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Math Help - How to do this algebra "trick"

  1. #1
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    How to do this algebra "trick"

    I'm trying to prove that 7|(3^{2n+1}+2^{n+2})

    Anyway, I've simplified the RHS to 3*9^n+4*9^n

    How can I "pull out" (3+4) here? ie, make it so that

    3*9^n+4*9^n=(3+4)*(???)

    Thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by paupsers View Post
    I'm trying to prove that 7|(3^{2n+1}+2^{n+2})

    Anyway, I've simplified the RHS to 3*9^n+4*9^n

    How can I "pull out" (3+4) here? ie, make it so that

    3*9^n+4*9^n=(3+4)*(???)

    Thanks!
    \displaystyle 3^{2n + 1} + 2^{n + 2} = 3 \cdot 9^n + 4 \cdot 2^n

    Note that 9 = 2 mod 7.

    -Dan
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  3. #3
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    (3+4).9^n no ? ( but i think what you made a mistake in simplifying RHS is wrong )
    You can actually get the result in one/two lines using congruence.
    so and since we get (by multupliying) the result.

    Topsquark is faster
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  4. #4
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    Quote Originally Posted by paupsers View Post
    I'm trying to prove that 7|(3^{2n+1}+2^{n+2})

    Anyway, I've simplified the RHS to 3*9^n+4*9^n

    How can I "pull out" (3+4) here? ie, make it so that

    3*9^n+4*9^n=(3+4)*(???)

    Thanks!
    That's the "distributive law": (ac+ bc)= (a+ b)c
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