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Math Help - Chinese Remainder Theorem

  1. #1
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    Chinese Remainder Theorem

    2x 1 (mod 7)
    5x 2 (mod 11)

    I know how to do the CRT, but the fact that this prob. has coefficients in front of the x is confusing me. How do I go about solving this?
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  2. #2
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    We can rewrite them as:

    2x\equiv 1\equiv 8 (mod \;\ 7)\Rightarrow \boxed{x\equiv 4(mod \;\ 7)}

    5x\equiv 2\equiv 13\equiv 24\equiv 35(mod 11)\Rightarrow \boxed{x\equiv 7(mod \;\ 11)}

    From these we can write:

    7a+4=11b+7...[1]

    Appyling modulo 7 to 7a+4=11b+7 and we get:

    b\equiv 1(mod \;\ 7)

    b=7c+1

    Sub into the right side of [1]:

    11(7c+1)+7=77c+18

    The solutions to the system are:

    x\equiv 18(mod \;\ 77)

    Which can be written as \boxed{x=77t+18}

    Let t=0,1,2,...... and they will satisfy the system of linear congruences.
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  3. #3
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    I see...thanks for your help!
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