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Thread: Prove by induction that our formula is divisible by 7.

  1. #1
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    Prove by induction that our formula is divisible by 7.

    Hi guys,

    the question is in two parts and I am stuck on the second part.

    In the first part we have to prove by induction that $6^n - 1$ is divisible by 5. This was done by setting $6^n - 1 = 5A (A > 0)$ and setting $6^{n+1} - 1 = B$ and showing that B was modified only by factors of 5 in the process of adding the (n+1) th term. I mention this only in case the solution to the second part of the question relies on the first part, which it doesn't seem to.

    The second part says Prove that $8^n - 7n + 6$ is divisible by 7 for all integral n.

    I tried the same method as for part 1 but I cannot get it to work.

    I have $8^k - 7k + 6 = 7A$ for n = k

    I have $8^{k+1} - 7(k+1) + 6 = B$ for n = k+1

    but I cannot reduce the k+1 term to an equation in the form of the k term as I could in part 1. Can anyone do better?
    Last edited by s_ingram; May 17th 2017 at 06:05 AM.
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  2. #2
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    Re: Prove by induction that our formula is divisible by 7.

    $8^{k+1}-7(k+1)+6 = 8\cdot 8^k -7k-7 + 6 = (8^k-7k+6)+7\cdot 8^k-7 = 7A+7(8^k-1)$
    Thanks from s_ingram
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  3. #3
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    Re: Prove by induction that our formula is divisible by 7.

    It's so easy when you see the answer! I posted the question and thought "that will take them a few hours" and it is done in minutes! many thanks as usual.
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