Cryptography MCQ set number 00556

1. “Rabin Cryptosystem is a variant of the Elgamal Cryptosystem”
a) True
b) False

Answer

Answer: b [Reason:] Rabin Cryptosystem is a variant of the RSA Cryptosystem.

2. Using Rabin cryptosystem with p=23 and q=7
Encrypt P=24 to find ciphertext. The Cipher text is
a) 42
b) 93
c) 74
d) 12

Answer

Answer: b [Reason:] Calculate n = p × q = 161
Plaintext P = 24
Ciphertext = C ≡ P2 (mod n)
= 242 mod 161 = 93 mod 161
Ciphertext transmitted = 93.

3. Which Cryptographic system uses C1 = (e1r) mod p and C1 = (e2r x P) mod p at the encryption side?
a) Elgamal
b) RSA
c) Rabin
d) Whirlpool

Answer

Answer: a [Reason:] The Elgamal cryptographic system uses the above formulae to compute the CT.

4. Sender chooses p = 107, e1 = 2, d = 67, and the random integer is r=45. Find the plaintext to be transmitted if the ciphertext is (28,9).
a) 45
b) 76
c) 66
d) 13

Answer

Answer: c [Reason:] P = [C2 (C1d)-1] mod p = 66.

5. Sender chooses p = 107, e1 = 2, d = 67, and the random integer is r=45. Find the plaintext to be transmitted if the ciphertext is (28,9).
a) 45
b) 76
c) 66
d) 13

Answer

Answer: c [Reason:] P = [C2 (C1d)-1] mod p = 66.

6. In Elgamal cryptosystem, given the prime p=31.
Choose e1= first primitive root of p and d=10, calculate e2.
a) 24
b) 36
c) 25
d) 62

Answer

Answer: c [Reason:] We choose e1=3 (a primitive root of p = 31) and d=10. Then we have e2 = 310 mod 31 = 25.

7. In Elgamal cryptosystem, given the prime p=31.
Encrypt the message “HELLO”; use 00 to 25 for encoding. The value of C2 for character ‘L’ is
a) 12
b) 07
c) 20
d) 27

Answer

Answer: d [Reason:]The Common factor for the calculation of C2 is e7 mod 31 = 257 mod 31 = 25.
P = “L” = 11; C1 = 37 mod 31 = 17; C2 = 11 x 25 mod 31 = 27; C = (17,27).

8. In Elgamal cryptosystem, given the prime p=31.
Encrypt the message “HELLO”; use 00 to 25 for encoding. The value of C2 for character ‘O’ is
a) 09
b) 07
c) 23
d) 27

Answer

Answer: a [Reason:]The Common factor for the calculation of C2 is e7 mod 31 = 257 mod 31 = 25.
P = “O” = 14; C1 = 37 mod 31 = 17; C2 = 14 x 25 mod 31 = 09; C = (17,09).

9. In Elgamal cryptosystem, given the prime p=31.
What is the respective plaintext character for C = (27, 20)?
a) H
b) L
c) O
d) M

Answer

Answer: a [Reason:] The Common factor for the calculation of C2 is e7 mod 31 = 257 mod 31 = 25.
C = 17, 20); P = 20 X (1710)-1 mod 31 = 07; “07” = “H”.

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