RSA cryptosystem = = Let p = 29 and q = 31. Then the RSA modulus n pq 899. We set the length of plaintext blocks to be 2 and the length of ciphertext blocks 3. Let the encryption exponent b = 29. If, for example, we want to encrypt the word live, we first divide the word into blocks of length 2, li and ve, and then find its corresponding number blocks 11 8 and 21 4. We consider these number blocks as numbers in base 26. Then we find the corresponding integers in base 10, that is 11 8 = 11 × 26 +8= 294 and 21 4 = 21 x 26 + 4 = 550. The integers 294 and 550 are regarded as modular numbers in Z899 and encrypted as 29429 (mod 899) = 91 and 5502⁹ (mod 899) = 492. Now write 91 and 492 in base 26 of length 3 we get 91 = 0 x 26² +3 x 26 +13 and 492 = 0 x 262 + 18 × 26 + 24 and the corresponding number blocks of length 3 are 0 3 13 and 0 18 24. Her the ciphertext for live is ADNASY as 0 3 13 = ADN and 0 18 24 = ASY.

RSA cryptosystem = = Let p = 29 and q = 31. Then the RSA modulus n pq 899. We set the length of plaintext blocks to be 2 and the length of ciphertext blocks 3. Let the encryption exponent b = 29. If, for example, we want to encrypt the word live, we first divide the word into blocks of length 2, li and ve, and then find its corresponding number blocks 11 8 and 21 4. We consider these number blocks as numbers in base 26. Then we find the corresponding integers in base 10, that is 11 8 = 11 × 26 +8= 294 and 21 4 = 21 x 26 + 4 = 550. The integers 294 and 550 are regarded as modular numbers in Z899 and encrypted as 29429 (mod 899) = 91 and 5502⁹ (mod 899) = 492. Now write 91 and 492 in base 26 of length 3 we get 91 = 0 x 26² +3 x 26 +13 and 492 = 0 x 262 + 18 × 26 + 24 and the corresponding number blocks of length 3 are 0 3 13 and 0 18 24. Her the ciphertext for live is ADNASY as 0 3 13 = ADN and 0 18 24 = ASY.

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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Question

Find the ciphertext for the plaintext well.
Find the decryption exponent a for the encryption exponen
b = 29.
Find the plaintext for the ciphertext ADOANE.

RSA cryptosystem
Let p = 29 and q = 31. Then the RSA modulus n = pq = 899. We set the
length of plaintext blocks to be 2 and the length of ciphertext blocks 3.
Let the encryption exponent b = 29. If, for example, we want to encrypt
the word live, we first divide the word into blocks of length 2, li and ve,
and then find its corresponding number blocks 11 8 and 21 4. We consider
these number blocks as numbers in base 26. Then we find the corresponding
integers in base 10, that is
11 8 = 11 × 26 + 8 = 294 and 21 4 = 21 × 26 + 4 = 550.
The integers 294 and 550 are regarded as modular numbers in Z899 and
encrypted as
29429 (mod 899) = 91 and 5502⁹ (mod 899) = 492.
Now write 91 and 492 in base 26 of length 3 we get
91 = 0 x 26² +3 x 26 +13 and 492 = 0 x 262 +18 × 26 +24
and the corresponding number blocks of length 3 are 0 3 13 and 0 18 24. Hen
the ciphertext for live is ADNASY as 0 3 13 = ADN and 0 18 24 = ASY.

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