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Thread: Modulo120

  1. #1
    Super Member dhiab's Avatar
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    Modulo120

    Prove that : $\displaystyle \forall n\in N:9n^{5}-5n^{3}-4n\equiv 0(mod120$)
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  2. #2
    Senior Member roninpro's Avatar
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    Note that $\displaystyle 120=8\cdot 3\cdot 5$ and $\displaystyle 9n^5-5n^3-4n=(n-1)n(n+1)(4+9n^2)$. It suffices to show that $\displaystyle 8, 3, \text{ and } 5$ divide $\displaystyle (n-1)n(n+1)(4+9n^2)$ for any $\displaystyle n$.

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  3. #3
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    $\displaystyle 9n^5 - 5n^3 - 4n = 9(n^5 - 5n^3 + 4n) - 5n^3 + 45n^3 - 4n - 36n $


    $\displaystyle = 9n(n^2-4)(n^2-1) + 40n^3 - 40n $

    $\displaystyle = 9(n-2)(n-1)(n)(n+1)(n+2) + 40n(n^2-1)$

    We know the product of five consective integers is divisible by $\displaystyle 5! = 120 $ so we only consider $\displaystyle 40n(n^2-1) $ .

    We need to show that $\displaystyle n(n^2-1)$ is the multiple of $\displaystyle 3 $

    $\displaystyle n(n^2-1) = n(n-1)(n+1) = (n-1)(n)(n+1) $ which is the multiple of $\displaystyle 3 $ .

    Therefore , we prove it .
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