Explanation Given information: Cipher text: 028 018 675 129 Concept used: The decryption key d = 307 since 307 is the inverse for 43 . Public key is n = 713 = 23.31 and e = 43 . Calculation: Now, decrypt the message to compute 28 307 mod 713 , 18 307 mod 713 , 675 307 mod 713 , 129 307 mod 713 Convert 307 to a sum of powers of 2 that is, 307 = 256 + 32 + 16 + 2 + 1 = 2 8 + 2 5 + 2 4 + 2 1 + 1 Perform 28 2 k for k = 1 , 2 , 3 , 4 . 28 = ¯ 28 ( mod 713 ) 28 2 = ¯ 784 ( mod 713 ) since 28 2 = 784 = 71 mod 713 since 78 4 mod 713 = 71 28 4 = ¯ ( 28 2 ) 2 mod 713 = ¯ 71 2 mod 713 since 28 2 mod 713 = 71 = 5041 mod 713 since 71 2 = 5041 = 50 mod 713 since 504 1 mod 713 = 50 28 8 = ¯ ( 28 4 ) 2 mod 713 = ¯ 50 2 mod 713 since 28 4 mod 713 = 50 = 2500 mod 713 since 50 2 = 2500 = 361 mod 713 since 25 00 mod 713 = 361 28 16 = ¯ ( 28 8 ) 2 mod 713 = 361 2 mod 713 since 28 2 mod 713 = 361 = ¯ 130321 mod 713 since 361 2 = 130321 = ¯ 555 mod 713 since 130321 mod 713 = 555 Perform 28 2 k for k = 5 , 6 , 7 , 8 28 32 = ¯ ( 28 16 ) 2 mod 713 = ¯ ( 555 ) 2 mod 713 since 28 16 mod 713 = 555 = ¯ 308025 mod 713 since 555 2 = 308025 = ¯ 9 mod 713 since 30 8025 mod 713 = 9 28 64 = ¯ ( 28 32 ) 2 mod 713 = ¯ 9 2 mod 713 since 28 32 mod 713 = 9 = ¯ 81 mod 713 since 9 2 = 81 28 128 = ¯ ( 28 64 ) 2 mod 713 = ¯ ( 81 ) 2 mod 713 since 28 64 mod 713 = 81 = ¯ 6561 mod 713 since 81 2 = 6561 = ¯ 144 mod 713 since 6561mod713=144 28 256 = ¯ ( 28 128 ) 2 mod 713 = ¯ ( 144 ) 2 mod 713 since 28 128 mod 713 = 144 = ¯ 20736 mod 713 since 144 2 = 20736 = ¯ 59 mod 713 since 20 736 mod 713 = 59 Use the fact that, 28 307 mod 713 = ( 28 256 + 32 + 16 + 2 + 1 ) mod 713 = ( 28 256 ⋅ 28 32 ⋅ 28 16 ⋅ 28 2 ⋅ 28 1 ) mod 713 = ( 59 ⋅ 9 ⋅ 555 ⋅ 71 ⋅ 28 ) mod 713 By algebra = 585873540 mod 713 Use a calculator = 14 Thus, the decryption for 28 is 14 , which corresponds to the letter N . Calculate the 28 2 k for k = 1 , 2 , 3 , 4 18 = ¯ 18 mod 713 since a = ¯ a ( mod n ) 18 2 = ¯ 324 mod 713 since18 2 = 324 18 4 = ¯ ( 18 2 ) 2 mod 713 = ¯ 324 2 mod 713 since 324 2 = 104976 = 104976 mod 713 and 18 2 mod 713 = 324 = 165 mod 713 104976 mod 713 = 165 18 8 = ¯ ( 18 4 ) 2 mod 713 = 165 2 mod 713 since 18 4 mod 713 = 165 = 27225 mod 713 since 165 2 = 27225 = 131 mod 713 27225 mod 713 = 131 18 16 = ¯ ( 18 8 ) 2 mod 713 = 131 2 mod 713 since 18 8 mod 713 = 131 = 17161 mod 713 since 131 2 = 17161 = 49 mod 713 since 17161 mod 713 = 49 Perform the 28 2 k for k = 5 , 6 , 7 , 8 18 32 = ¯ ( 18 16 ) 2 mod 713 = ( 49 ) 2 mod 713 since 18 16 mod 713 = 49 = 2401 mod 713 since 49 2 = 2401 = 262 mod 713 since 240 1 mod 713 = 262 18 64 = ¯ ( 18 32 ) 2 mod 713 = ( 262 ) 2 mod 713 since 18 32 mod 713 = 262 = 68644 mod 713 since 262 2 = 68644 = 196 mod 713 since 686 44 mod 713 = 196 18 128 = ¯ ( 18 64 ) 2 mod 713 = ( 196 ) 2 713 since 18 64 mod 713 = 196 = 38416 mod 713 since 196 2 = 38416 = 627 mod 713 since 384 16 mod 713 = 627 18 256 = ¯ ( 18 128 ) 2 mod 713 = ( 627 ) 2 mod 713 since 18 128 mod 713 = 627 = 393129 mod 713 since 627 2 = 393129 = 266 mod 713 since 393129 mod 713 = 266 Thus, the decryption for 18 we have to calculate is 18 307 mod 713 . 18 307 mod 713 = ( 18 256 + 32 + 16 + 2 + 1 ) mod 713 = ( 18 256 ⋅ 18 32 ⋅ 18 16 ⋅ 18 2 ⋅ 18 1 ) mod 713 = ( 266 ⋅ 262 ⋅ 49 ⋅ 324 ⋅ 18 ) mod 713 = ( 19915743456 ) mod 713 Using a calculator = 9 mod 713 by the definiton of mod . = 9 Thus, the decryption for 18 is 9 , which corresponds to the letter l . Perform 675 2 k for k = 1 , 2 , 3 , 4

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Author: EPP

Publisher: CENGAGE L

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Chapter 8.4, Problem 40ES

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028 018 675 129.

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Chapter 8.4, Problem 39ES

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To decrypt the given cipher text and calculate the original message: 028 018 675 129.

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To decrypt the given cipher text and calculate the original message: 028 018 675 129.

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Chapter 8 Solutions

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

028 018 675 129.