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Math Help - divisibility of numbers

  1. #1
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    divisibility of numbers

    Find all integers of the form 1 ≤ a < b < c, s.t.:
    1* ∃k a+b = kc
    2* ∃l a+c = lb
    3* ∃m b+c = ma

    Attempt: Well, I have answer that solution is of the form a, 2a, 3a. I can prove that if b = 2a, then c = 3a:
    From 2* we have c = a(2l - 1)
    Then from 1* and previous we have 3a = ka(2l - 1). So k = 1 or k = 3:
    If k = 3, then
    1 = 2l - 1, l = 1, a +c = 1.b = 2a what contradicts assumption that a < c.
    So k = 1. Then a + b = 3a = 1.c

    But I can't prove that b=2a. Any hints? Thanks!
    Last edited by jozou; August 15th 2011 at 09:45 AM.
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  2. #2
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    Re: divisibility of numbers

    Quote Originally Posted by jozou View Post
    Find all integers of the form 1 ≤ a < b < c, s.t.:
    1* ∃k a+b = kc
    2* ∃l a+c = lb
    3* ∃m b+c = ma

    Attempt: Well, I have answer that solution is of the form a, 2a, 3a. I can prove that if b = 2a, then c = 3a:
    From 2* we have c = a(2l - 1)
    Then from 1* and previous we have 3a = ka(2l - 1). So k = 1 or k = 3:
    If k = 3, then
    1 = 2l - 1, l = 1, a +c = 1.b = 2a what contradicts assumption that a < c.
    So k = 1. Then a + b = 3a = 1.c

    But I can't prove that b=2a. Any hints? Thanks!
    We can find that b=2a and c=3a without using divisibility properties, only with inequalities.

    None of k,l,m is 0, since a,b,c are positive.
    Spoiler:
    So the first inequality I've found is from 1*: kc=a+b<c+c=2c implying that k<2 and therefore k=1. Then a+b=c.

    We use the above equation to rewrite equation 2* in two ways replacing a or c: lb=a+c=2c-b>2b-b=b, then l>1; lb=a+c=2a+b<2b+b=3b, then l<3. From 1<l<3 it follows that l=2.

    Finallly, if you add equations 1* and 2* we find that 2a+b+c=kc+lb=c+2b, which gives b=2a. Substituting b in a+b=c we find that c=3a.


    Edit: I didn't even use equation 3*!
    Last edited by melese; August 15th 2011 at 01:48 PM.
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  3. #3
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    Re: divisibility of numbers

    Thanks melese, I like it, that is very elegant proof!
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