This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 2. Try to decrypt the following ciphertext decrypted by Caesar Cipher L DP KDSSB WR OHDUQ P= D(k, C) = (C  k) mod 26 So the decrypted text is: I am happy to learn. Cryptography and security CSC8419 ID: W0094835 Name: Tejas K Patel Page 3 Task 2: (8/100) Given p= 5 & q = 7 and (n) = (p1)*(q1), find the inverse of 17 mod (n), and demonstrate the inverse is correct. (n) = (p1)*(q1) = (51)*(71) = (4)*(6) = 24 24 = 1*17+7 17 = 2*7+3 7 = 2*3+1 So, 1 = 72*3 = 72*(172*7) = 5*72*17 = 5*(2417)2*17 = 5*247*17 (mod 24) = 7*17 mod 24 = 17*17 mod 24 So, here inverse is correct. The Inverse of 17 mod 24 is 17. Reference : Stallings, W 2006, Cryptography and network security principles and practices, 4 th edition, Pearson Education International, new jersey....
View
Full Document
 '10
 ronaddie
 Cryptography, Plaintext, Tejas K Patel, security CSC8419

Click to edit the document details