Explanation Given information: The given cipher text 675 089 089 048. Concept used: Inverse of a modulo n : For all integers a and n , if gcd ( a , n ) = 1 , then there exists an integer s such that a s ≡ 1 ( mod n ) . The integer s is called the inverse of a modulo n . Calculation: Consider the cipher text, 675 089 089 048 The objective is to decrypt the given cipher text and find the original message. The decryption key is d = 307 , since 307 is the inverse of 43 . Public key is n = 713 = 23 ⋅ 31 and e = 43 . Thus, to decrypt the message, compute the following: 675 307 mod 713 , 89 307 mod 713 , 89 307 mod 713 , 48 307 mod 713 Convert 307 to a sum of powers of 2. 307 = 256 + 32 + 16 + 2 + 1 = 2 8 + 2 5 + 2 4 + 2 1 + 1 Perform 675 2 k for k = 1 , 2 , 3 , 4 . 675 ≡ 675 ( mod 713 ) 675 2 ≡ 455625 ( mod 713 ) since 675 2 = 455625 = 18 mod 713 since 455625 mod 713 = 18 675 4 ≡ ( 675 2 ) 2 mod 713 ≡ 18 2 mod 713 since 675 2 mod 713 = 18 = 324 mod 713 since 18 2 = 324 675 8 ≡ ( 675 4 ) 2 mod 713 ≡ ( 324 ) 2 mod 713 since 675 4 mod 713 = 324 ≡ 104976 mod 713 since 324 2 = 104976 ≡ 165 mod 713 since 104976 mod 713 = 165 675 16 ≡ ( 675 8 ) 2 mod 713 ≡ ( 165 ) 2 mod 713 since 675 8 mod 713 = 165 ≡ 27225 mod 713 since 165 2 = 27225 ≡ 131 mod 713 since 27225 mod 713 = 131 Perform 675 2 k for k = 5 , 6 , 7 , 8 675 32 ≡ ( 675 16 ) 2 mod 713 ≡ ( 131 ) 2 mod 713 since 675 16 mod 713 = 131 ≡ 17161 mod 713 since 131 2 = 17161 ≡ 49 mod 713 since 17161 mod 713 = 49 675 64 ≡ ( 675 32 ) 2 mod 713 ≡ ( 49 ) 2 mod 713 since 675 32 mod 713 = 49 ≡ 2401 mod 713 since 49 2 = 2401 ≡ 262 mod 713 since 2401 mod 713 = 262 675 128 ≡ ( 675 64 ) 2 mod 713 ≡ ( 262 ) 2 mod 713 since 675 64 mod 713 = 262 ≡ 68644 mod 713 since 262 2 = 68644 ≡ 196 mod 713 since 68644 mod 713 = 196 675 256 ≡ ( 675 128 ) 2 mod 713 ≡ ( 196 ) 2 mod 713 since 675 128 = 196 mod 173 ≡ 38416 mod 713 since 196 2 = 38416 ≡ 627 mod 713 since 38416 mod 713 = 627 Use the fact that 675 307 mod 173 = ( 675 256 + 32 + 16 + 8 + 2 + 1 ) mod 173 = ( 675 256 ⋅ 675 32 ⋅ 675 16 ⋅ 675 2 ⋅ 675 1 ) mod 173 = ( 627 ⋅ 49 ⋅ 131 ⋅ 18 ⋅ 675 ) mod 173 = 48900262950 mod 173 = 3 mod 173 using calculator =3 Thus, 675 is decrypted as 3 , which corresponds to letter C . Calculate 89 2 k for k = 1 , 2 , 3 , 4 . 89 ≡ 89 mod 713 since by a ≡ a ( mod n ) 89 2 ≡ 89 2 mod 713 = 7921 mod 713 since 89 2 = 7921 = 78 mod 713 since 7921 mod 713 = 78 89 4 ≡ ( 89 2 ) 2 mod 713 ≡ ( 78 ) 2 mod 713 since 89 2 mod 713 = 78 ≡ 6084 mod 713 since 78 2 = 6084 = 380 mod 713 since 380 = 6084 mod 713 89 8 ≡ ( 89 4 ) 2 mod 713 ≡ ( 380 ) 2 mod 713 since 89 4 mod 713 = 380 ≡ 144400 mod 713 since 380 2 = 144400 = 374 mod 713 89 16 ≡ ( 89 8 ) 2 mod 713 ≡ ( 374 ) 2 mod 713 since 89 8 mod 713 = 374 ≡ 139876 mod 713 since 374 2 = 139876 = 128 mod 713 since 139876 mod 713 = 126 Perform 89 2 k for k = 5 , 6 , 7 , 8

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Author: EPP

Publisher: CENGAGE L

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Chapter 8.4, Problem 39ES

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To decrypt the given cipher text and calculate the original message:

675 089 089 048.

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Chapter 8.4, Problem 38ES

Chapter 8.4, Problem 40ES

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To decrypt the given cipher text and calculate the original message: 675 089 ​ 089 048.

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To decrypt the given cipher text and calculate the original message: 675 089​ 089 048.

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To decrypt the given cipher text and calculate the original message: 675 089 089 048.

To decrypt the given cipher text and calculate the original message:

675 089 089 048.