Posted by **harsh** on Wednesday, February 13, 2013 at 9:47am.

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

- math -
**Reiny**, Wednesday, February 13, 2013 at 11:27am
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

- math -
**harsh**, Wednesday, February 13, 2013 at 12:33pm
N=2^(1!)!+2^(2!)!+...+2^(1000!)!?

this is the actual series..

- math -
**hehe**, Friday, February 15, 2013 at 4:14pm
stop this.

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