Results 1 to 3 of 3

Math Help - Find all 4-tuples of distinct positive integers

  1. #1
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Find all 4-tuples of distinct positive integers

    Find all 4-tuples (a, b, c, d) of distinct positive integers so that a < b < c < d and 1/a + 1/b + 1/c + 1/d = 1.

    My attempt:

    If a = 1, 1/b + 1/c + 1/d = 0, which is impossible.

    If a >= 3, b >= 4, c >= 5, d >= 6 implies that 1/a + 1/b + 1/c + 1/d < 1/3 + 1/4 + 1/5 + 1/6 < (20 + 15 + 12 + 10)/60 = 57/60 < 1.

    So a = 2, 1/b + 1/c + 1/d = 1/2 and b >= 3.

    How do I proceed?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Find all 4-tuples of distinct positive integers

    How you solved that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Find all 4-tuples of distinct positive integers

    Quote Originally Posted by Also sprach Zarathustra View Post
    How you solved that?
    If a = 1, 1/b + 1/c + 1/d = 0, which is impossible.
    If a >= 3, b >= 4, c >= 5, d >= 6 implies that 1/a + 1/b + 1/c + 1/d <= 1/3 + 1/4 + 1/5 + 1/6 = (20 + 15 + 12 + 10)/60 = 57/60 < 1.
    So a = 2, 1/b + 1/c + 1/d = 1/2 and b >= 3.

    If b >= 6, c >= 7, d >= 8, 1/b + 1/c + 1/d <= 1/6 + 1/7 + 1/8 = (28 + 24 + 21)/168 = 73/168 < 1/2 - Contradiction!
    So b = 3, 4 or 5.

    Case 1: If b = 3, 1/c + 1/d = 1/2 - 1/3 = 1/6, c >= 4

    If c <= 6, 1/d = 1/6 - 1/c <= 0, which is impossible.
    If c >= 11, d >= 12, 1/c + 1/d <= 1/11 + 1/12 = 23/242 - Contradiction!
    So c = 7, 8, 9 or 10.

    If c = 7,
    1/7 + 1/d = 1/6
    1/d = 1/42
    d = 42
    If c = 8,
    1/8 + 1/d = 1/6
    1/d = (4 - 3) / 24
    d = 24
    If c = 9,
    1/9 + 1/d = 1/6
    1/d = (3 - 2)/18
    d = 18
    If c = 10,
    1/10 + 1/d = 1/6
    1/d = (5 - 3)/30
    d = 30/2 = 15

    Case 2: If b = 4, 1/c + 1/d = 1/4, c >= 5.
    If c >= 8, d >= 9, 1/c + 1/d <= 1/8 + 1/9 = 17/72 - Contradiction!
    So c = 5, 6 or 7.

    If c = 5,
    1/5 + 1/d = 1/4
    1/d = 1/20
    d = 20
    If c = 6,
    1/6 + 1/d = 1/4
    1/d = 1/12
    d = 12
    If c = 7,
    1/7 + 1/d = 1/4
    1/d = 3/28
    d = 28/3, which is not a positive integer.

    Case 3: If b = 5, 1/c + 1/d = 1/2 - 1/5 = 3/10, c >= 6.
    If c >= 7, d >= 8, 1/c + 1/d <= 1/7+1/8 = 15/56 < 3/10 - Contradiction!
    So c = 6.
    1/6 + 1/d = 3/10
    1/d = (9 - 5)/30 = 4/30 = 2/15
    d = 15/2, which is not a positive integer.

    Thus, there are 6 solutions: (2, 3, 7, 42), (2, 3, 8, 24), (2, 3, 9, 18), (2, 3, 10, 15), (2, 4, 5, 20) and (2, 4, 6, 12).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 3rd 2011, 03:40 AM
  2. find triples of positive integers.
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: September 17th 2009, 09:02 AM
  3. Find positive integers
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: April 22nd 2009, 02:37 PM
  4. Find all positive integers such that...?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 20th 2008, 11:20 AM
  5. Replies: 2
    Last Post: July 5th 2005, 06:20 PM

/mathhelpforum @mathhelpforum