2 **PROBLEMS 10.3**1. The message REPLY TODAY is to be encrypted in the ElGamal cryptosystem and forwarded to a user with public key (47, 5, 10) and private key \( k = 19 \). - (a) If the random integer chosen for encryption is \( j = 13 \), determine the ciphertext. - (b) Indicate how the ciphertext can be decrypted using the recipient’s private key.2. Suppose that the following ciphertext is received by a person having ElGamal public key (71, 7, 32) and private key \( k = 30 \): \[ (56, 45) \quad (56, 38) \quad (56, 29) \quad (56, 03) \quad (56, 67) \\ (56, 05) \quad (56, 27) \quad (56, 31) \quad (56, 38) \quad (56, 29) \] Obtain the plaintext message.3. The message NOT NOW (numerically 131419131422) is to be sent to a user of the ElGamal system who has public key (37, 2, 18) and private key \( k = 17 \). If the integer \( j \) used to construct the ciphertext is changed over successive four-digit blocks from \( j = 13 \) to \( j = 28 \) to \( j = 11 \), what is the encrypted message produced?4. Assume that a person has ElGamal public key (2633, 3, 1138) and private key \( k = 965 \). If the person selects the random integer \( j = 583 \) to encrypt the message BEWARE OF THEM, obtain the resulting ciphertext. *[Hint: \( 3^{583} \equiv 1424 \, (\text{mod} \, 2633), \quad 1138^{583} \equiv 97 \, (\text{mod} \, 2633) \).]*5. (a) A person with public key (31, 2, 22) and private key \( k = 17 \) wishes to sign a message whose first plaintext block is \( B = 14 \). If

Answer of two we use descryption formula to find plain text:…

Answered: key (71, 7, 32) and private key k = 30:…

key (71, 7, 32) and private key k = 30: (56, 45) (56, 38) (56, 29) (56, 03) (56, 67) (56, 05) (56, 27) (56, 31) (56,38) (56, 29)

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

**PROBLEMS 10.3**

1. The message REPLY TODAY is to be encrypted in the ElGamal cryptosystem and forwarded to a user with public key (47, 5, 10) and private key \( k = 19 \).
- (a) If the random integer chosen for encryption is \( j = 13 \), determine the ciphertext.
- (b) Indicate how the ciphertext can be decrypted using the recipient’s private key.

2. Suppose that the following ciphertext is received by a person having ElGamal public key (71, 7, 32) and private key \( k = 30 \):
\[
(56, 45) \quad (56, 38) \quad (56, 29) \quad (56, 03) \quad (56, 67) \\
(56, 05) \quad (56, 27) \quad (56, 31) \quad (56, 38) \quad (56, 29)
\]
Obtain the plaintext message.

3. The message NOT NOW (numerically 131419131422) is to be sent to a user of the ElGamal system who has public key (37, 2, 18) and private key \( k = 17 \). If the integer \( j \) used to construct the ciphertext is changed over successive four-digit blocks from \( j = 13 \) to \( j = 28 \) to \( j = 11 \), what is the encrypted message produced?

4. Assume that a person has ElGamal public key (2633, 3, 1138) and private key \( k = 965 \). If the person selects the random integer \( j = 583 \) to encrypt the message BEWARE OF THEM, obtain the resulting ciphertext.
*[Hint: \( 3^{583} \equiv 1424 \, (\text{mod} \, 2633), \quad 1138^{583} \equiv 97 \, (\text{mod} \, 2633) \).]*

5. (a) A person with public key (31, 2, 22) and private key \( k = 17 \) wishes to sign a message whose first plaintext block is \( B = 14 \). If

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