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Thread: 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.

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

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

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

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

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

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

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

    But plugging back x and y into the congruences, for example: $\displaystyle (85)-3(-4)\not\equiv 9 \bmod{12}$
    Last edited by Pn0yS0ld13r; Jul 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, $\displaystyle 13y \equiv -4 \bmod{12}$.
    Since $\displaystyle 13 \equiv 1 \bmod 12$ and $\displaystyle -4 \equiv 8 \bmod 12$, $\displaystyle y=8$. Plugging y in any of the congruences yields $\displaystyle x = 9$

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