1. Rabin Cryptosystem is a variant of the ElGamal Cryptosystem.
A. TRUE.
B. FALSE.
C. Nothing can be said.
D. None of the mentioned.
Answer= FALSE
2. Using Rabin cryptosystem with p=23 and q=7Encrypt P=24 to find ciphertext. The Cipher text is.
A. 42.
B. 93.
C. 74.
D. 12.
Answer= 93
3. Which Cryptographic system uses C1 = (e1r) mod p and C1 = (e2r x P) mod p at the encryption side?.
A. Elgamal.
B. RSA.
C. Rabin.
D. Whirlpool.
Answer= Elgamal
4. Sender chooses p = 107, e1 = 2, d = 67, and the random integer is r=45. Find the plaintext to be transmitted if the ciphertext is (28,9)..
A. 45.
B. 76.
C. 66.
D. 13.
Answer= 66
5. In Elgamal cryptosystem, given the prime p=31.Choose e1= first primitive root of p and d=10, calculate e2..
A. 24.
B. 36.
C. 25.
D. 62.
Answer= 25
6. 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. 7.
C. 20.
D. 27.
Answer= 27
7. In Elgamal cryptosystem, given the prime p=31.Encrypt the message "HELLO"; use 00 to 25 for encoding. The value of C2 for character 'O' is.
A. 9.
B. 7.
C. 23.
D. 27.
Answer= 9
8. 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= H
9. Imagine you had a set of weights {62, 93, 26, 52, 166, 48, 91, and 141}. Find subset that sums to V = 302..
A. {62, 48, 166, 52}.
B. {141, 26, 52, 48}.
C. {93, 26, 91, 48}.
D. {62, 26, 166, 48}.
Answer= {62, 26, 166, 48}
10. 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.
Answer= 40
11. For the Knapsack: {1 6 8 15 24}, find the plain text code if the ciphertext is 38..
A. 10010.
B. 1101.
C. 1001.
D. 1110.
Answer= 1101
12. Set {1, 2, 3, 9, 10, and 24} is superincreasing..
A. TRUE.
B. FALSE.
C. Nothing can be said.
D. None of the mentioned.
Answer= FALSE
13. A superincreasing knapsack problem is ____ to solve than a jumbled knapsack..
A. Easier.
B. Tougher.
C. Shorter.
D. Lengthier.
Answer= Easier
14. 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. 11111.
B. 10011.
C. 10111.
D. 10010.
Answer= 10011
15. Another name for Merkle-Hellman Cryptosystem is.
A. RC4.
B. Knapsack.
C. Rijndael.
D. Diffie-Hellman.
Answer= Knapsack
16. In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key..
A. TRUE.
B. FALSE.
C. Nothing can be said.
D. None of the mentioned.
Answer= FALSE
17. 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.
C. Nothing can be said.
D. None of the mentioned.
Answer= FALSE
18. 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= C= (148, 143, 50)
19. 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.
Answer= 11
20. For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where Cipher message=11 and thus find the plain text..
A. 88.
B. 122.
C. 143.
D. 111.
Answer= 88
21. 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.
Answer= 3031
22. Compute private key (d, p, q) given public key (e=23, n=233 ´ 241=56,153)..
A. 35212.
B. 12543.
C. 19367.
D. 32432.
Answer= 19367
23. Find the ciphertext for the message {100110101011011} using superincreasing sence { 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 ).
Answer= C = ( 47, 33, 65 )
24. Suppose that plaintext message units are single letters in the usual 26-letter alphabet with A-Z corresponding to 0-25. You receive the sence of ciphertext message units 14, 25, 89. The public key is the sence {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= INT
25. RSA is also a stream cipher like Merkel-Hellman..
A. TRUE.
B. FALSE.
C. Nothing can be said.
D. None of the mentioned.
Answer= TRUE
26. 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 ?(n).
B. p and q should be co-prime.
C. p and q should be prime.
D. p/q should give no remainder.
Answer= p and q should be prime
27. In RSA, ?(n) = _______ in terms of p and q..
A. (p)/(q).
B. (p)(q).
C. (p-1)(q-1).
D. (p+1)(q+1).
Answer= (p-1)(q-1)
28. In RSA, we select a value 'e' such that it lies between 0 and ?(n) and it is relatively prime to ?(n)..
A. TRUE.
B. FALSE.
C. Nothing can be said.
D. None of the mentioned.
Answer= FALSE
29. 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.
Answer= C=80
30. 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.
Answer= 5
31. USENET falls under which category of public key sharing?.
A. Public announcement.
B. Publicly available directory.
C. Public-key authority.
D. Public-key certificates.
Answer= Public announcement
32. n = 35; e = 5; C = 10. What is the plaintext (use RSA) ?.
A. 3.
B. 7.
C. 8.
D. 5.
Answer= 5