Question

Ciphertext WHKCJILXI was encrypted by affine cipher C = A*P + 8 (mod 26), where parameter A is uknown. You have a following additional information: plaintext letter L was encrypted by letter H. Find A and decrypt the message. You can use applet for decryption.

For your convenience, below are numeric values of ENGLISH letters:

A=0 B=1 C=2 D=3 E=4 F=5 G=6 H=7 I=8 J=9 K=10 L=11

M=12 N=13 O=14 P=15 Q=16 R=17 S=18 T=19 U=20 V=21

W=22 X=23 Y=24 Z=25

Answer 1

Since L is converted to H in cipher text.
Affine cipher works as C = A*P + 8 (mod 26)

Since L = 11 and H = 7.

so The cipher eqn becomes,

7 = A*11 + 8 (mod 26)

now we have to find out the value of A, we need to use hit and try method.

A = 1 : 19 mod 26 = 19 so False.
A = 2 : 2*11 + 8 (mod 26) = 30 mod 26 = 4 so false.
A = 3 : 3*11 + 8 (mod 26) = 41 mod 26 = 15 so false.
A = 4 : 4*11 + 8 (mod 26) = 52 mod 26 = 0 so false.
A = 5 : 5*11 + 8 (mod 26) = 63 mod 26 = 11 so false.
A = 6 : 6*11 + 8 (mod 26) = 74 mod 26 = 22 so false.
A = 7 : 7*11 + 8 (mod 26) = 85 mod 26 = 7 so True.

so A = 7,
now to decifer we need to find multiplicative inverse of 7.

let B is multiplicative inverse of 7

1 = 7*B mod 26.
Again we need to use hit and trial method for value of B.

B = 1 : 7 mod 26 = 7, so false
B = 2 : 14 mod 26 = 14, so false
B = 3 : 21 mod 26 = 21, so false
B = 4 : 28 mod 26 = 2, so false
B = 5 : 35 mod 26 = 9, so false
B = 6 : 42 mod 26 = 16, so false
B = 7 : 49 mod 26 = 23, so false
B = 8 : 56 mod 26 = 4, so false
B = 9 : 63 mod 26 = 11, so false
B = 10 : 70 mod 26 = 18, so false
B = 11 : 77 mod 26 = 25, so false
B = 12 : 84 mod 26 = 6, so false
B = 13 : 91 mod 26 = 13, so false
B = 14 : 98 mod 26 = 20, so false
B = 15 : 105 mod 26 = 1, so True.

so multiplicative inverse of 7 is 15.

A=0 B=1 C=2 D=3 E=4 F=5 G=6 H=7 I=8 J=9 K=10 L=11
M=12 N=13 O=14 P=15 Q=16 R=17 S=18 T=19 U=20 V=21
W=22 X=23 Y=24 Z=25

Now we have to decifer: W H K C J I L X I
P = (B*C - 8) mod 26, B is multiplicative inverse of 7.

Decifering W:
P = (15*22 - 8) mod 26 = 10 = K

Decifering H:
P = (7*22 - 8) mod 26 = 16 = Q

Decifering K:
P = (10*22 - 8) mod 26 = 4 = E

Decifering C:
P = (2*22 - 8) mod 26 = 10 = K

Decifering J:
P = (9*22 - 8) mod 26 = 8 = I

Decifering I:
P = (8*22 - 8) mod 26= 12 = M

Decifering L:
P = (11*22 - 8) mod 26 = 0 = A

Decifering X:
P = (23*22 - 8) mod 26 = 4 = E

Decifering I:
P = (8*22 - 8) mod 26 = 12 = M

So decifered text is: KQEKIMAEM