Explanation Given information: The pubic key e = 23 , m = 55 Cipher text message: 26 25 00 39 09 18 52 17 49 52 02 Formula used: 1) RSA Public Key Cryptosystem: Mapping f : A → A is defined by f ( x ) = x e mod m , then f has inverse mapping g : A → A , such that, g ( x ) = x d mod m . Suppose, A = { 0 , 1 , 2 , . . . , m − 1 } is an alphabet, consisting m “letters,” with m , p , q , e , and d where m = p q , p and q are two distinct prime numbers, e , such that, it should be relatively prime with ( p − 1 ) ( q − 1 ) , that is e defined by ( e , ( p − 1 ) ( q − 1 ) ) , d is defined by 1 = e d mod ( p − 1 ) ( q − 1 ) . In which the public keys are e and m , and the secret keys are p , q , and d . 2) The letter alphabet A - If alphabet consists of a to z in natural order followed by a blank, then have 27 “letters” that associates with the integer 0 , 1 , 2 , 3 , . . . , 26 as follows: Alphabet: a b c d e f . . . v w x y z " b l a n k " A : 0 1 2 3 4 5 . . . 21 22 23 24 25 26 3) Euclidean Algorithm: Let a and b are integers with b > 0 , then the procedure a = b q 0 + r 1 0 ≤ r 1 < b b = r 1 q 1 + r 2 0 ≤ r 2 < r 1 r 1 = r 2 q 2 + r 3 0 ≤ r 3 < r 2 ⋅ ⋅ ⋅ r k = r k + 1 q k + 1 + r k + 2 0 ≤ r k + 2 < r k + 1 is known as Euclidean Algorithm. Since the integers r 1 , r 2 , … , r k + 2 are decreasing and are all non-negative, there is a smallest integer n , such that r n + 1 = 0 : r n − 1 = r n q n + r n + 1 0 = r n + 1 If r 0 = b , then the last non-zero remainder r n is always greatest common divisor of a and b . 4) Theorem: Substitution Properties: Suppose, a ≡ b ( mod n ) and c ≡ d ( mod n ) . Then, a + c ≡ b + d ( mod n ) and a c ≡ b d ( mod n ) . Calculation: The pubic key m = 55 to find two distinct prime p and q , such that m = p q Choose p = 11 , q = 5 , where m = p q = 11 ⋅ 5 = 55 . To find d , d is defined by 1 = e d mod ( p − 1 ) ( q − 1 ) . Consider ( p − 1 ) ( q − 1 ) = ( 11 − 1 ) ( 5 − 1 ) = 10 ⋅ 4 = 40 and e = 23 . 1 = e d mod ( p − 1 ) ( q − 1 ) ⇒ 1 = 23 d mod 40 . . . ( 1 ) . To use Euclidean algorithm, to find d . 40 = 23 ( 1 ) + 17 23 = 17 + 6 17 = 6 ( 2 ) + 5 6 = 5 + 1 Thus, 1 = 6 − 5 = 6 − ( 17 − 6 ( 2 ) ) = 6 − 17 + 6 ( 2 ) = 6 ( 3 ) − 17 = ( 23 − 17 ) ( 3 ) − 17 = 23 ( 3 ) − 17 ( 3 ) − 17 = 23 ( 3 ) − 17 ( 4 ) = 23 ( 3 ) − ( 40 − 23 ) ( 4 ) = 23 ( 3 ) − 40 ( 4 ) + 23 ( 4 ) = 23 ( 7 ) − 40 ( 4 ) Thus, 1 = 23 ( 7 ) − 40 ( 4 ) ⇒ 1 = 23 ( 7 ) mod 40 Thus, from (1), d = 7

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ISBN: 9781285463230

Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Chapter 2.8, Problem 21E

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The message that was sent in RSA Public Key Cryptosystem, when the public key is e = 23, m = 55 and the cipher text message is

26 25 00 39 09 18 52 17 49 52 02 and use 27− letter alphabet.

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Chapter 2.8, Problem 20E

Chapter 2.8, Problem 22E

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The message that was sent in RSA Public Key Cryptosystem, when the public key is e = 23 , m = 55 and the cipher text message is 26 25 00 39 09 18 52 17 49 52 02 and use 27 − letter alphabet.

Solution Summary: The author explains the cipher message in RSA Public Key Cryptosystem.

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The message that was sent in RSA Public Key Cryptosystem, when the public key is e = 23, m = 55 and the cipher text message is

26 25 00 39 09 18 52 17 49 52 02 and use 27− letter alphabet.

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Chapter 2 Solutions

The message that was sent in RSA Public Key Cryptosystem, when the public key is e = 23 , m = 55 and the cipher text message is 26 25 00 39 09 18 52 17 49 52 02 and use 27 − letter alphabet.

Solution Summary: The author explains the cipher message in RSA Public Key Cryptosystem.

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The message that was sent in RSA Public Key Cryptosystem, when the public key is e = 23, m = 55 and the cipher text message is 26 25 00 39 09 18 52 17 49 52 02 and use 27− letter alphabet.

Want to see the full answer?

Chapter 2 Solutions

The message that was sent in RSA Public Key Cryptosystem, when the public key is e = 23, m = 55 and the cipher text message is

26 25 00 39 09 18 52 17 49 52 02 and use 27− letter alphabet.