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Math Help - Find x,y,z

  1. #1
    Member great_math's Avatar
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    Find x,y,z

    Find natural numbers x,y,z such that

    \frac1{x}+\frac1{y}+\frac1{z}=\frac1{5}

    where x<y<z
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  2. #2
    Super Member
    Joined
    May 2009
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    x = 10, y = 15, z = 30

    \begin{aligned}<br />
\frac{1}{10} + \frac{1}{15} + \frac{1}{30} &= \frac{1}{5} \\<br />
\frac{3}{30} + \frac{2}{30} + \frac{1}{30} &= \frac{1}{5} \\<br />
\frac{6}{30} &= \frac{1}{5}<br />
\end{aligned}

    I used guess-and-check, though...

    Here's another solution:
    x = 8, y = 16, z = 80

    \begin{aligned}<br />
\frac{1}{8} + \frac{1}{16} + \frac{1}{80} &= \frac{1}{5} \\<br />
\frac{10}{80} + \frac{5}{80} + \frac{1}{80} &= \frac{1}{5} \\<br />
\frac{16}{80} &= \frac{1}{5}<br />
\end{aligned}

    Here's a third solution (again, using guess-and-check):
    x = 10, y = 12, z = 60

    \begin{aligned}<br />
\frac{1}{10} + \frac{1}{12} + \frac{1}{60} &= \frac{1}{5} \\<br />
\frac{6}{60} + \frac{5}{60} + \frac{1}{60} &= \frac{1}{5} \\<br />
\frac{12}{60} &= \frac{1}{5}<br />
\end{aligned}

    ...and a fourth (this is fun ):
    x = 12, y = 15, z = 20

    \begin{aligned}<br />
\frac{1}{12} + \frac{1}{15} + \frac{1}{20} &= \frac{1}{5} \\<br />
\frac{5}{60} + \frac{4}{60} + \frac{3}{60} &= \frac{1}{5} \\<br />
\frac{12}{60} &= \frac{1}{5}<br />
\end{aligned}


    01
    Last edited by yeongil; June 23rd 2009 at 06:01 AM. Reason: Consolidating posts
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  3. #3
    Super Member
    Joined
    May 2009
    Posts
    527
    Solution #5 (I'm on a roll ):
    x = 8, y = 20, z = 40

    \begin{aligned}<br />
\frac{1}{8} + \frac{1}{20} + \frac{1}{40} &= \frac{1}{5} \\<br />
\frac{10}{80} + \frac{4}{80} + \frac{2}{80} &= \frac{1}{5} \\<br />
\frac{16}{80} &= \frac{1}{5}<br />
\end{aligned}

    Solution #6:
    x = 9, y = 15, z = 45

    \begin{aligned}<br />
\frac{1}{9} + \frac{1}{15} + \frac{1}{45} &= \frac{1}{5} \\<br />
\frac{5}{45} + \frac{3}{45} + \frac{1}{45} &= \frac{1}{5} \\<br />
\frac{9}{45} &= \frac{1}{5}<br />
\end{aligned}

    Solution #7:
    x = 10, y = 14, z = 35

    \begin{aligned}<br />
\frac{1}{10} + \frac{1}{14} + \frac{1}{35} &= \frac{1}{5} \\<br />
\frac{7}{70} + \frac{5}{70} + \frac{2}{70} &= \frac{1}{5} \\<br />
\frac{14}{70} &= \frac{1}{5}<br />
\end{aligned}

    Solution #8 (this is getting ridiculous):
    x = 6, y = 40, z = 120

    \begin{aligned}<br />
\frac{1}{6} + \frac{1}{40} + \frac{1}{120} &= \frac{1}{5} \\<br />
\frac{20}{120} + \frac{3}{120} + \frac{1}{120} &= \frac{1}{5} \\<br />
\frac{24}{120} &= \frac{1}{5}<br />
\end{aligned}


    01
    Last edited by yeongil; June 23rd 2009 at 06:22 AM. Reason: Consolidating posts; adding more solutions
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