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Thread: a question......

  1. #1
    Junior Member
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    a question......

    $\displaystyle a.b-c=4$

    $\displaystyle b.c-a=3$

    in equality, a, b, c are positive integers.

    What is total of $\displaystyle a+b+c $ minumum?
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  2. #2
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    $\displaystyle a.b-c=4$

    $\displaystyle b.c-a=3$

    add the two equations together:

    $\displaystyle ab+bc-a-c=7$

    move things around:

    $\displaystyle ab-a+bc-c=7$

    factor:

    $\displaystyle a(b-1)+c(b-1)=7$

    factor again:

    $\displaystyle (a+c)(b-1)=7$

    hmm, 7 is a prime number though and so cannot have two integer factors, unless $\displaystyle b-1 = 1$ and $\displaystyle a+c=7$ or $\displaystyle b-1=7$ and $\displaystyle a+c=1$ (which means that in this case either a or c must be zero, I don't know if you consider that a positive integer)

    In the first case: $\displaystyle b = 2$ and $\displaystyle a+c=7$

    So if we add those two together: $\displaystyle a+b+c=9$


    In the second case: $\displaystyle b = 8$ and $\displaystyle a+c=1$

    So if we add those two together: $\displaystyle a+b+c=9$

    So the answer is: $\displaystyle \boxed{a+b+c=9}$
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