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Unformatted text preview: COMP 170 – Fall 2008 Midterm 2 Solution Q1. Bob is constructing an RSA keypair. He first chooses p = 11 , q = 19 and sets n = 11 · 19 = 209 . He then constructs his public key e and private key d and publishes the ( n, e ) pair. (a) Bob’s private key is d = 7 . What is the value of his public key e ? . Q1. Bob is constructing an RSA keypair. He first chooses p = 11 , q = 19 and sets n = 11 · 19 = 209 . He then constructs his public key e and private key d and publishes the ( n, e ) pair. (a) Bob’s private key is d = 7 . What is the value of his public key e ? . By the definition of the RSA algorithm d · e mod T = 1 where T = ( p 1)( q 1) = 10 · 18 = 180 . Using, e.g., the extended GCD algorithm, we find that the multiplicative inverse of 7 mod T is e = 103 . Q1. Bob is constructing an RSA keypair. He first chooses p = 11 , q = 19 and sets n = 11 · 19 = 209 . He then constructs his public key e and private key d and publishes the ( n, e ) pair. (b) Alice wants to send Bob a message M, &lt; M &lt; n. She calculates X = M e mod n to send Bob and finds that X = 15 . What is the value of the original message M ? Q1. Bob is constructing an RSA keypair. He first chooses p = 11 , q = 19 and sets n = 11 · 19 = 209 . He then constructs his public key e and private key d and publishes the ( n, e ) pair. (b) Alice wants to send Bob a message M, &lt; M &lt; n. She calculates X = M e mod n to send Bob and finds that X = 15 . What is the value of the original message M ? M = X d mod n = 15 7 mod 209 = 203 . (The last equality can be derived any of multiple ways) Q2(a) Is (15 60 mod 61) = (15 62 mod 63)? Q2(b) Is (100 440 mod 89) = (100 1320 mod 89)? Q2(c) Evaluate 3 1052 mod 60 . Q2.(a) Is ( 15 60 mod 61 ) = ( 15 62 mod 63 ) ? No. 61 is a prime number so, by Fermat’s little theorem, 15 60 mod 61 = 1 . ....
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This note was uploaded on 08/25/2010 for the course COMP COMP170 taught by Professor M.j.golin during the Spring '10 term at HKUST.
 Spring '10
 M.J.Golin
 Computer Science

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