ECE-103-1075-Quiz2_solutions

ECE-103-1075-Quiz2_solutions - ID NUMBER: NAME: Solutions...

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ID NUMBER: NAME: Solutions ECE 103 Quiz #2 Monday, June 25, 2007. Non-programmable calculators are permitted, but show your work! Problem Value Mark Awarded 1 6 2 5 3 6 4 8 TOTAL 25 1. (6 points) This concerns the RSA public-key cryptosystem, and a user with public (encryption) key ( e,n ) = (13 , 943) . (a) Encrypt the plain-text M = 2 . 2 2 4 (mod 943) 2 4 16 (mod 943) 2 8 256 (mod (943) 2 13 2 8 · 2 4 · 2 256 · 16 · 2 1024 · 8 81 · 8 648 (mod 943) . The encrypted text is C M 13 648 (mod 943). (b) Factor n = 943 into primes and calculate φ (943) . After some guesswork and rough computation, you find that 943 = 23 · 41 . Therefore, φ (943) = (23 - 1)(41 - 1) = 22 · 40 = 880. (c) Find the decryption key ( d,n ) corresponding to (13 , 943) . To do this we solve 13 d 1 (mod 880) using the linear Diophantine equation 13 d +880 y = 1 and the Extended Euclidean Algorithm. q
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This note was uploaded on 09/28/2011 for the course ECE 103 taught by Professor Nayak during the Spring '11 term at Waterloo.

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ECE-103-1075-Quiz2_solutions - ID NUMBER: NAME: Solutions...

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