# Cipher using Mod

• April 24th 2012, 07:19 PM
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???
• April 24th 2012, 07:45 PM
Soroban
Re: Cipher using Mod

That's what I got, too . . .

• April 25th 2012, 01:53 AM
a tutor
Re: Cipher using Mod
But you should be deciphering PSRUOPVU. The whole process is quite abstract.
• April 25th 2012, 02:12 PM
Re: Cipher using Mod
Quote:

Originally Posted by a tutor
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
• April 25th 2012, 02:31 PM
a tutor
Re: Cipher using Mod
Well, I'll start you off.

A -> P
B -> S
S -> R
.
.
.
.
.
• April 26th 2012, 08:20 AM
Re: Cipher using Mod
Abstract
• April 26th 2012, 05:13 PM
Soroban
Re: Cipher using Mod

Quote:

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 . . .

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

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