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Math Help - Cipher using Mod

  1. #1
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    Cipher using Mod

    I am trying to solve a cipher using f(x) = 3x+15 (mod 26). Going in order with A=0 all the way to Z=25. So the encoding for the letter A was A -> 0 -> 3*0+15 (mod 26) = 15 -> P.

    so im getting A=15=P, B=18=S, C=21=V, D=24=Y...

    i was trying to cipher ' PSRUOPVU '

    and im getting: IROXFIAX.

    Is this right???
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  2. #2
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    Re: Cipher using Mod

    Hello, Aquameatwad!

    That's what I got, too . . .

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  3. #3
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    Re: Cipher using Mod

    But you should be deciphering PSRUOPVU. The whole process is quite abstract.
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    Re: Cipher using Mod

    Quote Originally Posted by a tutor View Post
    But you should be deciphering PSRUOPVU. The whole process is quite abstract.
    yes i meant i was trying to decipher. and its from an abstract algebra class
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  5. #5
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    Re: Cipher using Mod

    Well, I'll start you off.

    A -> P
    B -> S
    S -> R
    .
    .
    .
    .
    .
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  6. #6
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    Re: Cipher using Mod

    Abstract
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  7. #7
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    Re: Cipher using Mod

    Hello, Aquameatwad!

    I am trying to solve a cipher using: f(x) \:=\: 3x+15 \text{ (mod 26)}.
    Going in order with A=0 all the way to Z=25.

    So the encoding for the letter A is:
    . . A\;=\;0\quad\Rightarrow\quad3\cdot0+15\text{ (mod 26)}\quad\Rightarrow\quad 15\;-\;P

    So im getting : . A=15=P,\;B=18=S,\;C=21=V,\;D=24=Y,\;\hdots

    i was trying to decipher PSRUOPVU.

    We need a decipering table . . .

    . . \begin{array}{cc}\text{Code} & \text{Clear} \\ \hline A & V \\ B & E \\ C & N \\ D & W \\ E & F \\ F & O \\ G & X \\ H & G \\ I & P \\ J & Y \\ K & H \\ L & Q \\ M & Z \end{array} \qquad \begin{array}{cc}\text{Code} & \text{Clear} \\ \hline N & I \\ O & R \\ P & A \\ Q & J \\ R & S \\ S & B \\ T & K \\ U & T \\ V & C \\ W & L \\ X & U \\ Y & D \\ Z & M \end{array}


    \text{Therefore: }\;\boxed{\begin{array}{cccccccc}P&S&R&U&O&P&V&U \\ \hline A&B&S&T&R&A&C&T \end{array}}

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