Explore topic-wise MCQs in Cryptography and Network Security.

This section includes 22 Mcqs, each offering curated multiple-choice questions to sharpen your Cryptography and Network Security knowledge and support exam preparation. Choose a topic below to get started.

1.

Compute private key (d, p, q) given public key (e=23, n=233 ´ 241=56,153).

A. 35212
B. 12543
C. 19367
D. 32432
Answer» D. 32432
Discussion
2.

In an RSA system the public key of a given user is e = 31, n = 3599. What is the private key of this user?

A. 3031
B. 2412
C. 2432
D. 1023
Answer» B. 2412
Discussion
3.

For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where PT message=88 and thus find the CT.

A. 23
B. 64
C. 11
D. 54
Answer» D. 54
Discussion
4.

The plaintext message consist of single letters with 5-bit numerical equivalents from (00000)2 to (11001)2. The secret deciphering key is the superincreasing 5-tuple (2, 3, 7, 15, 31), m = 61 and a = 17. Find the ciphertext for the message “WHY”.

A. C= (148, 143, 50)
B. C= (148, 143, 56)
C. C= (143, 148, 92)
D. C= (148, 132,92)
Answer» B. C= (148, 143, 56)
Discussion
5.

In Merkle-Hellman Cryptosystem, the public key can be used to decrypt messages, but cannot be used to decrypt messages. The private key encrypts the messages.

A. True
B. False
Answer» C.
Discussion
6.

In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key.

A. True
B. False
Answer» C.
Discussion
7.

Consider knapsack that weighs 23 that has been made from the weights of the superincreasing series {1, 2, 4, 9, 20, and 38}. Find the ‘n’.

A. 011111
B. 010011
C. 010111
D. 010010
Answer» C. 010111
Discussion
8.

THE_PLAINTEXT_MESSAGE_CONSIST_OF_SINGLE_LETTERS_WITH_5-BIT_NUMERICAL_EQUIVALENTS_FROM_(00000)2_TO_(11001)2._THE_SECRET_DECIPHERING_KEY_IS_THE_SUPERINCREASING_5-TUPLE_(2,_3,_7,_15,_31),_M_=_61_AND_A_=_17._FIND_THE_CIPHERTEXT_FOR_THE_MESSAGE_‚ÄÖ√Ñ√∂‚ÀÖ√Ë‚ÀÖ‚À´WHY‚ÄÖ√Ñ√∂‚ÀÖ√Ë‚ÀÖŒÄ.?$#

A. C= (148, 143, 50)
B. C= (148, 143, 56)
C. C= (143, 148, 92)
D. C= (148, 132,92)
Answer» B. C= (148, 143, 56)
Discussion
9.

IN_MERKLE-HELLMAN_CRYPTOSYSTEM,_THE_PUBLIC_KEY_CAN_BE_USED_TO_DECRYPT_MESSAGES,_BUT_CANNOT_BE_USED_TO_DECRYPT_MESSAGES._THE_PRIVATE_KEY_ENCRYPTS_THE_MESSAGES.?$

A. True
B. False
Answer» C.
Discussion
10.

For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where PT message=88 and thus find the CT.$

A. 23
B. 64
C. 11
D. 54
Answer» D. 54
Discussion
11.

The plaintext message consist of single letters with 5-bit numerical equivalents from (00000)2 to (11001)2. The secret deciphering key is the superincreasing 5-tuple (2, 3, 7, 15, 31), m = 61 and a = 17. Find the ciphertext for the message “WHY”.$#

A. C= (148, 143, 50)
B. C= (148, 143, 56)
C. C= (143, 148, 92)
D. C= (148, 132,92)
Answer» B. C= (148, 143, 56)
Discussion
12.

Compute private key (d, p, q) given public key (e=23, n=233 ´ 241=56,153).$

A. 35212
B. 12543
C. 19367
D. 32432
Answer» D. 32432
Discussion
13.

In an RSA system the public key of a given user is e = 31, n = 3599. What is the private key of this user?

A. 3031
B. 2412
C. 2432
D. 1023
Answer» B. 2412
Discussion
14.

For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where Cipher message=11 and thus find the plain text.

A. 88
B. 122
C. 143
D. 111
Answer» B. 122
Discussion
15.

In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key?

A. True
B. False
Answer» C.
Discussion
16.

Another name for Merkle-Hellman Cryptosystem is

A. RC4
B. Knapsack
C. Rijndael
D. Diffie-Hellman
Answer» C. Rijndael
Discussion
17.

Consider knapsack that weighs 23 that has been made from the weights of the superincreasing series {1, 2, 4, 9, 20, and 38}. Find the ‘n’.$

A. 011111
B. 010011
C. 010111
D. 010010
Answer» C. 010111
Discussion
18.

A superincreasing knapsack problem is ____ to solve than a jumbled knapsack.

A. Easier
B. Tougher
C. Shorter
D. Lengthier
Answer» B. Tougher
Discussion
19.

Set {1, 2, 3, 9, 10, and 24} is superincreasing.

A. True
B. False
Answer» C.
Discussion
20.

For the Knapsack: {1 6 8 15 24}, find the plain text code if the ciphertext is 38.

A. 10010
B. 01101
C. 01001
D. 01110
Answer» C. 01001
Discussion
21.

For the Knapsack: {1 6 8 15 24}, Find the cipher text value for the plain text 10011.

A. 40
B. 22
C. 31
D. 47
Answer» B. 22
Discussion
22.

Imagine you had a set of weights {62, 93, 26, 52, 166, 48, 91, and 141}. Find subset that sums to V = 302.

A. {62, 48, 166, 52}
B. {141, 26, 52, 48}
C. {93, 26, 91, 48}
D. {62, 26, 166, 48}
Answer» E.
Discussion