4, 2013 at 12:07am.

N=(2^(1!)!) + (2^(2!)!) + ... + (2^(1000!)!)?

please help..

- math -
**drwls**, Sunday, August 4, 2013 at 12:27amN = 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 5:44amSince any number n! where n>=5 ends in zero,

(2^1)! + (2^2)! + ...

= 2 + 24 + ...

ends in 6

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