Alice received the following ciphertext from Bob, "08 01 08 05". Bob had encrypted it using the RSA cypher with Alice's public key (pq, e) = (55, 3), where p = 5 and q = 11. Note that (p-1)(9-1) = 40. The value for d in Alice's private key, (pq, d) is a positive inverse for 3 modulo (p-1)(91). It was found to be 27 in Examples 8.4.8(b) and 8.4.10. What is Bob's message after Alice decrypts it? (Assume Bob encoded one letter at a time using the encoding A = 01, 8 = 02, C = 03, ..., Z = 26.) To decrypt Bob's message, Alice uses the decryption formula mod where M is the code for a letter of the message, C is the encrypted version of the letter, (pq, e) = (55, 3) is the public key, and (pq, d) = (55, 27) is the private key. (a) To begin, Alice computes the values of a, b, c, d and e that are indicated below. 08¹a (mod 55) 082b (mod 55) 084 c (mod 55) 088 Ed (mod 55) 0816e (mod 55) She finds that a = |, C = [ ],d= and e = Because 27 = 16 + 8 + 2 + 1, 0827= 0816 +8+2+1 0816. 088.082.081, she uses the values of a, b, d, and e to compute 0827 mod 55 = (a.b.de) mod 55 = [ Thus, the first letter in Bob's message is 27 mod 55 = (b) Alice finds the second letter of Bob's message by computing (c) What is Bob's message after Alice finishes decrypting it?
Alice received the following ciphertext from Bob, "08 01 08 05". Bob had encrypted it using the RSA cypher with Alice's public key (pq, e) = (55, 3), where p = 5 and q = 11. Note that (p-1)(9-1) = 40. The value for d in Alice's private key, (pq, d) is a positive inverse for 3 modulo (p-1)(91). It was found to be 27 in Examples 8.4.8(b) and 8.4.10. What is Bob's message after Alice decrypts it? (Assume Bob encoded one letter at a time using the encoding A = 01, 8 = 02, C = 03, ..., Z = 26.) To decrypt Bob's message, Alice uses the decryption formula mod where M is the code for a letter of the message, C is the encrypted version of the letter, (pq, e) = (55, 3) is the public key, and (pq, d) = (55, 27) is the private key. (a) To begin, Alice computes the values of a, b, c, d and e that are indicated below. 08¹a (mod 55) 082b (mod 55) 084 c (mod 55) 088 Ed (mod 55) 0816e (mod 55) She finds that a = |, C = [ ],d= and e = Because 27 = 16 + 8 + 2 + 1, 0827= 0816 +8+2+1 0816. 088.082.081, she uses the values of a, b, d, and e to compute 0827 mod 55 = (a.b.de) mod 55 = [ Thus, the first letter in Bob's message is 27 mod 55 = (b) Alice finds the second letter of Bob's message by computing (c) What is Bob's message after Alice finishes decrypting it?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:Alice received the following ciphertext from Bob, "08 01 08 05". Bob had encrypted it using the RSA cypher with Alice's public key (pq, e) = (55, 3), where p = 5 and q = 11. Note that (p − 1)(q − 1) = 40. The value for d in Alice's private key, (pq, d) is a positive
inverse for 3 modulo (p − 1)(q − 1). It was found to be 27 in Examples 8.4.8(b) and 8.4.10. What is Bob's message after Alice decrypts it? (Assume Bob encoded one letter at a time using the encoding A = 01, 8 = 02, C = 03, ..., Z = 26.)
To decrypt Bob's message, Alice uses the decryption formula
M = C
mod
where M is the code for a letter of the message, C is the encrypted version of the letter, (pq, e) = (55, 3) is the public key, and (pq, d) = (55, 27) is the private key.
(a) To begin, Alice computes the values of a, b, c, d and e that are indicated below.
084 c (mod 55)
08¹ = a (mod 55)
088 d (mod 55)
She finds that a =
082 b (mod 55)
0816e (mod 55)
b =
, C =
d=
and e =
Because
27 = 16 + 8 + 2 + 1, 0827= 0816 +8+2+1 = 0816. 088.082.08¹,
she uses the values of a, b, d, and e to compute 0827 mod 55 = (a.b.de) mod 55 =
Thus, the first letter in Bob's message is
27 mod 55 =
(b) Alice finds the second letter of Bob's message by computing
(c) What is Bob's message after Alice finishes decrypting it?
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
