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        riddles >> easy >> Prove [4^(2n+1)] + [3^(n+2)] mod 13 =  0
        
(Message started by: wonderful on May 10^th, 2008, 4:24pm)

Title: Prove [4^(2n+1)] + [3^(n+2)] mod 13 = 0
Post by wonderful on May 10^th, 2008, 4:24pm

Show that [4^(2n+1)] + [3^(n+2)] is divisible by 13 n belongs to Z+.

Have A Great Day!

Title: Re: Prove [4^(2n+1)] + [3^(n+2)] mod 13 = 0
Post by black_death on May 10^th, 2008, 5:42pm

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Can be proved by induction
For n=1 F(n) = 91 which is divisible by 13
Let F(n) = 13k
Now for n+1 F(n+1) = 4^[2(n+1) +1] + 3^[(n+1] +2]
= 16*4^(2n+1) + 3*3^(n+2)
= 16 *[13k - 3^(n+2)] + 3*3^(n+2)
=16*13k - 13*3^(n+2)

and that expression is divisible by 13

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