SOLUTION: Is 2^n3^2n-1 always divisible by 17?

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Question 4611: Is 2^n3^2n-1 always divisible by 17?
Answer by khwang(438)   (Show Source): You can put this solution on YOUR website!
Yes, proof as below.
2^n3^2n -1 = (2*3^2)^n - 1 = (18^n) - 1
Use a^n-b^n = (a-b)(a^(n-1) + a^(n-2)b+...+ ab^(n-2)+ b^(n-1))
Now a= 18, b = 1,
we have (18^n) - 1 = (18-1)(18^17+18^16+...+18+1)
= 17*(18^17+18^16+...+18+1)
This shows 17 is a divisor of 2^n3^2n -1
Or proved by Mathematical Induction.
Kenny
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