Results 1 to 8 of 8

Math Help - solve in Z

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    547
    Thanks
    1

    solve in Z

    solve in Z :
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,078
    Thanks
    7
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    547
    Thanks
    1
    please : more of détailes on resolution
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,078
    Thanks
    7
    Which part of the solution didn't you understand?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    547
    Thanks
    1
    Quote Originally Posted by alexmahone View Post
    Which part of the solution didn't you understand?
    resolution génerale
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Oct 2008
    From
    Guernsey
    Posts
    69
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    547
    Thanks
    1

    Offered resolution

    OFFERED RESOLUTION
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    547
    Thanks
    1
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 10:29 PM
  2. how do i solve this IVP: y'=y^2 -4
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 24th 2010, 11:14 AM
  3. Replies: 1
    Last Post: June 9th 2009, 10:37 PM
  4. how do i solve this?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 2nd 2008, 02:58 PM
  5. how to solve ..
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 2nd 2008, 08:17 AM

Search Tags


/mathhelpforum @mathhelpforum