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Math Help - Exponential cipher

  1. #1
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    Exponential cipher

    I tried to work through this ex in my book which is not typical of the ex I have done and I am confused after first part
    for part a I worked out that inverse of 5 in Z36 is 29. hope this is correct
    but I have no clue as to where to and how to start part b - which is attached here. Can someone please guide me in detail as I am studying this for the first type.
    Last edited by vidhi96; August 11th 2011 at 12:28 AM. Reason: answer no longer needed
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  2. #2
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    Re: Exponential cipher

    Quote Originally Posted by vidhi96 View Post
    I tried to work through this ex in my book which is not typical of the ex I have done and I am confused after first part
    for part a I worked out that inverse of 5 in Z36 is 29. hope this is correct
    This is correct : 5\times 29 = 145 = 4\times 36 + 1

    Quote Originally Posted by vidhi96 View Post
    but I have no clue as to where to and how to start part b - which is attached here. Can someone please guide me in detail as I am studying this for the first type.
    The table gives you a correspondence C : \{A,\dots,Z\} \rightarrow \{1,\dots,26\}. So for a letter \alpha, the cipher letter is C^{-1}(E(C(\alpha))) (I didn't check, but in order to apply C^{-1}, one must have m^5 (\mathrm{mod}\, 37) \in \{1,\dots,26\},\ \forall m\in\{1,\dots,26\}).

    When you have a word, it's a sequence of letter : you encrypt every letter of this sequence and have this way a sequence of cipher letter. It's your cipher text.
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  3. #3
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    Re: Exponential cipher

    i am more confused. What I figured out by now is message ext is <23 8 5 14> spelling WHEN.

    so now I need to find c - which is chipher text is that correct




    Quote Originally Posted by pece View Post
    This is correct : 5\times 29 = 145 = 4\times 36 + 1



    The table gives you a correspondence C : \{A,\dots,Z\} \rightarrow \{1,\dots,26\}. So for a letter \alpha, the cipher letter is C^{-1}(E(C(\alpha))) (I didn't check, but in order to apply C^{-1}, one must have m^5 (\mathrm{mod}\, 37) \in \{1,\dots,26\},\ \forall m\in\{1,\dots,26\}).

    When you have a word, it's a sequence of letter : you encrypt every letter of this sequence and have this way a sequence of cipher letter. It's your cipher text.
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  4. #4
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    Re: Exponential cipher

    Quote Originally Posted by vidhi96 View Post
    i am more confused. What I figured out by now is message ext is <23 8 5 14> spelling WHEN.
    Now you have to apply E to each of these numbers and then retranslate the result into letters.
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  5. #5
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    Re: Exponential cipher

    Quote Originally Posted by pece View Post
    Now you have to apply E to each of these numbers and then retranslate the result into letters.
    Ok I'll try and type it and send it to you tomorrow
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