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 )

View Answer

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

View Answer

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

View Answer

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

View Answer

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)

View Answer

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

View Answer

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

View Answer

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

View Answer

Explanation: n = pq = 11 × 19 = 209.

C=M

^{e}mod n ; C=5

^{17}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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

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

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