Cryptography MCQ set number 00536

1. In the elliptic curve group defined by y2= x3- 17x + 16 over real numbers, what is P + Q if P = (0,-4) and Q = (1, 0)?
a) (15, -56)
b) (-23, -43)
c) (69, 26)
d) (12, -86)

Answer

Answer: a [Reason:] P=(x1, y1)= (0,-4)
Q=(x2, y2)= (1,0)
From the Addition formulae:
λ= (0-(-4)) / (1-0) = 4
x3= = 16 – 0 – 1 = 15 and
y3= 4(0 – 15) –(-4) = -56
Thus R=P + Q = (15, -56).

2. Bob selects E67(2, 3) as the elliptic curve over GF(p). He selects e1 = (2, 22) and d = 4.
Then he calculates e2 = d × e1 and the publicly announces the tuple (E, e1, e2). Now, Alice wants to send the plaintext P = (24, 26) to Bob and she selects r = 2. What are C1 and C2?
a) C1=(35,1) ; C2 =(21,44)
b) C1=(44,21) ; C2 =(1,35)
c) C1=(44,21) ; C2 =(44,21)
d) C1=(21,44); C2 =(35,1)

Answer

Answer: a [Reason:] Alice finds the points C1= r × e1 ; C1= (35, 1),
C2=P + r × e2 ; C2= (21, 44).

3. For the point P (7, 0) defined in the curve E13(1, 1). What is –P?
a) (7,1)
b) (8,12)
c) (8,1)
d) (7,0)

Answer

Answer: d [Reason:] The inverse of P(11,2) is (11,11) or (11,-2).

4. Consider knapsack that weighs 23 that has been made from the weights of the superincreasing series {1, 2, 4, 9, 20, and 38}. Find the ‘n’.
a) 011111
b) 010011
c) 010111
d) 010010

Answer

Answer: b [Reason:] 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.

5. 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)

Answer

Answer: a [Reason:] {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).

6. 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

Answer

Answer: b [Reason:] 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.

7. How many primitive roots does Z<19> have?
a) 5
b) 8
c) 7
d) 6

Answer

Answer: d [Reason:] Z<19> has the primitive roots as 2,3,10,13,14 and 15.

8. Find the order of group G= <Z7*, x>
a) 6
b) 4
c) 3
d) 5

Answer

Answer: a [Reason:] |G| = f(7) = (71-70) = 6
G = = { 1, 2, 3, 4, 5, 6 }

9. Which among the following values: 17, 20, 38, and 50, does not have primitive roots in the group G = <Zn∗, ×>?
a) 17
b) 20
c) 38
d) 50

Answer

Answer: b [Reason:] The group G = <Zn*, ×> has primitive roots only if n is 2, 4, pt, or 2pt
‘p’ is an odd prime and‘t’ is an integer.
G = <Z17∗, ×> has primitive roots, 17 is a prime.
G = <Z20∗, ×> has no primitive roots.
G = <Z38∗, ×> has primitive roots, 38 = 2 × 19 prime.
G = <Z50∗, ×> has primitive roots, 50 = 2 × 52 and 5 is a prime.

10.Find the solution of x²≡3 mod 23
a) x≡±16 mod 23
b) x≡±13 mod 23
c) x≡±22 mod 23
d) x≡±7 mod 23

Answer

Answer: a [Reason:] a=3 3^((23+1)/4)≡3⁶≡1 (QR and there is solution)
x ≡ ±3(23 + 1)/4 (mod 23) ≡±16 i.e. x = 7 and 16

11. In Elgamal cryptosystem, given the prime p=31.
Encrypt the message “HELLO”; use 00 to 25 for encoding. The value of C2 for character ‘L’ is
a) 12
b) 07
c) 20
d) 27

Answer

Answer: d [Reason:]The Common factor for the calculation of C2 is e⁷ mod 31 = 25⁷ mod 31 = 25.
P = “L” = 11; C1 = 3⁷ mod 31 = 17; C2 = 11 x 25 mod 31 = 27; C = (17,27).

12. In Elgamal cryptosystem, given the prime p=31.
What is the respective plaintext character for C = (27, 20)?
a) H
b) L
c) O
d) M

Answer

Answer: a [Reason:] The Common factor for the calculation of C2 is e⁷ mod 31 = 25⁷ mod 31 = 25.
C = 17, 20); P = 20 X (17¹⁰)^-1 mod 31 = 07; “07” = “H”.

13. For 1000 nodes in IP level, how many keys would be required?
a) 499000
b) 499500
c) 500500
d) 500000

Answer

Answer: b [Reason:] Use N(N-1)/2 where N=1000.

14. “Meet in the middle attack” is an attack
a) where the timing required for the attack via brute force is drastically reduced
b) where the adversary uses 2 or more machines to decrypt thus trying to reduce the time
c) where messages are intercepted and then either relayed or substituted with another message
d) where cryptanalysis takes lesser time than the brute force decryption

Answer

Answer: c [Reason:] “Meet in the middle attack” is an attack where messages are intercepted and then either relayed or substituted with another message.

15. Which systems use a timestamp?
i) Public-Key Certificates
ii) Public announcements
iii) Publicly available directories
iv) Public-Key authority

a) i) and ii)
b) iii) and iv)
c) i) and iv)
d) iv) only

Answer

Answer: c [Reason:] Public announcements and Public Certificates involve the use of timestamps.