This set of Cryptography Multiple Choice Questions & Answers (MCQs) focuses on “Knapsack/ Merkle – Hellman/ RSA Cryptosystem”.

a) {62, 48, 166, 52}

b) {141, 26, 52, 48}

c) {93, 26, 91, 48}

d) {62, 26, 166, 48}

View Answer

Explanation: {62, 26, 166, 48} =302.

2. 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

View Answer

Explanation: 1+15+24 = 40.

3. 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

View Answer

Explanation: If someone sends you the code 38 this can only have come from the plain text 01101.

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

a) True

b) False

View Answer

Explanation: It is not because 10 < 1+2+3+9.

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

a) Easier

b) Tougher

c) Shorter

d) Lengthier

View Answer

Explanation: A superincreasing knapsack is chosen to make computations easier while manual calculations of knapsack problems.

a) 011111

b) 010011

c) 010111

d) 010010

View Answer

Explanation: v0=1, v1=2, v2=4, v3=9, v4=20, v5=38

K=6, V=23

Starting from largest number:

v5 > V then ϵ_5=0

v4 < V then V = V – v4 = 23 – 20 = 3 ϵ_4=1

v3 > V then ϵ_3=0

v2> V then ϵ_2=0

v1 < V then V = V – v1= 3 – 2 = 1 ϵ_1=1

v0 =1 then V = V – v0= 1 – 1 = 0 ϵ_0=1

n= ϵ_5 ϵ_4 ϵ_3 ϵ_2 ϵ_1 ϵ_0 = 010011.

7. Another name for Merkle-Hellman Cryptosystem is

a) RC4

b) Knapsack

c) Rijndael

d) Diffie-Hellman

View Answer

Explanation: Knapsack is another name for Merkel-Hellman Cryptosystem.

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

a) True

b) False

View Answer

Explanation: The hard knapsack becomes the public key and the easy knapsack becomes the private key.

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

View Answer

Explanation: The public key can be used to encrypt messages, but cannot be used to decrypt messages. The private key decrypts the messages.

10. 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)

View Answer

Explanation: {wi }= {a vi mod m}

{wi} = { 17×2 mod 61, 17×3 mod 61, 17×7 mod 61, 17×15 mod 61, 17×31 mod 61}

{wi} = {34, 51, 58, 11, and 39}

PlainText In binary Ci

W- 22 10110 148

H – 7 00111 143

Y – 24 11000 50

So that the ciphertext sent will be C= (148, 143, 50).

11. 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

View Answer

Explanation: n = pq = 11 × 19 = 187.

C=M

^{e}mod n ; C=88

^{7}mod 187 ; C = 11 mod 187.

a) 88

b) 122

c) 143

d) 111

View Answer

Explanation: n = pq = 11 × 19 = 187.

C=M

^{e}mod n ; C=11

^{23}mod 187 ; C = 88 mod 187.

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

View Answer

Explanation: By trail and error, we determine that p = 59 and q = 61. Hence f(n) = 58 x 60 = 3480.

Then, using the extended Euclidean algorithm, we find that the multiplicative

inverse of 31 modulo f(n) is 3031.

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

a) 35212

b) 12543

c) 19367

d) 32432

View Answer

Explanation: Since n=233 ´ 241=56,153, p=233 and q=241

f(n) = (p – 1)(q – 1) = 55,680

Using Extended Euclidean algorithm, we obtain

d = 23–1 mod 55680 = 19,367.

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