Cryptography Decryption Questions and Answers

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

1. Suppose that there are two primes, P₁ = 229 and p₂ = 61. Find the value of z and Φ.
a) 13969, 13680
b) 5853, 23452
c) 7793, 34565
d) 17146, 69262
View Answer

Answer: a
Explanation: We know that, z = p₁*p₂ = 229*61 = 13969 and Φ = (p₁ – 1)(p₂ – 1) = (229 – 1)(61 – 1) = 228*60 = 13680.

2. ________ can decrypt traffic to make it available to all other network security functions such as web proxies.
a) SSL visibility appliances
b) RSA appliances
c) Rodriguez cipher system
d) Standard cipher system
View Answer

Answer: a
Explanation: In the data loss prevention systems, Web proxies and antivirus network security functions, SSL visibility appliances decrypt traffic to make it available for all networks.

3. The ROT13 caesar cipher system has an offset of ___________
a) 13
b) 45
c) 71
d) 37
View Answer

Answer: a
Explanation: The ROT13 Caesar cipher system has an offset of 13 and it is one of the comprehensive cipher scheme. However, the Vigenere cipher employs Caesar cipher as one element of the encryption process.

4. In a public key system, the cipher text received is C = 10 if RSA encryption used with a public key(e = 11, n = 77) to deduce the plain text. Determine the value of ϕ(n)?
a) 49
b) 60
c) 123
d) 70
View Answer

Answer: b
Explanation: Given n = 77, that means p and q must be 7 and 11 that is they must be co-prime to each other. Now we know that ϕ(n) = (p – 1) (q – 1)
ϕ(n) = (7 – 1) (11 – 1)
ϕ(n) = 6*10
ϕ(n) = 60.

5. To encrypt a message _______ is used on the character’s positions.
a) boolean algebra
b) bijective function
c) inverse function
d) surjective function
View Answer

Answer: b
Explanation: We have a mathematical notion that a bijective function can be used on the characters’ positions to encrypt a message and an inverse function is used to decrypt the message.

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6. The public key of given user, in an RSA encryption system is e = 57 and n = 3901. What is the value of Euler’s totient function ϕ(n) for calculating the private key of the user?
a) 4369
b) 3772
c) 871
d) 7892
View Answer

Answer: b
Explanation: Given that n=3901 and e=31. We know that n = p∗q where p and q are prime numbers, which gives 3901 = 47*83. Now, ϕ(n) is Euler’s totient function i.e., ϕ(n) = (p−1)∗(q−1)
ϕ(n) = (47−1)∗(83−1)
ϕ(n) = 46*82 = 3772.

7. Using RSA algorithm what is the value of cipher test c if the plain text e = 7 and P = 5, q = 16 & n = 832. Determine the Euler’s totient function for the plain text?
a) 47
b) 584
c) 428
d) 60
View Answer

Answer: d
Explanation: Given plain text (m) = 7, P = 5, Q = 16, where P and Q are two prime integer
n = P * Q ⇒ n = 5*16 = 80 ⇒ Z = (P-1)*(Q-1) ⇒ Z = (5-1)*(16-1) = 4*15 = 60.

8. There are 67 people in a company where they are using secret key encryption and decryption system for privacy purpose. Determine the number of secret keys required for this purpose?
a) 887
b) 6529
c) 2211
d) 834
View Answer

Answer: c
Explanation: Since every two employee have their own secret key encryption and decryption. Both users have to agree on a secret key to communicate using symmetric cryptography. After that, each message is encrypted with that key it is transmitted and decrypted with the same key. Here, key distribution must be secret. For n = 67 we would need \(\frac{n(n-1)}{2} = \frac{67(67-1)}{2}\) = 2211 keys.

9. In a transposition cipher, the plaintext is constructed by the ________ of the ciphertext.
a) permutation
b) combination
c) sequence
d) series
View Answer

Answer: a
Explanation: In cryptography, a method of encryption where the positions of plaintext held by units are shifted according to a regular system so that the ciphertext constructs a permutation of the plaintext is termed as transposition cipher.

10. How many bits of message does the Secure Hash Algorithm produce?
a) 160 bits
b) 1035 bits
c) 621 bits
d) 3761 bits
View Answer

Answer: a
Explanation: The Secure Hash Algorithm or SHA is based on MD4 encryption system. This algorithm gives an output of a longer 160-bit message that is why it is harder to construct another message that yields the same resultant message.

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