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Math Help - Quickie #14

  1. #1
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    Quickie #14

    This is not really a "Quickie".
    The solution is a bit longer, but it's quite clever.


    In base A, the fraction F_1 = 0.373737\hdots and the fraction F_2 = 0.737373\hdots

    In base B, the fraction F_1 = 0.252525\hdots and the fraction F_2 = 0525252\hdots

    Find A,\,B,\,F_1,\,F_2.

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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    This is not really a "Quickie".
    The solution is a bit longer, but it's quite clever.


    In base A, the fraction F_1 = 0.373737\hdots and the fraction F_2 = 0.737373\hdots

    In base B, the fraction F_1 = 0.252525\hdots and the fraction F_2 = 0525252\hdots

    Find A,\,B,\,F_1,\,F_2.
    Observe that in any base that contains digits 1 through 7:

    F_1=A F_2 - 7
    F_1=B F_2 - 5

    F_2=A F_1 - 3
    F_2=B F_1 - 2

    and solve.

    The solution is F_1=1/3, F_2=2/3, A=11, B=8

    RonL
    Last edited by CaptainBlack; January 20th 2007 at 01:30 AM.
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  3. #3
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    Quote Originally Posted by Soroban View Post
    This is not really a "Quickie".
    The solution is a bit longer, but it's quite clever.


    In base A, the fraction F_1 = 0.373737\hdots and the fraction F_2 = 0.737373\hdots

    In base B, the fraction F_1 = 0.252525\hdots and the fraction F_2 = 0525252\hdots

    Find A,\,B,\,F_1,\,F_2.

    Hello Soroban,

    if you multiply a fraction in "decimal" notation by the base you move the "decimal" point one digit to the right. (It's a more general rule to the SHR-command in assembler).

    If you multiply F_2 by A then the digits after the "decimal" point are the same as at F_1. If you subtract F_1 from A*F_2 then you get an integer. If you multiply F_1 by A then the digits after the "decimal" point are the same as at F_2. If you subtract F_2 from A*F_1 then you get an integer.
    This method will give 4 linear equations:

    \begin {array}{c} A \cdot F_2-F_1=7  \\ B \cdot F_2-F_1=5  \\ A \cdot F_1-F_2= 3 \\ B \cdot F_1-F_2=2  \end {array}

    I've got: A = 11, B= 8, F_1=1/3, F_3=2/3 (this time base 10)

    EB

    PS: As usual CaptainBlack is much faster than I
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  4. #4
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    Hello, Captain Black and earboth!

    I thought my book had a clever solution.
    . . But yours are much shorter and more elegant.

    I still want to show everyone the book's solution.
    . . It has an interesting (unexpected) approach.


    F_1 \;=\;\frac{3A+7}{A^2} + \frac{3A+7}{A^4} + \frac{3A+7}{A^6} + \hdots\;=\;\frac{3A + 7}{A^2-1}

    F_2\;=\;\frac{2B+5}{B^2} + \frac{2B+5}{B^4} + \frac{2B+5}{B^6} + \hdots\;=\;\frac{2B + 5}{B^2-1}

    . . Hence: . F_1 \;=\;\frac{3A+7}{A^2-1}\;=\;\frac{2B+5}{B^2-1} [1]


    Similarly: . F_2\;=\;\frac{7A + 3}{A^2-1} \;=\;\frac{5B + 2}{B^2-1} [2]


    Add [1] and [2]: . F_1 + F_2 \;=\;\frac{3A + 7}{A^2-1} + \frac{7A + 3}{A^2-1} \;=\;\frac{2B+5}{B^2-1} + \frac{5B + 2}{B^2-1}

    . . \frac{10A + 10}{A^2-1} \;=\;\frac{7B+7}{B^2-1}\quad\Rightarrow\quad\frac{10(A + 1)}{(A - 1)(A + 1)} \;=\;\frac{7(B+1)}{(B-1)(B+1)}

    . . \frac{10}{A-1} \:=\:\frac{7}{B-1}\quad\Rightarrow\quad7A - 10B\:=\:-3 [3]


    Subtract [1] from [2]: . F_2 - F_1\;=\;\frac{7A + 3}{A^2 - 1} - \frac{3A + 7}{A^2 - 1} \;=\;\frac{5B + 2}{B^2-1} - \frac{2B+5}{B^2-1}

    . . \frac{4A - 4}{A^2-1} \;=\;\frac{3B - 3}{B^2-1}\quad\Rightarrow\quad\frac{4(A-1)}{(A-1)(A + 1)} \;=\;\frac{3(B-1)}{(B-1)(B+1)}

    . . \frac{4}{A + 1} \;=\;\frac{3}{B + 1}\quad\Rightarrow\quad 3A - 4B \:=\:1 [4]


    [3] and [4] give us a system of equations: . \begin{array}{cc}7A - 10B & = \:-3 \\ 3A - 4B & = \; 1 \end{array}

    . . with solutions: . A = 11,\;B = 8

    Then: . F_1 = \frac{1}{3},\;F_2 = \frac{2}{3}


    I agree . . . This solution is much longer.
    But there are some interesting techniques involved
    . . which I've added to my arsenal.

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