Stuck need help! The class I'm taking is computer science discrete structures. Problem is attached. please view both attachments before answering. **** I need to decrypt the cipher text (13 20 20 09 ) using the RSA cipher I have attached to this question.
Stuck need help! The class I'm taking is computer science discrete structures. Problem is attached. please view both attachments before answering. **** I need to decrypt the cipher text (13 20 20 09 ) using the RSA cipher I have attached to this question.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Stuck need help!
The class I'm taking is computer science discrete structures.
Problem is attached. please view both attachments before answering.
**** I need to decrypt the cipher text (13 20 20 09 ) using the RSA cipher I have attached to this question.
Really struggling with this concept. Thank you so much.
![Example 8.4.1o
Decrypting a Message Using RSA Cryptography
Imagine that Alice has hired you to help her decrypt messages and has shared with you the
values of p and q. Compute Alice's private key (pq, d) and use the formula M = C“ mod pq
to decrypt the following ciphertext for her: 17 14.
Solution Because p = 5 and q = 11, (p – 1)(q – 1) = 40, the decryption key d is a posi-
tive inverse for 3 modulo 40. Knowing that you would need this number, we computed it
in Example 8.4.8(b) and found it to be 27. Thus to decrypt the ciphertext 17, you need to
compute
M = 17 mod
1727 mod 55.
%3D
pq =
To do so, note that
27 = 16+8+2+1.
Next, find the residues obtained when 17 is raised to successively higher powers of 2, up
to 24:
= 16:
17 mod 55 = 17 mod 55
= 17
172 mod 55 =
172 mod 55
= 14
17* mod 55 = (172)² mod 55 = 14² mod 55 = 31
%3D
17° mod 55 = (17*)² mod 55 = 31² mod 55 = 26
%3D
1716 mod 55 = (17°)² mod 55 = 262 mod 55 = 16
Then use the fact that
1727 = 1716+8+2+1 = 1716-178-172-17'
to write
1727 mod 55 = (17'6.178. 17².17) mod 55
= [(17° mod 55)(17° mod 55)(17² mod 55)(17 mod 55)] (mod 55)
by Corollary 8.4.4
= (16.26 14.17) (mod 55)
= 99008 (mod 55)
= 8 (mod 55).
Hence 172" mod 55 = 8, and thus the plaintext of the first part of Bob's message is 8, or 08.
In the last step, you find the letter corresponding to 08, which is H. In exercises 14 and 15
at the end of this section, you are asked to show that when you decrypt 14, the result is 9,
which corresponds to the letter I, so you can tell Alice that Bob's message is HI.
SO
Figure 8.4.1 illustrates the process of sending and receiving a message using RSA
cryptography.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F282904df-64d2-44b1-a59e-7b2a86d0de90%2F5e45e913-9032-46cf-b8b7-41fcb6fd9099%2Fhr2pent_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 8.4.1o
Decrypting a Message Using RSA Cryptography
Imagine that Alice has hired you to help her decrypt messages and has shared with you the
values of p and q. Compute Alice's private key (pq, d) and use the formula M = C“ mod pq
to decrypt the following ciphertext for her: 17 14.
Solution Because p = 5 and q = 11, (p – 1)(q – 1) = 40, the decryption key d is a posi-
tive inverse for 3 modulo 40. Knowing that you would need this number, we computed it
in Example 8.4.8(b) and found it to be 27. Thus to decrypt the ciphertext 17, you need to
compute
M = 17 mod
1727 mod 55.
%3D
pq =
To do so, note that
27 = 16+8+2+1.
Next, find the residues obtained when 17 is raised to successively higher powers of 2, up
to 24:
= 16:
17 mod 55 = 17 mod 55
= 17
172 mod 55 =
172 mod 55
= 14
17* mod 55 = (172)² mod 55 = 14² mod 55 = 31
%3D
17° mod 55 = (17*)² mod 55 = 31² mod 55 = 26
%3D
1716 mod 55 = (17°)² mod 55 = 262 mod 55 = 16
Then use the fact that
1727 = 1716+8+2+1 = 1716-178-172-17'
to write
1727 mod 55 = (17'6.178. 17².17) mod 55
= [(17° mod 55)(17° mod 55)(17² mod 55)(17 mod 55)] (mod 55)
by Corollary 8.4.4
= (16.26 14.17) (mod 55)
= 99008 (mod 55)
= 8 (mod 55).
Hence 172" mod 55 = 8, and thus the plaintext of the first part of Bob's message is 8, or 08.
In the last step, you find the letter corresponding to 08, which is H. In exercises 14 and 15
at the end of this section, you are asked to show that when you decrypt 14, the result is 9,
which corresponds to the letter I, so you can tell Alice that Bob's message is HI.
SO
Figure 8.4.1 illustrates the process of sending and receiving a message using RSA
cryptography.
Transcribed Image Text:decrypt the cipher-
text and translate the result into letters of the alphabet to
discover the message.
13 20 20 09
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