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Thread: Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

  1. #1
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    Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

    without using mathematical induction and also explicitly not using the application of mod in your demonstration.


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    $\displaystyle Show \ \ 24 \ \ divides \ \ (2*7^n \ + \ 3*5^n \ - \ 5) \ \ $ for all positive integers n.

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    Re: Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

    We can observe that:

    $\displaystyle \left(2\cdot7^{1}+3\cdot5^{n}-5\right)=24$

    $\displaystyle \left(2\cdot7^{n+1}+3\cdot5^{n+1}-5\right)-\left(2\cdot7^{n}+3\cdot5^{n}-5\right)=12\left(5^n+7^n\right)$

    So, we need to show that $\displaystyle 5^n+7^n$ is even. The sum of two odd numbers is even.
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    Re: Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

    Quote Originally Posted by MarkFL View Post
    We can observe that:

    $\displaystyle \left(2\cdot7^{1}+3\cdot5^{n}-5\right)=24$
    .
    .
    .
    You had a typo for the second exponent at that moment:

    $\displaystyle \left(2\cdot7^{1}+3\cdot5^{1}-5\right)=24$
    Last edited by greg1313; Aug 1st 2018 at 01:17 PM.
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    MHF Contributor MarkFL's Avatar
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    Re: Show 24 divides (2*7^n + 3*5^n - 5) for all positive integers n, but ...

    Quote Originally Posted by greg1313 View Post
    You had a typo for the second exponent at that moment:

    $\displaystyle \left(2\cdot7^{1}+3\cdot5^{1}-5\right)=24$
    Yes, and I can no longer edit, but you got my meaning.
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