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)

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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)

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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)

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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

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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)

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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

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^{(-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

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8. Find the order of group G= <Z7*, x>

a) 6

b) 4

c) 3

d) 5

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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

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10.Find the solution of x^{2}≡3 mod 23

a) x≡±16 mod 23

b) x≡±13 mod 23

c) x≡±22 mod 23

d) x≡±7 mod 23

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^{((23+1)/4)}≡3

^{6}≡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

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^{7}mod 31 = 25

^{7}mod 31 = 25. P = “L” = 11; C1 = 3

^{7}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

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^{7}mod 31 = 25

^{7}mod 31 = 25. C = 17, 20); P = 20 X (17

^{10})

^{-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

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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

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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