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Thread: 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; Aug 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 $\displaystyle b=2a$ and $\displaystyle c=3a$ without using divisibility properties, only with inequalities.

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

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

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


    Edit: I didn't even use equation 3*!
    Last edited by melese; Aug 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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