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Thread: Show that it is divisible by 7 for all positive integers...

  1. #1
    Junior Member Bartimaeus's Avatar
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    Show that it is divisible by 7 for all positive integers...

    Show that $\displaystyle 43^n + 83 * 92 (to the power of) (3n-1)$ is divisible by 7 for all positive integers $\displaystyle n$



    Thanks for your time guys, really appreciate it and sorry about the dodgy math after the 92, I'm not very good when it comes to this computer stuff.
    Last edited by Bartimaeus; Sep 22nd 2007 at 01:46 PM. Reason: mistake in problem
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Bartimaeus View Post
    Show that $\displaystyle 43^n + 83 * 92 (to the power of) (3n-1)$ is divisible by 7 for all positive integers $\displaystyle n$



    Thanks for your time guys, really appreciate it and sorry about the dodgy math after the 92, I'm not very good when it comes to this computer stuff.
    If you know modular arithmetic, then
    $\displaystyle 43 \equiv 1~\text{(mod 7)}$

    $\displaystyle 83 \equiv -1~\text{(mod 7)}$

    and
    $\displaystyle 92 \equiv 1~\text{(mod 7)}$

    So
    $\displaystyle 43^n + 83 \cdot 92^{3n-1} \equiv 1^n + (-1) \cdot 1^{3n - 1} ~\text{(mod 7)}$

    $\displaystyle \equiv 1 - 1~\text{(mod 7)} ~ \forall ~n$

    $\displaystyle \equiv 0 ~\text{(mod 7)}$

    So $\displaystyle 43^n + 83 \cdot 92^{3n-1}$ is divisible by 7 for all values of n.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Bartimaeus View Post
    Show that $\displaystyle 43^n + 83 * 92 (to the power of) (3n-1)$ is divisible by 7 for all positive integers $\displaystyle n$



    Thanks for your time guys, really appreciate it and sorry about the dodgy math after the 92, I'm not very good when it comes to this computer stuff.
    If you don't know modular arithmetic, then consider n = 1:
    $\displaystyle 43^1 + 83 \cdot 92^{3 \cdot 1 - 1} = 43 + 83 \cdot 92^2 = 702555$
    which is divisible by 7.

    Now assume the theorem to be true for some n = k.
    Assume that
    $\displaystyle 43^k + 83 \cdot 92^{3k - 1} = 7x$
    where x is some integer.

    Consider the case for n = k + 1:
    $\displaystyle 43^{k + 1} + 83 \cdot 92^{3(k + 1) - 1}$

    $\displaystyle = 43 \cdot 43^k + 92^3 \cdot 83 \cdot 92^{3k - 1}$
    This should be divisible by 7.

    Now, consider again:
    $\displaystyle 43^k + 83 \cdot 92^{3k - 1} = 7x$

    Solving this for $\displaystyle 83 \cdot 92^{3k - 1}$ gives:
    $\displaystyle 83 \cdot 92^{3k - 1} = 7x - 43^k$

    Inserting this into the k + 1 expression gives:
    $\displaystyle 43 \cdot 43^k + 92^3 \cdot (7x - 43^k)$

    $\displaystyle = 43 \cdot 43^k + 7 \cdot (92^3x) - 92^3 \cdot 43^k$

    $\displaystyle = (43 - 92^3) \cdot 43^k + 7(92^3x)$

    The last term is manifestly divisible by 7. Let's look at the first term:
    $\displaystyle (43 - 92^3) \cdot 43^k$

    The coefficient of the $\displaystyle 43^k$ is
    $\displaystyle 43 - 92^3 = -778645 = 7(-111235)$

    So
    $\displaystyle 43^{k + 1} + 83 \cdot 92^{3(k + 1) - 1} = 7(-111235) \cdot 43^k + 7(92^3x)$
    where x is an integer. This is obviously divisible by 7.

    Thus $\displaystyle 43^n + 83 \cdot 92^{3n - 1}$ is divisible by 7 for n = 1, thus for n = 2, thus etc.

    -Dan
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