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

## Answer this Question

## Related Questions

- math - What are the last three digits of the number N=2(1!)!+2(2!)!+...+2(1000...
- math - Posted by nicole on Monday, September 22, 2008 at 4:32pm in response to ...
- math - Posted by nicole on Monday, September 22, 2008 at 4:32pm in response to ...
- math - Can u please help me with this..this is my sons 6th grade work and im ...
- Math - Three digits are randomly selected without replacement from 1, 9, 9 and 6...
- Math - 120 Are the last two digits divisible by 4 Are the last three digits ...
- Math - 120 Are the last two digits divisible by 4 Are the last three digits ...
- Computer Science - public class RandomPhoneNumber { public static void main(...
- Math - What am I? I am a decimal fraction with three digits I am bigger than ...
- math - A locker combination has three nonzero digits, and digits cannot be ...

More Related Questions