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Math Help - finite series

  1. #1
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    finite series

    Ahoy all late nighters. looking for a some clarity on a problem I have. Please go easy. Im sure to screw something up when writing this.

    Ok so heres the problem as I know it. It is an algorithm with the following behavior:
    {t}_{0} = 0 \\ {t}_{1} = {t}_{0}*17 + {x}_{0} \\ {t}_{2} = {t}_{1}*17 + {x}_{1} \\ {t}_{3} = {t}_{2}*17 + {x}_{2} \\ \vdots \\ {t}_{n} = {t}_{n-1}*17 + {x}_{n-1} \\

    which I believe is the following series

    {t}_{n} = 17^{n-1}{x}_{0} + 17^{n-2}{x}_{1} + \hdots + 17^{1}{x}_{n-2} + 17^{0}{x}_{n-1}

    Now all the {x}_{n}'s are unknown and are an integer in the range 1-26
    Given a value for {t}_{n} can a unique sum be found? I suspect there will be multiple solutions. Could someone help me simplify this?

    If it helps, assume {t}_{n} = 22640
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by sixstringartist View Post
    Ahoy all late nighters. looking for a some clarity on a problem I have. Please go easy. Im sure to screw something up when writing this.

    Ok so heres the problem as I know it. It is an algorithm with the following behavior:
    {t}_{0} = 0 \\ {t}_{1} = {t}_{0}*17 + {x}_{0} \\ {t}_{2} = {t}_{1}*17 + {x}_{1} \\ {t}_{3} = {t}_{2}*17 + {x}_{2} \\ \vdots \\ {t}_{n} = {t}_{n-1}*17 + {x}_{n-1} \\

    which I believe is the following series

    {t}_{n} = 17^{n-1}{x}_{0} + 17^{n-2}{x}_{1} + \hdots + 17^{1}{x}_{n-2} + 17^{0}{x}_{n-1}

    Now all the {x}_{n}'s are unknown and are an integer in the range 1-26
    Given a value for {t}_{n} can a unique sum be found? I suspect there will be multiple solutions. Could someone help me simplify this?

    If it helps, assume {t}_{n} = 22640
    Is...

    t_{n}= 17^{n-1}\ x_{0} + 17^{n-3}\ x_{1} + ... + 17\ x_{n-2} + x_{n-1} = 22640 (1)

    Now is...

    \text {int} (\frac{22640}{17})= 1331 \implies x_{n-1}= 22640-1331*17=13

    \text {int} (\frac{1331}{17})= 78 \implies x_{n-2}= 1331-78*17=5

    \text {int} (\frac{78}{17})= 4 \implies x_{n-3}= 78-4*17=10

    \text {int} (\frac{4}{17})= 0 \implies x_{n-4}= 4-0*17=4

    ... so that is n=4 and...

    t_{n}= 17^{3}*4 + 17^{2}*10 + 17*5 + 13 = 22640 (2)

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor chisigma's Avatar
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    The solution we found in the last post requires all the x_{n} in the range 1-16... if the range is 1-26, then another possible value of x_{2} is 22 and that produces the solution x_{1}=9 and x_{0}=4 so that is...

    t_{n}= 17^{3}*4 + 17^{2}*9+17*22+13= 22640

    ... and from that it derives the other solution x_{1}=26 and x_{0}=3 so that is...

    t_{n}= 17^{3}*3 + 17^{2}*26+17*22+13= 22640

    No more solution exists...

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 21st 2011 at 10:28 PM. Reason: another solution found...
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  4. #4
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    you are doing a modulus operation no?

    I understand now. Thanks for the help
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