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Math Help - Congruence

  1. #1
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    Congruence

    Hi, Masters!

    Can I have some questions?

    1) Why is it that (the inverse of 3)^3 will be -33 * 3 * 11^2 * 3 = 121 * 3 = 21 * 3 = 63 mod 100?

    I assume 3*33 = 99 so it is -1 mod 100. So -33 came from here. But I don't know where 3* 11^2 * 3 come from..

    2) y^31 = 3 mod 100.

    I just memorised it should be y = 3^31^-1 mod 100. And 31^-1 is -9, so the equation should be 3^-9 mod 100.
    How can I calculate 3^-9 mod 100...?

    I guess I am poor at calculating the inverse.
    Please help me out masters.

    It is really difficult to get any information about higher congruence like y^32...
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  2. #2
    MHF Contributor

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    Re: Congruence

    Well if 3*33 = -1 (mod 100), then it stands to reason that:

    3*(-33) = 1 (mod 100), that is, the inverse of 3 (mod 100) is -33 = 67.

    If we cube this, we get:

    (67)3 = (-33)3 = (-33)(33)2 = (-33)(32)(112) = (-33)(9)(121) = (-33)(3)(3)(121) = (3)(121) = (3)(21) = 63 (mod 100).

    Your second question is harder to answer.

    Assuming the answer IS 3-9 mod 100, we calculate:

    3-9 = (3-1)9 = 679 = [(67)2]4(67) = (89)4(67) = (-11)4(67) = (11)4(67) = (21)2(67) = (41)(67) = 47.
    Thanks from yanirose
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