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Math Help - Finding the triplets

  1. #1
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    Finding the triplets

    Find the all arranged triplets ,from positive real numbers ( a,b,c)

    that: (a)bc=3,a(b)c=4,ab(c)=5,which (x) is the greatest integer less than or equal to x
    Last edited by Mhmh96; June 21st 2012 at 02:12 PM.
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    Re: Finding the triplets

    Label the original equations (1), (2), and (3) in that order.

    First, note that a,b,c \ge 1 (otherwise \lfloor a \rfloor = 0, etc., contradiction).

    Multiplying all three equations, we get a^2b^2c^2 = \frac{60}{\lfloor a \rfloor \lfloor b \rfloor \lfloor c \rfloor} \Rightarrow abc = \frac{\sqrt{60}}{\sqrt{\lfloor a \rfloor \lfloor b \rfloor \lfloor c \rfloor}}.

    We can prove that a,b,c < 3. Assume that \lfloor b \rfloor = 3. In order to satisfy (1), b must equal 3, otherwise c would have to be less than 1. If b = 3, \lfloor a \rfloor = c = 1. Solving for a via (2) and (3), we obtain a = \frac{4}{3} and a = \frac{5}{3}, no solution. Similarly, if \lfloor c \rfloor = 3, we get the same contradiction. If \lfloor a \rfloor = 3, b = c = 1, no solutions.

    Therefore, \lfloor a \rfloor, \lfloor b \rfloor, \lfloor c \rfloor \in {1,2}. It follows that \lfloor a \rfloor \lfloor b \rfloor \lfloor c \rfloor \in {1,2,4}. I'll let you do the casework (I obtained at least two solutions btw).

    (There should be {} with 1,2 and 1,2,4, but LaTeX interprets the brackets differently, and idk how to input them otherwise).
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    Re: Finding the triplets

    Quote Originally Posted by richard1234 View Post
    (There should be {} with 1,2 and 1,2,4, but LaTeX interprets the brackets differently, and idk how to input them otherwise).
    Nice solution. You can get braces with \{ and \}: \lfloor a\rfloor\lfloor b\rfloor\lfloor c\rfloor\in\{1,2,4\}
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    Re: Finding the triplets

    Quote Originally Posted by Reckoner View Post
    Nice solution. You can get braces with \{ and \}: \lfloor a\rfloor\lfloor b\rfloor\lfloor c\rfloor\in\{1,2,4\}
    Ah, thanks.
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