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Thread: Tricky Problem

  1. #1
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    Tricky Problem

    Ok so you are given; a, b, c are integers

    abc = 729
    and a + b + c = 91

    Find a^2 + b^2 + c^2

    How could I do this algebraically?
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  2. #2
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    Hint: 1+9+81=91
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  3. #3
    Super Member Bacterius's Avatar
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    Reply

    Say $\displaystyle c = 1$. Therefore, we have :

    $\displaystyle ab = 729$
    $\displaystyle a + b = 90$

    Substitute :

    $\displaystyle a(90 - a) = 729$
    $\displaystyle 90a - a^2 = 729$
    $\displaystyle a^2 - 90a + 729 = 0$

    Solve for $\displaystyle a$.

    Facts for the lazy :
    Spoiler:
    You sure ?
    Spoiler:

    $\displaystyle \triangle = 90^2 - 2916 = 5184$

    $\displaystyle a = \frac{\ {90 - \sqrt{\triangle}}}{2} = 9$

    $\displaystyle b = \frac{\ {90 + \sqrt{\triangle}}}{2} = 81$

    And for the curious :
    Spoiler:
    Since you would get a polynomial with the same coefficients if you had substituted $\displaystyle b$ instead of $\displaystyle a$, the two solutions of the quadratic equation are $\displaystyle a$ and $\displaystyle b$.
    Oh, and by the way, you just factored 729 into smaller factors


    Finding the sum of the squares shouldn't be too hard now (remember $\displaystyle c = 1$)
    Last edited by Bacterius; Nov 6th 2009 at 06:31 PM.
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