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Math Help - Linear Congruence Equation, what am I doing wrong?

  1. #1
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    Linear Congruence Equation, what am I doing wrong?

    What Am I doing wrong?

    Find x and y which solve both congruence equations simultaneously.

    x-3y\equiv 9 \bmod{12}
    x+10y\equiv 5 \bmod{12}

    cx+ey\equiv a \bmod{n}
    dx+fy\equiv b \bmod{n}

    So a=9, b=5, c=1, d=1, e=-3, f=10, n=12.

    There is number z such that (cf-de)z\equiv 1 \bmod{n}

    So 13z \equiv 1 \bmod{12}, thus z=1 (right?)

    x \equiv z(af-be) \bmod{n}
    y \equiv z(bc-ad) \bmod{n}

    So x \equiv 1(85) \bmod{12} and y \equiv 1(-4) \bmod{12}

    But plugging back x and y into the congruences, for example: (85)-3(-4)\not\equiv 9 \bmod{12}
    Last edited by Pn0yS0ld13r; July 21st 2008 at 01:29 AM. Reason: Typo, as pointed out by ThePerfectHacker
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  2. #2
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    Nevermind, I answered my own question.

    Subtracting the congruences, 13y \equiv -4 \bmod{12}.
    Since 13 \equiv 1 \bmod 12 and -4 \equiv 8 \bmod 12, y=8. Plugging y in any of the congruences yields x = 9

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  3. #3
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    Quote Originally Posted by Pn0yS0ld13r View Post
    x-3y\equiv 9 \bmod{12}
    x+10\equiv 5 \bmod{12}
    I think you mean,
    x-3y \equiv 9 ~ (12)
    x+10y\equiv 5 ~ (12)
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