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(256^n)-1 is always divisible by 17?

Guest May 5, 2017
 #1
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yes you can use recursivity to demonstrate it 256^(n+1) -1
Guest May 5, 2017
 #2
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Proof by induction:

 

n = 1:     256^1 - 1 → 255 → 15*17  hence divisible by 17

 

n → n+1:    256^(n+1) - 1 - (256^n - 1) → 256^n*(256 - 1) → 256^n*15*17  hence divisible by 17

Alan  May 6, 2017

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