Consider a Diffie-Hellman scheme with a common prime q = 11 and a primitive root a = 2Do not use calculator! If user A has public key Ya = 9 what is As private key Xa Ya = a^Xa mod q9 = 2^Xa mod 11whe

• If user B has public key Yb = 7, what is B's private key Xb? Yb = a^Xb mod q 7 = 2^Xb mod 11 when Xb = 0, 2^0 mod 11 = 1 ¹ 7 when Xb = 1, 2^1 mod 11 = 2 ¹ 7 when Xb = 2, 2^2 mod 11 = 4 ¹ 7 when Xb = 3, 2^3 mod 11 = 8 ¹ 7 when Xb = 4, 2^4 mod 11 = 5 ¹ 7 when Xb = 5, 2^5 mod 11 = 10 ¹ 7 when Xb = 6, 2^6 mod 11 = 9 ¹ 7 when Xb = 7, 2^7 mod 11 = 7 = 7. ß found it! So Xb = 7