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Math Help - Quickie #7

  1. #1
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    Quickie #7

    Show that for all integer values of x,

    . . x^9 - 6x^7 + 9x^5 - 4x^3 is divisible by 8640.

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  2. #2
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    Soroban, am I on the right track?.

    We have to show that x^{9}-6x^{7}+9x^{5}-4x^{4} is divisible by 8640=2^{6}\cdot{3^{3}}\cdot{5}

    Factor: x^{3}(x-2)(x-1)^{2}(x+1)^{2}(x+2)


    I would say because 5 is a factor of x(x-1)(x-2)(x+2)

    3 is a factor of x(x-1)(x-2), \;\ x(x-1)(x+2), \;\ x(x+1)(x+2)

    8 is a factor of x(x-2)(x-1)(x+1), \;\ x(x-1)(x+1)(x+2)
    Last edited by galactus; January 4th 2007 at 04:23 AM.
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  3. #3
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    Hello, Galactus!

    Of course, you meant: 5 is a factor of x(x-1)(x+1)(x-2)(x+2)

    Then your solution is correct.


    The "Quickie" solution goes like this:

    We know that the product of n consecutive integers is divisible by n
    . . and the product of four consecutive integers is divisible by 2^3.

    We have: . 8640 \:=\:2^6\!\cdot\!3^3\!\cdot\!5

    . . and: . N \;=\;(x-2)\ (x-1)^2\ x^3\ (x+1)^2\ (x+2)


    \text{Since }N \;=\;\underbrace{[(x-2)(x-1)x]}_{\text{3 consecutive}}\ \underbrace{[(x-1)x(x+1)]}_{\text{3 consecutive}}\ \underbrace{[x(x+1)(x+2)]}_{\text{3 consecutive}}
    . . then N is divisible by 3^3.


    \text{Since }N \;=\;\underbrace{[(x-2)(x-1)x(x+1)(x+2)]}_{\text{5 consecutive}}\ [(x-1)x^2(x+1)]
    . . then N is divisible by 5.


    \text{Since }N \;=\;x\ \underbrace{[(x-2)(x-1)x(x+1)]}_{\text{div by }2^3}\ \underbrace{[(x-1)x(x+1)(x+2)]}_{\text{div by }2^3}
    . . then N is divisible by 2^6.


    Therefore, N is divisible by 2^6\!\cdot\!3^3\!\cdot\!5\;=\;8640.

    Last edited by Soroban; January 5th 2007 at 08:10 AM.
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