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Math Help - Fraction Conundrum

  1. #1
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    Fraction Conundrum

    How to solve this Question?
    :

    1/z+ 1/y+ 1/x + 1/w + 1/v + 1/t + 1/s = 1

    1/z means 1 over z a fraction.
    The letters can be any number.
    How to find seven numbers that fit into the equation?

    (the numerator is always 1)

    What if its 8 sets or 9 sets of fraction? is there any technique?
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by hotgal24 View Post
    How to solve this Question?
    :

    1/z+ 1/y+ 1/x + 1/w + 1/v + 1/t + 1/s = 1

    1/z means 1 over z a fraction.
    The letters can be any number.
    How to find seven numbers that fit into the equation?

    (the numerator is always 1)

    What if its 8 sets or 9 sets of fraction? is there any technique?
    In my opinion the letters can be just about any numbers, unless there is something special about them. Such as a ratio between them. Is this actually a real problem or did you make it up?
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  3. #3
    Senior Member Peritus's Avatar
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    In my opinion the letters can be just about any numbers, unless there is something special about them. Such as a ratio between them. Is this actually a real problem or did you make it up?
    janvdl is right, in this specific case you have only one constraint and seven unknowns which means that you have 6 D.O.F.'s --> you can choose six of the variables to be whatever you want and the last variable is determined by the equation at hand.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by hotgal24 View Post
    How to solve this Question?
    :

    1/z+ 1/y+ 1/x + 1/w + 1/v + 1/t + 1/s = 1

    1/z means 1 over z a fraction.
    The letters can be any number.
    How to find seven numbers that fit into the equation?

    (the numerator is always 1)

    What if its 8 sets or 9 sets of fraction? is there any technique?
    How about
    s = t = v = w = x = y = z = 7?

    -Dan
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  5. #5
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    Hello, hotgal24!

    What is the exact wording of the problem?


    Solve: . \frac{1}{s} + \frac{1}{t} + \frac{1}{v} + \frac{1}{w} + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \:=\:1

    The letters can be any number. . . . . Really?

    Then we can have: . \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}\;=\;1 . . . duh!


    Or how about:

    . . \left(\frac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}\right) + \left(\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6}\right)

    . . \frac{1}{2} + \left(\dfrac{1}{8}+\dfrac{1}{8}\right) + \left(\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}\ri  ght)

    . . \left(\frac{1}{5\frac{1}{3}} + \frac{1}{5\frac{1}{3}} + \frac{1}{5\frac{1}{3}} + \frac{1}{5\frac{1}{3}}\right) + \frac{1}{9\frac{1}{7}} + \frac{1}{7\frac{1}{9}}


    I would assume the denominators must be distinct positive integers.
    . . If so, I have a solution for it.

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  6. #6
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    The different letters must be different positive integer number lowest possible.
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