Q8.Paper3,d12. Using RSA algorithm, what is the value of cipher text C, if the plain text M = 5 and p = 3, q = 11 & d = 7 ?

(A) 33

(B) 5

(C) 25

(D) 26

Explanation.

Answer(D)

• Choose p = 3 and q = 11
• Compute n = p * q = 3 * 11 = 33
• Compute omega (n) = (p - 1) * (q - 1) = 2 * 10 = 20
• Choose e such that 1 < e < omega(n) and e and n are co prime. Given D=7
• Compute a value for e such that (d * e) % omega(n) = 1. One solution is e = 3  because[(e * 7) % 20 = 1]  
• Public key is (e, n) => (3, 33)
• Private key is (d, n) => (7, 33) 
• • The encryption of m = 5 is c = 5³ % 33 = (125%33)=26.
  The decryption of c = 26 is m = 26⁷ % 33 =8031810176. -8031810171=5 the original message, we are right