1. Suppose pseudo-random numbers are produced by using: xn+1 = (3xn + 11) mod 13. If x3=5, find x2 and x4.2. Suppose pseudo-random numbers are produced by using: xn+1 = (2xn + 7) mod 9.a) If x0 = 1, find x2 and x3 b) If x3 = 3, find x2 and x4.3. Using the function f(x) = (x + 10) mod 26 to encrypt messages. Answer each of thesequestions.a) Encrypt the message STOP b) Decrypt the message LEI4. Which memory locations are assigned by the hashing function h(k) = k mod 101 to therecords of insurance company customers with these Social Security Numbers?a) 104578690 b) 4322221875. Use the Euclidean algorithm to finda) gcd(14, 28) b) gcd(8, 28) c) gcd(100, 101) d) gcd(28,35)e) lcm(7, 28) f) lcm(12, 28) g) lcm(100, 101) h) lcm(28,35)

a) To find x2 when x3=5, we need to solve the equation 5 = (3x2 + 11) mod 13. However, this equation has multiple solutions and without additional information, we cannot find a unique value for x2. b) To find x4, we substitute x3=5 into the equation to get x4 = (3*5 + 11) mod 13 = 4.

a) To find x2 and x3 when x0=1, we first find x1 = (21 + 7) mod 9 = 9 mod 9 = 0. Then, x2 = (20 + 7) mod 9 = 7 and x3 = (27 + 7) mod 9 = 2. b) To find x2 and x4 when x3=3, we need to solve the equation 3 = (2x2 + 7) mod 9 for x2. Again, this equation has multiple solutions and without additional information, we cannot find a unique value for x2. To find x4, we substitute x3=3 into the equation to get x4 = (23 + 7) mod 9 = 13 mod 9 = 4.

a) To encrypt the message STOP using the function f(x) = (x + 10) mod 26, we convert each letter to its corresponding number (S=19, T=20, O=15, P=16), apply the function, and convert back to letters. The encrypted message is "ABYZ". b) To decrypt the message LEI, we convert each letter to its corresponding number (L=12, E=5, I=9), apply the inverse function f^-1(x) = (x - 10) mod 26, and convert back to letters. The decrypted message is "CAB".

The memory locations assigned by the hashing function h(k) = k mod 101 to the records of insurance company customers with these Social Security Numbers are: a) 104578690 mod 101 = 87 b) 432222187 mod 101 = 80

a) gcd(14, 28) = 14 b) gcd(8, 28) = 4 c) gcd(100, 101) = 1 d) gcd(28,35) = 7 e) lcm(7, 28) = 28 f) lcm(12, 28) = 84 g) lcm(100, 101) = 10100 h) lcm(28,35) = 140