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Math Help - Inverse mod problem

  1. #1
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    Inverse mod problem

    I know I must be the stupidest person alive but to save my life I can not figure out:

    inverse of 5 mod 7!!!! I know it is 3 but I CAN NOT figure out the way to get this and I really want to pull all my hair out!!! If someone can help I would really appreciate it!
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Frostking View Post
    I know I must be the stupidest person alive but to save my life I can not figure out:

    inverse of 5 mod 7!!!! I know it is 3 but I CAN NOT figure out the way to get this and I really want to pull all my hair out!!! If someone can help I would really appreciate it!
    You're looking for d such that 5d=1 mod 7

    For this, use the Euclidian algorithm :

    {\color{red}7}={\color{blue}5}*1+{\color{magenta}2  } -> {\color{magenta}2}={\color{red}7}-{\color{blue}5}

    {\color{blue}5}={\color{magenta}2}*2+1

    -> 1={\color{blue}5}-{\color{magenta}2}*2={\color{blue}5}-2({\color{red}7}-{\color{blue}5})=3*{\color{blue}5}-2*{\color{red}7}

    Hence 3*5=1+2*7=1 \mod 7

    d=3
    Last edited by Moo; May 20th 2008 at 01:06 PM. Reason: fixed colours
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  3. #3
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    Quote Originally Posted by Frostking View Post
    I know I must be the stupidest person alive but to save my life I can not figure out:

    inverse of 5 mod 7!!!! I know it is 3 but I CAN NOT figure out the way to get this and I really want to pull all my hair out!!! If someone can help I would really appreciate it!

    You can use the Euclidean algorithm

    7=1(5)+2 \iff 2=7-1(5)
    5=2(2)+1 \iff 1=5-2(2)

    Now subbing the first equation into the 2nd for 2 gives

    1=5-2(7-1(5)) \iff 1=3(5)-2(7)

    The last shows that 3(5)=1 \mod 7

    I hope this helps
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