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Math Help - find x,y,z

  1. #1
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    find x,y,z

    x.y .z
    integers such that

    \frac{24}{5}=x+\frac{1}{y +\frac{1}{z+1}}
    Last edited by CaptainBlack; June 25th 2007 at 04:11 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by perash View Post
    x.y .z
    integers such that

    \frac{24}{5}=x+\frac{1}{y +\frac{1}{z+1}}
    First put x=4, does that help?

    RonL
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    First put x=4, does that help?

    RonL
    Why "4" Captain?
    Could you please explain a bit, i'd like to know how to do this too...
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  4. #4
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    Quote Originally Posted by janvdl View Post
    Why "4" Captain?
    Could you please explain a bit, i'd like to know how to do this too...
    Because,
    \frac{25}{4} = 6 + \frac{1}{4}

    I think it is a mistake.
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Because,
    \frac{25}{4} = 6 + \frac{1}{4}

    I think it is a mistake.
    I dont agree, i dont think we can assume just 1 set of numbers...
    I might be wrong, but i dont think there can be only one solution...
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by janvdl View Post
    Why "4" Captain?
    Could you please explain a bit, i'd like to know how to do this too...
    Because if:

    <br />
\frac{24}{5}=x+\frac{1}{y +\frac{1}{z+1}}<br />

    and x, y, z>0 then x = \lfloor 24/5\rfloor, and I have a solution in my notebook for x=4.

    RonL
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by perash View Post
    x.y .z
    integers such that

    \frac{24}{5}=x+\frac{1}{y +\frac{1}{z+1}}
    CB's hint was not a typo. Let's try x = 4:
    \frac{24}{5} - 4 = \frac{1}{y + \frac{1}{z + 1}}

    \frac{5}{4} = y + \frac{1}{z + 1}

    1 + \frac{1}{4} = y + \frac{1}{z + 1}

    Thus y = 1 and z = 3.

    Quite clever really. Applying the same logic that gave CaptainBlack x = 4, we can also do x = 5. If there are any more solutions out there I can't find them with this scheme in mind, anyway.

    -Dan
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  8. #8
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    Hello, perash!

    \text{Find integers }x,\,y,\,z\text{ such that: }\;\frac{24}{5}\;=\;x+\frac{1}{y +\frac{1}{z+1}}
    This is what CaptainBlack suggested . . .


    Since \frac{24}{5} \:=\:4 + \frac{4}{5}, . let x = 4

    Then: . \frac{1}{y + \frac{1}{z+1}} \:=\:\frac{4}{5}\quad\Rightarrow\quad y + \frac{1}{z+1} \:=\:\frac{5}{4}\quad\Rightarrow\quad y + \frac{1}{z+1}\:=\:1 + \frac{1}{4}

    Then it's obvious that: . y = 1,\:z = 3

    Got it?

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