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Math Help - Congruency

  1. #1
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    Congruency

    I'm having trouble solving this problem.

    Solve the congruence 2x is equivalent to 7 (mod 17)

    Can someone work it out for me in step by step form with some explinations
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  2. #2
    o_O
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    Add the modulus to 7 :

    \begin{array}{rcll} 2x & \equiv & 7 & \left(\text{mod } 17\right) \\ 2x & \equiv & 7 {\color{red}\ + \ 17} & \left(\text{mod } 17\right) \\ 2x & \equiv & 24 & \left(\text{mod } 17\right) \\ 2x & \equiv & 2 \cdot 12 & \left(\text{mod } 17\right) \end{array}

    Since \gcd (2,17) = 1, we can cancel both sides by 2 without worrying about the modulus.
    Last edited by o_O; March 7th 2009 at 12:51 PM.
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  3. #3
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    Quote Originally Posted by o_O View Post
    \begin{array}{rcll}2x & \equiv & 24 & \left(\text{mod } 17\right) \\ 2x & \equiv & 2 \cdot 17 & \left(\text{mod } 17\right) \end{array}
    how does this follow
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  4. #4
    o_O
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    Sorry that was a typo. Fixed it now.
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  5. #5
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    Congruence ax = b (mod c) for some a, b, c in Z can be written as a linear diophantine equation ax - cy = b. And that is easy to solve using the box method.

    For this problem, 2x = 7 (mod 17) can be written as 2x - 17y = 7.
    Using the box method, we found the Bezout relation: -56.2 + 7.17 = 7
    Therefore, x = -56 = 12 (mod 17).
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