Discrete Mathematics Questions and Answers – Cryptography – Encryption


This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Cryptography – Encryption”.

1. How many bytes of the secret key is generated using Diffie-Hellman encryption/decryption scheme?
a) 256
b) 871
c) 1024
d) 962
View Answer

Answer: a
Explanation: Diffie-Hellman encryption/decryption scheme generates 256 bytes shared a secret key. This secret key then is used by AES key to encrypt this data.

2. In which of the following systems, encryption slower than decryption?
a) elliptic curve cryptography
b) parabolic curve cryptography
c) symmetric cryptography
d) antisymmetric cryptography
View Answer

Answer: b
Explanation: It is known that performing encryption using the public key takes more time than performing decryption using the private key in elliptic curve cryptography (ECC) and the key consists of 60 bytes.

3. If there are 256 cipher texts per plain text and a total of 218 plaintexts of length 18 exists. Then determine the number of distinct ciphertexts?
a) 761
b) 274
c) 186
d) 289
View Answer

Answer: b
Explanation: If there are 256 cipher texts per plain text and a total of 218 plaintexts of length 18 exists which will all decrypt to the same plaintext, and this holds for every plaintext. There are a total of 256 plaintexts of length 56. Now, there must be 256. 218 = 274 distinct ciphertexts which all decrypt to plaintexts of length 56. If all those ciphertexts are the same length, they must be at least 74 bits long.

4. TEA cipher uses which of the following structure?
a) standard cipher structure
b) pseudo random structure
c) feistel structure
d) block structure
View Answer

Answer: c
Explanation: The Feistel structure system TEA operates on two 32-bit unsigned integer numbers. It uses a 128-bit key that can be used to build a simple key schedule by mixing all of the key elements.

5. Let A’s public key is n=6, 736, 180, 7817, 961, 456, 267 and e = 5 and B sends the ciphertext. c = 456, 871, 122, 391, 882, 538 to A. Determine B’s message in numeric format?
a) 235813
b) 57971.89
c) 770190.04
d) 687651.9
View Answer

Answer: c
Explanation: It is known that to get original message m after decrypting we can have the formula m=c1/e. In this case: (456,871,122,391,882,538)1/3 = 770190.04 and this is the required answer.

6. In encryption, which of the following is the best text encoding technique?
a) ASCII encoding
b) Hex-encoding
c) Unicode technique
d) Base64 encoding
View Answer

Answer: c
Explanation: Base64 and hex encoding scheme encode characters(or only bytes). First, we need to encode the characters as bytes and after that encode the bytes. In terms of compactness and simplicity, the best technique is Unicode scheme.

7. _______ are used as the base of the Public Key Infrastructure.
a) SSL certificates
b) TLS certificates
c) X.509 certificates
d) HAS certificates
View Answer

Answer: c
Explanation: The X.509 certificates may be used as a base of the Public Key Infrastructure. PKIX is a tree structure where a Certificate Authority can be used to give trust to end entity certificates. X.509 certificates cannot directly use symmetric cryptography.

8. The default key size of RC2 Feistel cipher is _______
a) 64GB
b) 64 bits
c) 64 bytes
d) 64KB
View Answer

Answer: c
Explanation: RC2 is a 64-bit source-heavy Feistel cipher system with a default key size of 64 bits. It is a complex cipher which uses secret indices and performs bitwise rotations, logical operations(AND, NOT, and OR) and modular addition.

9. How many combinations of keys can be constructed from a 72 ciphertext stream cipher?
a) 4271
b) 7345
c) 3291
d) 2556
View Answer

Answer: d
Explanation: For stream cipher, if there are n ciphertexts then there are n*(n−1)/2 combination of keys to be made.
= \(72*\frac{72-1}{2}\)
= 72*35.5
= 2556.

10. What is the block size of RC6 Feistel block cipher?
a) 5013 bits
b) 128 bits
c) 596 bits
d) 1768 bits
View Answer

Answer: b
Explanation: The RC6 Feistel block cipher is a 20-round cipher scheme which includes a fixed block size of 128 bits and it supports 128, 192, and 256-bit keys for encryption of messages.

Sanfoundry Global Education & Learning Series – Discrete Mathematics.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn