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Math Help - Modular arithmetic

  1. #1
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    Modular arithmetic

    How do I solve:



    tr^(-a) ≡ m (mod p)


    where:


    t = 6
    r = 7
    a = 6
    p = 17



    6*7^(-6) ≡ 0.000050999 (mod 17)


    This obviously isn't right?


    Edit: Sorry, I typed t = 7, it should have been t = 6, now corrected, and the -a should be ^(-a) due to a copy-paste which I neglected to check if copied itself correctly, apparently it did not, sigh.
    Last edited by posix_memalign; November 18th 2008 at 04:43 PM.
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  2. #2
    o_O
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    \begin{array}{rcll} {\color{red}t}{\color{blue}r} - {\color{magenta}a} & \equiv & m & (\text{mod } {\color [cmyk] {0,.35,1,0}p}) \\ {\color{red}(7)}{\color{blue}(7)} - {\color{magenta}(6)} & \equiv & m & (\text{mod } {\color [cmyk] {0,.35,1,0}17}) \\ 43 & \equiv & m & (\text{mod }17)  \end{array}

    You accidentally plugged in 6 instead of 7 for t.

    Now that last line is essentially asking: "What is the remainder after 43 is divided by 17?"
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  3. #3
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    Quote Originally Posted by o_O View Post
    \begin{array}{rcll} {\color{red}t}{\color{blue}r} - {\color{magenta}a} & \equiv & m & (\text{mod } {\color [cmyk] {0,.35,1,0}p}) \\ {\color{red}(7)}{\color{blue}(7)} - {\color{magenta}(6)} & \equiv & m & (\text{mod } {\color [cmyk] {0,.35,1,0}17}) \\ 43 & \equiv & m & (\text{mod }17)  \end{array}

    You accidentally plugged in 6 instead of 7 for t.

    Now that last line is essentially asking: "What is the remainder after 43 is divided by 17?"
    Thanks but I'm sorry I copy pasted the problem from Openoffice where I was working on it and thus some formatting was changed, I also managed to enter a typo from the part I didn't copy and paste -- incredible -- however, now it should be correct.
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  4. #4
    o_O
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    \begin{array}{rcll} tr^{-a} & \equiv & m & (\text{mod } p) \\ (6)(7)^{-6} & \equiv & m & (\text{mod } 17) \\ 6 \cdot \displaystyle  \frac{1}{7^6} & \equiv & m & (\text{mod } 17) \\ 6 & \equiv &  7^6m & (\text{mod } 17) \end{array}

    One can show that: 7^6 \equiv 9 \ (\text{mod } 17)...and... 6 \equiv -45 \ (\text{mod }17)

    So that last line becomes: -45 \equiv 9m \ (\text{mod } 17)
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