Jse the RSA cipher with public key (pq, e) = (23 · 31, 43) = (713, 43) and private key (pq, d), where d = 307, to decrypt the given ciphertext and find the original message. (Assume the letters of the alphabet are encoded as follows: 1 = 01, B = 02, C = 03, . .., Z = 26.) 675 001 423

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter2: The Integers
Section2.8: Introduction To Cryptography (optional)
Problem 19E: Suppose that in an RSA Public Key Cryptosystem. Encrypt the message "algebra" using the -letter...
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please find the value for a,b,c,d,e,f,g,h,i

Use the RSA cipher with public key (pq, e) = (23 · 31, 43) = (713, 43) and private key (pq, d), where d = 307, to decrypt the
given ciphertext and find the original message. (Assume the letters of the alphabet are encoded as follows:
A = 01, B = 02, C = 03, ..., z = 26.)
675 001 423
Since d = 307 = 256 + 32 + 16 + 2 + 1, find the first letter of the decrypted message by computing 675° mod 713. Now
6751 = a (mod 713)
6758 =d (mod 713)
67564 =g (mod 713)
6752 =b (mod 713)
6754 =c (mod 713)
67532 =f (mod 713)
675256 =i (mod 713).
675
16
=e (mod 713)
675128
=h (mod 713)
Transcribed Image Text:Use the RSA cipher with public key (pq, e) = (23 · 31, 43) = (713, 43) and private key (pq, d), where d = 307, to decrypt the given ciphertext and find the original message. (Assume the letters of the alphabet are encoded as follows: A = 01, B = 02, C = 03, ..., z = 26.) 675 001 423 Since d = 307 = 256 + 32 + 16 + 2 + 1, find the first letter of the decrypted message by computing 675° mod 713. Now 6751 = a (mod 713) 6758 =d (mod 713) 67564 =g (mod 713) 6752 =b (mod 713) 6754 =c (mod 713) 67532 =f (mod 713) 675256 =i (mod 713). 675 16 =e (mod 713) 675128 =h (mod 713)
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