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$y \equiv 7x\pmod{26}$ to encrypt the plaintext message “WATCH OUT”.

$A=0$

$Z=25$

I would assume that I would need to solve the $y\equiv 7x \pmod{26}$ equation firstly, before I would then be able to encrypt the "watch out" material. At present, I am not sure how to do this.

N. F. Taussig
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Jack White
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  • Did you mean $A = 0$ and $Z = 25$ or $A = 1$ and $Z = 26$? – N. F. Taussig Mar 30 '19 at 19:25
  • A=0 and Z=25. This has thrown me, because I calculated the first value (W) as being Y.

    I did by 7(x) with x being the plain text value. W=22, ergo, 7(22) is 154,

    154/26=5.9 (drop the fraction) 526=130 526 remainder of 24 So....answer is....24

    Making it Y.

    However, I am stuck with A. Because A is 0, this suggests that the value is 26, but 26 is not present, as Z=25

     – Jack White Mar 30 '19 at 20:06
  • You are correct that $W$ is mapped to $Y$. Since $7 \cdot 0 = 0 = 0 \cdot 26 + 0$, $A$ is mapped to itself. – N. F. Taussig Mar 31 '19 at 00:42

1 Answers1

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If the clear-text digit is $x$ and the encrypted digit is $y\equiv 7x\mod{26}$, then for example $E=4$ maps to $y = 7\cdot 4 = 28\equiv 2\mod{26}$, so the encrypted value is C.

If you mean for the clear-text value to be $y$ and the encrypted value for $x$, then you must first solve $y\equiv 7x\mod{26}$ for $x$: $x = 7^{-1}\cdot y\mod{26}$, and now you must find the multiplicative inverse of $7$ modulo $26$.

rogerl
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