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Math Help - Divisible by 7

  1. #16
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    I might as well add to this.
    Suppose that it is true for K, i.e. 3^{2K+1}+2^{K+2}} is divisible by 7.
    Look at 3^{2(K+1)+1}+2^{(K+1)+2}=3^{2K+3}+2^{K+3}
    So  3^{2K+3}+2^{K+3}=3^{2K+3}+3^2<br />
\cdot 2^{K+2}-3^2\cdot 2^{K+2}+2^{K+2} =3^{2}\left(3^{2K+1}+2^{K+3}\right)-2^{K+2}\left(3^2-2\right).
    That is clearly divisible by 7.
    Last edited by Plato; August 21st 2010 at 03:28 PM.
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  2. #17
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    Quote Originally Posted by Plato View Post
    I might as well add to this.
    Suppose that it is true for K, i.e. 3^{2K+1}+2^{K+1}} is divisible by 7.
    I think you've meant 3^{2K+1}+2^{K+2}}...
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  3. #18
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    Quote Originally Posted by Soroban View Post
    Hello, yuud!

    Here is a sneaky method . . . if you know modulo arithmetic.



    3^{2n+1} + 2^{n+2} \;\;=\;\left(3^3\right)^\frac{2n+1}{3}} + \left(2^3\right)^{\frac{n+2}{3}}

    . . . . . . . . . . . =\;(27)^{\frac{2n+1}{3}} + (8)^{\frac{n+2}{2}}

    . . . . . . . . . . . \equiv\; (\text{-}1)^{\frac{2n+1}{3}} + (1)^{\frac{n+2}{2}}\text{ (mod 7)}

    . . . . . . . . . . . \equiv\;\left(\sqrt[3]{\text{-}1}\right)^{2n+1} + \left(\sqrt[3]{1}\right)^{n+2}\text{ (mod 7)}

    . . . . . . . . . . . \equiv\; (\text{-}1)^{2n+1} + (1)^{n+2}\text{ (mod 7)}

    . . . . . . . . . . . \equiv\; -1 + 1\text{ (mod 7)}


    3^{2n+1} + 2^{2n+1} \;\equiv\;0 \text{ (mod 7)}


    It is divisible by 7 for all positive intergers n.

    I haven't studied modular arithmetic... but THANKS
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