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

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Q. 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.

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Q. 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”.

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Q. For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where PT message=88 and thus find the CT.

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Q. For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where Cipher message=11 and thus find the plain text.

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Q. In an RSA system the public key of a given user is e = 31, n = 3599. What is the private key of this user?

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

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Q. Find the ciphertext for the message {100110101011011} using superincreasing sequence { 1, 3, 5, 11, 35 } and private keys a = 5 and m=37.

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Q. 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

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Q. RSA is also a stream cipher like Merkel-Hellman.

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Q. In the RSA algorithm, we select 2 random large values ‘p’ and ‘q’. Which of the following is the property of ‘p’ and ‘q’?

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Q. In RSA, Ф(n) = _______ in terms of p and q.

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Q. In RSA, we select a value ‘e’ such that it lies between 0 and Ф(n) and it is relatively prime to Ф(n).

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Q. For p = 11 and q = 19 and choose e=17. Apply RSA algorithm where message=5 and find the cipher text.

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Q. For p = 11 and q = 19 and choose d=17. Apply RSA algorithm where Cipher message=80 and thus find the plain text.

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Q. USENET falls under which category of public key sharing?

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