Explanation Given information: The given cipher text is “51 14 49 15”. Formula used: Division algorithm: Let a be an integer and d be a positive integer. Then, there are unique integers q and r with 0 ≤ r < d such that a = d q + r , here, q is the quotient and r is the remainder. So, q = a div d and r = a mod d . Know that the technique is x m ( mod y ) = [ x ( mod y ) ] m ( mod y ) . RSA Cipher Work: The formula of the RSA cryptography is M = C d mod p q . To decrypt the numbers, use the formula M = C d mod p q . …… (1) Where, d = 27 and p q = 55 . To decrypt 51 , use the formula (1). M = 51 27 mod 55 Its known that, 27 as a sum of powers of 2. 27 = 2 4 + 2 3 + 2 1 + 2 0 = 16 + 8 + 2 + 1 Rewrite the statement as: M = 51 ( 16 + 8 + 2 + 1 ) mod 55 = ( 51 16 ⋅ 51 8 ⋅ 51 2 ⋅ 51 1 ) mod 55 = ( 51 16 mod 55 ⋅ 51 8 mod 55 ⋅ 51 2 mod 55 ⋅ 51 1 mod 55 ) mod 55 = ( 51 16 mod 55 ⋅ ( 51 4 ) 2 mod 55 ⋅ 2601 mod 55 ⋅ 51 ) mod 55 = ( 51 16 mod 55 ⋅ ( 51 4 mod 55 ) 2 mod 55 ⋅ 16 ⋅ 51 ) mod 55 Simplify the above equation. ( 51 16 mod 55 ⋅ ( 51 4 mod 55 ) 2 mod 55 ⋅ 16 ⋅ 51 ) mod 55 = ( 51 16 mod 55 ⋅ ( ( 51 2 ) 2 mod 55 ) 2 mod 55 ⋅ 16 ⋅ 51 ) mod 55 = ( 51 16 mod 55 ⋅ ( ( 51 2 mod 55 ) 2 mod 55 ) 2 mod 55 ⋅ 16 ⋅ 51 ) mod 55 = ( 51 16 mod 55 ⋅ ( 16 2 mod 55 ) 2 mod 55 ⋅ 16 ⋅ 51 ) mod 55 = ( ( 51 8 ) 2 mod 55 ⋅ 36 2 mod 55 ⋅ 16 ⋅ 51 ) mod 55 Again, simplify the above equation. ( ( 51 8 ) 2 mod 55 ⋅ 36 2 mod 55 ⋅ 16 ⋅ 51 ) mod 55 = ( 31 2 mod 55 ⋅ 31 ⋅ 16 ⋅ 51 ) mod 55 = ( 961 mod 55 ⋅ 31 ⋅ 16 ⋅ 51 ) mod 55 = ( 26 ⋅ 31 ⋅ 16 ⋅ 51 ) mod 55 = 657696 mod 55 = 6 Thus, 51 27 mod 55 = 6 . To decrypt 14 , use the formula (1). M = 14 27 mod 55 Its known that, 27 as a sum of powers of 2. 27 = 2 4 + 2 3 + 2 1 + 2 0 = 16 + 8 + 2 + 1 Rewrite the statement as: M = 14 ( 16 + 8 + 2 + 1 ) mod 55 = ( 14 16 ⋅ 14 8 ⋅ 14 2 ⋅ 14 1 ) mod 55 = ( 14 16 mod 55 ⋅ 14 8 mod 55 ⋅ 14 2 mod 55 ⋅ 14 1 mod 55 ) mod 55 = ( 14 16 mod 55 ⋅ ( 14 4 ) 2 mod 55 ⋅ 196 mod 55 ⋅ 14 ) mod 55 = ( 14 16 mod 55 ⋅ ( 14 4 mod 55 ) 2 mod 55 ⋅ 31 ⋅ 14 ) mod 55 Simplify the above equation. ( 14 16 mod 55 ⋅ ( 14 4 mod 55 ) 2 mod 55 ⋅ 31 ⋅ 14 ) mod 55 = ( 14 16 mod 55 ⋅ ( ( 14 2 ) 2 mod 55 ) 2 mod 55 ⋅ 31 ⋅ 14 ) mod 55 = ( 14 16 mod 55 ⋅ ( ( 14 2 mod 55 ) 2 mod 55 ) 2 mod 55 ⋅ 31 ⋅ 14 ) mod 55 = ( 14 16 mod 55 ⋅ ( 31 2 mod 55 ) 2 mod 55 ⋅ 31 ⋅ 14 ) mod 55 = ( ( 14 8 ) 2 mod 55 ⋅ 26 2 mod 55 ⋅ 31 ⋅ 14 ) mod 55 Again, simplify the above equation. ( ( 14 8 ) 2 mod 55 ⋅ 26 2 mod 55 ⋅ 31 ⋅ 14 ) mod 55 = ( 16 2 mod 55 ⋅ 16 ⋅ 31 ⋅ 14 ) mod 55 = ( 256 mod 55 ⋅ 16 ⋅ 31 ⋅ 14 ) mod 55 = ( 36 ⋅ 16 ⋅ 31 ⋅ 14 ) mod 55 = 249984 mod 55 = 9 Thus, 14 27 mod 55 = 9

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Chapter 8 Solutions

Discrete Mathematics With Applications

Ch. 8.1 - If R is a relation from A to B, xA , and yB , the...Ch. 8.1 - Prob. 2TY Ch. 8.1 - Prob. 3TY Ch. 8.1 - Prob. 4TY Ch. 8.1 - If R is a relation on a set A, the directed graph...Ch. 8.1 - As in Example 8.1.2, the congruence modulo 2...Ch. 8.1 - Prove that for all integers m and n,m-n is even...Ch. 8.1 - The congruence modulo 3 relation, T, is defined...Ch. 8.1 - Define a relation P on Z as follows: For every...Ch. 8.1 - Prob. 5ES

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