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Math Help - congruency questions

  1. #1
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    congruency questions

    Find all integers x with 0\leq x\leq621which are the solutions of 255x\equiv12(mod621)<br />

    I got the gcd(255,621) to be 3.

    And my linear congruency was 3=621*23+255*(-56)

    But i am not sure on how solve the for all values.
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  2. #2
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    Hello cooltowns
    Quote Originally Posted by cooltowns View Post
    Find all integers x with 0\leq x\leq621which are the solutions of 255x\equiv12(mod621)<br />

    I got the gcd(255,621) to be 3.

    And my linear congruency was 3=621*23+255*(-56)

    But i am not sure on how solve the for all values.
    OK, so you can now say:
    255\times(-56) = 621\times(-23)+3

    \Rightarrow 255\times(-56) \equiv3\mod621

    \Rightarrow 255\times(-224) \equiv12\mod621
    We now note that 255=85\times3 and 621=207\times3.
    \Rightarrow 85\times3(-224+207n)\equiv12\mod621, n = 1, 2, 3, ...
    So the general solution of
    255x\equiv12\mod621
    is
    x=-224+207n
    And in particular, taking n = 2, 3, 4, we get the solutions for 0\le x< 621:
    x = 190, 397, 604
    Grandad
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