Results 1 to 10 of 10

Math Help - Little number theory based maths problem

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    179
    Thanks
    1

    Little number theory based maths problem

    Hi!

    I have chosen a positive integer. On division by 14 my integer gives remainder 6. On division by 15 my integer gives remainder 2. On division by 13 my integer gives remainder 12. What is the smallest possible positive integer I could have chosen?

    Good luck!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Go through my method in this post.

    Edit: Unless you meant this as a challenge problem of course, in which case this thread is under the wrong sub-forum.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2009
    Posts
    179
    Thanks
    1
    Thanks alexmahone.... i can follow that to a point, but then in your example, you are given an equation whereby the remainder is 0, so then you are able make the equation where d=0 and get your answer.... i am not given this, so how do i get around it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by sirellwood View Post
    Thanks alexmahone.... i can follow that to a point, but then in your example, you are given an equation whereby the remainder is 0, so then you are able make the equation where d=0 and get your answer.... i am not given this, so how do i get around it?
    How about posting some working?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2009
    Posts
    179
    Thanks
    1
    Quote Originally Posted by alexmahone View Post
    How about posting some working?
    x = 6(mod14)
    x=14a +6

    x=2(mod15)
    14a+6 = 2(mod15)
    14a=-4(mod15)
    14a=11(mod15)
    14a=15b+11

    combining my equations.... x=15b+17

    15b+17=12(mod13)
    15b=-5(mod13)
    3b=-1(mod13)
    3b=13c-1

    x=15b+17
    x=5(13c-1)+17
    x=65c+12

    .....from here i am stuck, i seem to be going roundin circles?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by sirellwood View Post
    combining my equations.... x=15b+17
    This merely implies that x=2 (mod 15). But you already knew that! Which means that you haven't used the previous part of the problem.

    14a=-4 (mod 15)
    7a=-2 (mod 15)
    a=4 (mod 15)
    a=15b+4

    x=14a+6=14(15b+4)+6=210b+62

    Can you proceed?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Mar 2009
    Posts
    179
    Thanks
    1
    210b+62 = 12mod13
    210b=-50mod13
    21b=-5mod13
    b=-25mod13
    b=1mod13
    b=13c+1

    x=210(13c+1)+62
    x=2730c +272

    now i know at this point, we are meant to imply that c=0 and infer that x=272.... But i dont fully understand why we can say clearly c=0?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by sirellwood View Post
    But i dont fully understand why we can say clearly c=0?
    Because the problem asks for the smallest possible positive integer.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Mar 2009
    Posts
    179
    Thanks
    1
    oh ok, thank you, basically i was confused as to why it was only at this point that we put c=0, rather than say take b=0 in x= 210b+62.... Then i realised that because at that point, not all parts of the question had been considered, i.e. x=12(mod13) hadnt been processed.... this makes sense now :-)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Dec 2009
    Posts
    1
    oinw
    Last edited by hmmm; May 4th 2011 at 11:30 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. LINUX based free Maths software
    Posted in the Math Software Forum
    Replies: 3
    Last Post: September 9th 2009, 11:19 PM
  2. Maths Based Economics
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: March 31st 2009, 03:08 AM
  3. Number Theory problem
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: December 30th 2008, 12:07 PM
  4. Replies: 2
    Last Post: December 18th 2008, 05:28 PM
  5. 12 Maths based Physics problems
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: April 22nd 2007, 05:00 AM

Search Tags


/mathhelpforum @mathhelpforum