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Divisible by 7 (Posted on 2007-02-15) Difficulty: 3 of 5
Prove that if A = ½(15+√197), then 7 is a factor of [A^n] for all positive integer values of n, where [w] denotes the greatest integer less than or equal to w.

See The Solution Submitted by Dennis    
Rating: 4.0000 (1 votes)

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Solution Comment 5 of 5 |
Let A=(15+sqrt197)/2 and B=(15-sqrt197)/2.
(A+B)^n=sum_i=0...n{(n over i)*A^i*B^(n-i)} = A^n + B^n + C.
From the first post we know that C = 7*k for some k, because it's a sum of numbers of the form A*B*d (for some integer d), and A*B is divisible by 7.
On the other hand we know that (A+B)^n=15^n, and 15%7=1, hence (15^n)%7=1 for all n>=0.
This shows that (A^n+B^n)%7=1. The last step is too notice, that 0<B<1, so 0 Nice one.
  Posted by Arek on 2007-03-05 07:26:24
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