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Posers and Puzzles

Posers and Puzzles

  1. 04 May '05 06:59
    Find all pairs of positive integers x and y that safisfy the equation:

    1/x - 1/y = 1/2005
  2. Standard member PBE6
    Bananarama
    04 May '05 15:57
    OK, here's what I got:

    x = 1600; y = 8020
    x = 1980; y = 158796
    x = 2000; y = 802000

    x = 2004; y = 0 is another pair, but 0 is not a positive integer.

    Sadly, I solved this by trial and error. :'( If anyone knows how to do this analytically, I'd love to find out how.
  3. Donation Acolyte
    Now With Added BA
    04 May '05 17:34
    Originally posted by PBE6
    OK, here's what I got:

    x = 1600; y = 8020
    x = 1980; y = 158796
    x = 2000; y = 802000

    x = 2004; y = 0 is another pair, but 0 is not a positive integer.

    Sadly, I solved this by trial and error. :'( If anyone knows how to do this analytically, I'd love to find out how.
    Anayltically? I don't think there's a magic method of doing this, but it's easy to show rigorously that you've found all the solutions: all you need to do is remark that for all solutions in positive integers,
    1/x < 1/2005, so x < 2005,
    and find y for each x in turn, confirming that the solutions you listed below are the only ones in positive integers.

    1/2004 -1/0 is -infinity though, so that's not a solution at all.
  4. Standard member PBE6
    Bananarama
    04 May '05 17:52
    Originally posted by Acolyte
    Anayltically? I don't think there's a magic method of doing this, but it's easy to show rigorously that you've found all the solutions: all you need to do is remark that for all solutions in positive integers,
    1/x < 1/2005, so x < 2005,
    and find y for each x in turn, confirming that the solutions you listed below are the only ones in positive integers.

    1/2004 -1/0 is -infinity though, so that's not a solution at all.
    Oops, my mistake (with the 0).
  5. 04 May '05 19:17 / 2 edits
    Originally posted by PBE6
    OK, here's what I got:

    x = 1600; y = 8020
    x = 1980; y = 158796
    x = 2000; y = 802000

    x = 2004; y = 0 is another pair, but 0 is not a positive integer.

    Sadly, I solved this by trial and error. :'( If anyone knows how to do this analytically, I'd love to find out how.
    I think your 1600 should be 1604.

    x must be < 2005; so we need to find those values of x such that y = 2005x/(2005 - x) is also an integer.

    Here's what I get for (x,y), and I think this is all such pairs:

    (1604, 8020)
    (1980, 158796)
    (2000, 802000)
    (2004, 4018020)

    EDIT: I don't see any magical analytical solution either, although the problem can be whittled down quite a bit.
  6. Standard member PBE6
    Bananarama
    04 May '05 20:04
    Originally posted by davegage
    I think your 1600 should be 1604.

    x must be < 2005; so we need to find those values of x such that y = 2005x/(2005 - x) is also an integer.

    Here's what I get for (x,y), and I think this is all such pairs:

    (1604, 8020)
    (1980, 158796)
    (2000, 802000)
    (2004, 4018020)

    EDIT: I don't see any magical analytical solution either, although the problem can be whittled down quite a bit.
    Oops again. I think my blood sugar is low. I need a cookie.

    Acolyte, bring me your head again.
  7. 04 May '05 22:29
    Originally posted by PBE6
    Oops again. I think my blood sugar is low. I need a cookie.

    Acolyte, bring me your head again.
    Let a,b be relatively prime integers , and x=ka and y=kb, b>a

    then k=c(b-a) where abc=2005 will give all the solutions:

    x=(b-a)ca, y=(b-a)cb.

    When c=1 there are two sets of factors to use for ab;
    when c=5 or 401 there is just one set of factors for ab.
    c cannot be 2005 as x=y not possible.

    There are just these four since the only factors of 2005
    are 1, 5, 401, and 2005.

    [ 2004(1) , 2004(2005) ] , [ 396(5) , 396(401) ]

    [400(5)(1) , 400(5)(401) ] , [ 4(401)(1) , 4(401)(5) ]