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Unformatted text preview: d leaves upon division by N ; this is M. Proof: R d ≡ (M e ) d (mod N ) ≡ M ed (mod N ) ≡ M k N ± N ) + 1 (mod N) ≡ M (mod N ) Thus the original message M is recovered. Example of RSA message sending: p = 11, q = 7, N = 77, N ± N ) = 60. Let e = 13, d = 37 (37.13 = 481) 13.37 = 481 – 1 + 8.60 (see how to find 37 later) Let M = 71 M e = (71) 13 71 ≡ 6 mod 77 So M e ≡ 6 13 (mod 77) 6 3 ≡ 216 ≡15 (mod 77) 6 6 ≡ (15) 2 ≡ 225 (mod 77 ) ≡ 6 (mod 77) ∴ 6 12 ≡ (15) 2 ≡ 36 (mod 77 ) 6 13 ≡ 216 ≡15 ∴ M e ≡ 6 13 (mod 77) ≡ 15 (mod 77) So, send 15 as the message. Decode 15 37 ≡ x (mod 77) 15 2 ≡6 (mod 77) ∴ 15 26 ≡ ( 6 ) 13 (mod 77) ≡ 15 (mod 77) 15 4 ≡ 36 (mod 77) 15 8 ≡ 36 2 ≡ 1296 15 37 ≡ 15 26 36 8 36 3 ≡ 71 (mod 77) © P. Rosenthal , University of Toronto, Department of Mathematics typed by A. Ku Ong...
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This note was uploaded on 04/26/2011 for the course MATH 246 taught by Professor Applebaugh during the Spring '10 term at University of Toronto.
 Spring '10
 Applebaugh
 Math

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