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Math Help - solve in Z

  1. #1
    Super Member dhiab's Avatar
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    solve in Z

    solve in Z :
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Note that 8 is a solution to both equations.

    11*9=99

    8+99 is also a solution. (Why?)
    Similarly 8+2*99 is also a solution.

    Conclusion: 8+99k is the general solution, where k is any integer.
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  3. #3
    Super Member dhiab's Avatar
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    please : more of détailes on resolution
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Which part of the solution didn't you understand?
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  5. #5
    Super Member dhiab's Avatar
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    Quote Originally Posted by alexmahone View Post
    Which part of the solution didn't you understand?
    resolution génerale
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  6. #6
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    The conclusion is valid, the argument less so. If once cites the Chinese Remainder theorem, that argument would be sufficient

    Alternatively:

    x \equiv -3 \pmod{11} \Rightarrow x = 11k-3

    Therefore,

    11k-3 \equiv -1 \pmod{9} \Rightarrow 11k \equiv 2 \pmod{9} \Rightarrow k \equiv 2(11)^{-1} \pmod{9}
    \Rightarrow k \equiv 1 \pmod{9} \Rightarrow k = 9l + 1 \Rightarrow x = 99l+8
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  7. #7
    Super Member dhiab's Avatar
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    Offered resolution

    OFFERED RESOLUTION
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  8. #8
    Super Member dhiab's Avatar
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    Quote Originally Posted by SimonM View Post
    The conclusion is valid, the argument less so. If once cites the Chinese Remainder theorem, that argument would be sufficient

    Alternatively:

    x \equiv -3 \pmod{11} \Rightarrow x = 11k-3

    Therefore,

    11k-3 \equiv -1 \pmod{9} \Rightarrow 11k \equiv 2 \pmod{9} \Rightarrow k \equiv 2(11)^{-1} \pmod{9}
    \Rightarrow k \equiv 1 \pmod{9} \Rightarrow k = 9l + 1 \Rightarrow x = 99l+8
    HELLO : k differe of k ' thank you
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