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Thread: Chinese Remainder Thm/Simult. congruences: X ≡ 2 mod 3, X ≡ 2 mod 4, X ≡ 1 mod 5

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    Chinese Remainder Thm/Simult. congruences: X ≡ 2 mod 3, X ≡ 2 mod 4, X ≡ 1 mod 5

    "Find an integer X such that


    X ≡ 2 mod 3,


    X ≡ 2 mod 4, and

    X ≡ 1 mod 5."


    So this is what I did for the first two congruences:


    x = 3a + 2


    x = 4b +2


    3a + 2 = 4b +2


    Possible value for a is 1 and b is 3/4.


    Therefore x = 3(1) + 2 = 4(3/4) + 2 = 5 (mod 12).


    So x = 12c + 5


    And the 4th congruence has equation:


    x = 5d +1


    12c + 5 = 5d + 1


    5d - 12c = 4.


    Possible value of d is 4 and c is 4/3.


    12(4/3) + 5 = 5(4) + 1 = 21 (mod 60)


    But 21 is clearly wrong.


    It seems like I even got the calculation for the first two congruences wrong.


    I have seen this method work before so I'm not sure where I've gone wrong here.


    Any help?
    Last edited by sktsasus; 01-29-2018 at 09:24 PM.

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