The plaintext message consist of single letters with 5-bit numerical equivalents from (00000)2 to (11001)2. The secret deciphering key is the superincreasing 5-tuple (2, 3, 7, 15, 31), m = 61 and a = 17. Find the ciphertext for the message “WHY”.(a) C= (148, 143, 50)(b) C= (148, 143, 56)(c) C= (143, 148, 92)(d) C= (148, 132,92)I have been asked this question during an online interview.Question is from Overview in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

The plaintext message consist of single letters with 5-bit numerical e

1.	The plaintext message consist of single letters with 5-bit numerical equivalents from (00000)2 to (11001)2. The secret deciphering key is the superincreasing 5-tuple (2, 3, 7, 15, 31), m = 61 and a = 17. Find the ciphertext for the message “WHY”.(a) C= (148, 143, 50)(b) C= (148, 143, 56)(c) C= (143, 148, 92)(d) C= (148, 132,92)I have been asked this question during an online interview.Question is from Overview in chapter Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security
Answer» Right choice is (a) C= (148, 143, 50) Easy explanation: {WI }= {a VI mod m} {wi} = { 17×2 mod 61, 17×3 mod 61, 17×7 mod 61, 17×15 mod 61, 17×31 mod 61} {wi} = {34, 51, 58, 11, and 39} PlainTextIn binaryCi W- 2210110148 H – 700111143 Y – 241100050 So that the ciphertext sent will be C= (148, 143, 50).

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