# Asymmetric Ciphers Questions and Answers – Knapsack/ Merkle – Hellman/ RSA Cryptosystem – II

This set of Cryptography online test focuses on “Knapsack/ Merkle – Hellman/ RSA Cryptosystem”.

1. Find the ciphertext for the message {100110101011011} using superincreasing sequence { 1, 3, 5, 11, 35 } and private keys a = 5 and m=37.
a) C = ( 33, 47, 65 )
b) C = ( 65, 33, 47 )
c) C = ( 47, 33, 65 )
d) C = ( 47, 65, 33 )

Explanation: {vi} = { 1, 3, 5, 11, 35 }
a = 5 and m = 37
Public key generation:
{wi} = avi mod m
wi = {5, 15, 25, 18, 27}
Break the message into k-bit tuple i.e. 5-bit tuple
10011 01010 11011
Encoding of M as follows:
M Ci
10011 47
01010 33
11011 65

Ciphertext sent will be: C = (47, 33, and 65).

2. Suppose that plaintext message units are single letters in the usual 26-letter alphabet with A-Z corresponding to 0-25. You receive the sequence of ciphertext message units 14, 25, 89. The public key is the sequence {57, 14, 3, 24, 8} and the secret key is b = 23, m = 61.
Decipher the message. The Plain text is
a) TIN
b) INT
c) KIN
d) INK

Explanation: Solve using Knapsack Cryptosystem.
Wi = {57, 14, 3, 24, 8}
b = 23 and m = 61
a = b-1 mod m
61 = 2 x23 + 15
23 = 1x 15 + 8 Therefore 1= 8 x 23 – 3 x 61
15 = 1x 8 + 7 b-1 = 23-1= 8
8 = 1x 7 + 1 a = 8
v_i=a^(-1) w_i mod m
=bw_i mod m
v_i={ 30, 17, 8, 3, 1}

Cipher text V = bC mod m Plaintext
14 23 x 14 mod 61 = 17 01000 = 8 = I
25 23 x 25 mod 61 = 26 01101 = 13 = N
89 23 x 89 mod 61 = 34 10011 = 19 = T.

3. RSA is also a stream cipher like Merkel-Hellman.
a) True
b) False

Explanation: RSA is a block cipher system.

4. In the RSA algorithm, we select 2 random large values ‘p’ and ‘q’. Which of the following is the property of ‘p’ and ‘q’?
a) p and q should be divisible by Ф(n)
b) p and q should be co-prime
c) p and q should be prime
d) p/q should give no remainder

Explanation: ‘p’ and ‘q’ should have large random values which are both prime numbers.

5. In RSA, Ф(n) = _______ in terms of p and q.
a) (p)/(q)
b) (p)(q)
c) (p-1)(q-1)
d) (p+1)(q+1)

Explanation: Ф(n) = (p-1)(q-1).

6. In RSA, we select a value ‘e’ such that it lies between 0 and Ф(n) and it is relatively prime to Ф(n).
a) True
b) False

Explanation: gcd(e, Ф(n))=1; and 1 < e < Ф(n).

7. For p = 11 and q = 19 and choose e=17. Apply RSA algorithm where message=5 and find the cipher text.
a) C=80
b) C=92
c) C=56
d) C=23

Explanation: n = pq = 11 × 19 = 209.

8. For p = 11 and q = 19 and choose d=17. Apply RSA algorithm where Cipher message=80 and thus find the plain text.
a) 54
b) 43
c) 5
d) 24

Explanation: n = pq = 11 × 19 = 209.
C=Me mod n ; C=517 mod 209 ; C = 80 mod 209.

9. USENET falls under which category of public key sharing?
a) Public announcement
b) Publicly available directory
c) Public-key authority
d) Public-key certificates

Explanation: Many users have adopted the practice of appending their public key to messages that they send to public forums, such as USENET newsgroups and Internet mailing lists.

Perform encryption on the following PT using RSA and find the CT.

10. p = 3; q = 11; M = 5
a) 28
b) 26
c) 18
d) 12

Explanation: n = 33; f(n) = 20; d = 3; C = 26.

11. p = 5; q = 11; M = 9
a) 43
b) 14
c) 26
d) 37

Explanation: n = 55; f(n) = 40; d = 27; C = 14.

12. p = 7; q = 11; M = 8
a) 19
b) 57
c) 76
d) 59

Explanation: n = 77; f(n) = 60; d = 53; C = 57.

13. p = 11; q = 13; M = 7
a) 84
b) 124
c) 106
d) 76

Explanation: n = 143; f(n) = 120; d = 11; C = 106.

14. p = 17; q = 31; M = 2
a) 254
b) 423
c) 128
d) 523

Explanation: n = 527; f(n) = 480; d = 343; C = 128.

15. n = 35; e = 5; C = 10. What is the plaintext (use RSA) ?
a) 3
b) 7
c) 8
d) 5

Explanation: Use RSA system to decrypt and get PT = 5.

Sanfoundry Global Education & Learning Series – Cryptography and Network Security.

To practice all areas of Cryptography for online test, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]