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Math Help - Discrete Log Problem

  1. #1
    Newbie cryptocrow's Avatar
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    Discrete Log Problem

    Good morning, I had trouble with this:

    3^6 = 44 (mod 137)
    3^10 = 2 (mod 137)

    Find x 0 <= x <= 135 such that 3^x = 11 (mod 137)

    Thank you...
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  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by cryptocrow View Post
    Good morning, I had trouble with this:

    3^6 = 44 (mod 137)
    3^10 = 2 (mod 137)

    Find x 0 <= x <= 135 such that 3^x = 11 (mod 137)

    Thank you...
     3^{10} = 2 \text{ \, mod \, 137}

     3^{20} = 4  \text{ \, mod \, 137}

     3^{-20} = 4^{-1} \text{ \, mod\, 137}

     3^6 = 44 \text{ \, mod \, 137} \Rightarrow  (3^6)(3^{-20}) = (44)(4^{-1}) \text{ \, mod\, 137}

    3^{-14} = 11 \text{ \,mod \, 137}

    By Fermats little theorem 3^{136} = 1 \text{ \, mod \, 137}

    3^{122} = 11 \text{ \, mod\, 137}

    So x = 122
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  3. #3
    Newbie cryptocrow's Avatar
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    Ah, I see where I went wrong... thanks for your help!
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