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Thread: Congrences

  1. #1
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    Congrences

    Could someone please explain to me how to do this question:

    3x=5(mod 7)
    The answer is 4 but I'm not sure how to get there.

    Help would be appreciated.
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    Hello,

    Quote Originally Posted by free_to_fly View Post
    Could someone please explain to me how to do this question:

    3x=5(mod 7)
    The answer is 4 but I'm not sure how to get there.

    Help would be appreciated.
    You have to perform a division :

    $\displaystyle x=5 \cdot 3^{-1} \bmod 7$

    But you have to calculate $\displaystyle a=3^{-1}$, which is a such that $\displaystyle 3a=1 \bmod 7$

    One can see that $\displaystyle 15=3*5=1 \bmod 7$

    So $\displaystyle a=5 \bmod 7$


    Therefore :

    $\displaystyle 3x=5 \bmod 7 \Longleftrightarrow \overbrace{3a}^{1}x=5a \bmod 7 $

    $\displaystyle \implies x=5a \bmod 7 \implies x=25 \bmod 7 \implies x=4 \bmod 7 $
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    Ok, I think I more or less follow the previous explanation, but for this question I'm a little confused still:
    Q: 6x=2 (mod 8)

    $\displaystyle x=2 \cdot 6^{-1} \bmod 8 $

    but this cancels to:
    $\displaystyle
    x=1 \cdot 3^{-1} \bmod 8$ (not sure if I can cancel like that)
    So$\displaystyle
    a=3^{-1}
    $
    $\displaystyle
    3a=1 \bmod 8$
    But 3*3=9=1 (mod 8) so
    $\displaystyle
    a=3 \bmod 8
    $
    This gives x=3 (mod 8)
    I know the answer is suppose to be x=3(mod 8) please can you show me where I went wrong.
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  4. #4
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    Yes, I forgot to mention it. There is an inverse if and only if they are coprime.

    What I mean is that a has an inverse modulo n, if and only if a and n are coprime.

    Why ?

    Here, $\displaystyle 6x=2 \bmod 8$

    Going back to the definition of the congruence, we have : $\displaystyle 6x-2=8k$, where $\displaystyle k \in \mathbb{Z}$.

    That is to say $\displaystyle 3x-1=4k$.
    So this is the same as $\displaystyle 3x=1 \bmod 4$.

    When there is a common factor in $\displaystyle a=b \bmod n$, divide a,b, and n by this factor.

    ---------------------------
    In a general case, how to find an inverse ?

    - observe : for example, what number is divisible by 3 and such that it is 1 added to a multiple of 4 ? Answer is 9. It's easy by trial and error when it's small number.

    - use the euclidian algorithm... I can give you links where I've done such things, or you can read that : Extended Euclidean algorithm - Wikipedia, the free encyclopedia

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