PGP Encryption a formerly non-exportable weapon

I need help with a Algebra question. All explanations and answers will be used to help me learn.

PGP Encryption

Here is the Set up of this Certificate of Authority:

e= 129, n= 4897,

And the two (mod n)inverses: 129, 1401.

To receive encrypted messages, Olivia posts these values of n and e to a public key encryption service, called a certificate authority, which guarantees the integrity of her public key.

If Henry wants to send Olivia a message, he codes his message as a number using a number for each letter of the alphabet. For instance, he might use

A = 11, B = 12, C = 13, …, and Z = 36.

For Henry to send the message MATH, he would use the numbers 23 11 30 18 and code those numbers using Olivia’s public key, N 129 mod 4897 = M, where N is the plaintext (23 11 30 18), and M is the ciphertext, the result of using the modular equation.

This is shown below. (Can you calculate 23^129 on your calculator?! it has 175 digits! and we need all of them )
23^129 mod 4897 = 3065

11^129 mod 4897 = 2001

30^129 mod 4897 = 957 (Can you calculate 23^129 on your calculator?! it has 190 digits!)

18^129 mod 4897 = 2753

Thus Henry sends Olivia the ciphertext numbers 3065, 2001, 957, and 2753.

When Olivia receives the message, she uses her private key, M^1401 mod 4897 = N,
where M is the ciphertext she received from Henry and N is the original plaintext, to decode the message. Olivia calculates
3065^1401 mod 4897 = 23,

2001^1401 mod 4897 = 11,

957^1401 mod 4897 = 30,

2753^1401 mod 4897 = 18.

Olivia decodes the message as 23 11 30 18 or MATH.

For the following, use the values for n, e, and d that we used above.

Encrypt the word “Entscheidungsproblem ” using
A= 11, B = 12, C = 13, …, and Z = 36.

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