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Question 2
We use the provided information that
• TH encrypted gives GW and
• HE encrypted gives R3.
We also use the decryption function P =a*C + b (mod 37*37), where P is the plaintext digraph and C is the ciphertext digraph.
P =a*C +b (mod 1369)
TH =37*29 + 17 = 1090
GW = 37* 16 + 32 = 624...

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