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38s and Bunches of 1s (Posted on 2017-06-30) Difficulty: 3 of 5
Prove every number in the sequence 38, 381, 3811, 38111, 381111, ... is composite.

See The Solution Submitted by Brian Smith    
Rating: 4.5000 (2 votes)

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Solution Nice problem | Comment 4 of 7 |

The trick is that there are 3 closely related, but slightly different cases, all composite, to be considered:        
1/900 (7^3*1000^n - 100) = 3((7^3*1000^n - 100)/2700) =  | 381 | 381111 | 381111111 | 381111111111 | 381111111111111, etc.             
1/90 (7^3*1000^n - 10) = 37(7^3*1000^n - 10)/3330 =  | 3811 | 3811111 | 3811111111 | 3811111111111 | 3811111111111111, etc.             
1/9 (7^3*1000^n - 1) = 1/9(7*10^n - 1)(7*10^n + 7^2*10^(2n) + 1) =  | 38111 | 38111111 | 38111111111 | 38111111111111 | 38111111111111111, etc.       

For obvious reasons, some of the factors of these numbers have interesting repetitive features of their own.            

Edited on June 30, 2017, 10:39 am
  Posted by broll on 2017-06-30 10:38:00

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