HW1_sol - 1.15(a) 2 5 det 9 5 = (2 5 5 9) mod 26 = 35 mod...

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= = = = = = = = = = = = = = 1 24 17 4 13 10 22 11 25 105 336 147 420 273 504 126 63 441 5 16 7 20 13 24 6 3 21 21 9 15 17 2 23 4 12 11 1 21 ) 26 (mod 5 5 26 mod 21 26 mod 3817 9 15 17 2 23 4 12 11 1 det 1.15(b) 20 1 15 11 2 17 21 5 23 2 9 5 5 23 5 9 5 2 23 ) 26 (mod 17 17 26 mod 35 26 mod ) 9 5 5 2 ( 5 9 5 2 det 1.15(a) 1 1 1 1
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() () 21 5 8 5 14 6 23 15 4 20 21 3 6 13 8 : properly encrypt characters plaintext 3 last that the ng by verifyi check this can we desired, If 5 14 6 23 15 4 20 21 3 14 19 13 4 19 14 15 20 17 1 17 7 1 2 7 5 2 10 14 19 13 4 19 14 15 20 17 10 0 19 7 19 0 4 17 1 K exists. K ), 1 ) 19 , 26 gcd( ( in Z 11 inverse unique a has ... 19 10 0 19 7 19 0 4 17 1 det 3, m Case exist. not does K ), 1 ) 10 , 26 gcd( ( in Z inverse no has ... 10 0 4 17 1 det 14 15 20 17 0 4 17 1 14 15 20 17 0 4 17 1 2, m Case exist. not does K .... 15 ) 4 ( e , 20 ) 17 ( e , 17 ) 1 ( e 1, m Case
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This note was uploaded on 12/25/2010 for the course ALL 0204 taught by Professor 79979 during the Spring '10 term at National Chiao Tung University.

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HW1_sol - 1.15(a) 2 5 det 9 5 = (2 5 5 9) mod 26 = 35 mod...

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