M135F08A9S

M135F08A9S - MATH 135 Fall 2008 Assignment #9 Solutions...

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Unformatted text preview: MATH 135 Fall 2008 Assignment #9 Solutions Hand-In Problems 1. (a) To encrypt, we use the public key. Note that 21 = 16 + 4 + 1 = (10101) 2 , so 2762 21 = 2762 16 2762 4 2762 1 . Using Square and Multiply, 2762 2 (2762 1 ) 2 2762 2 7628644 8019 (mod 12193) 2762 4 (2762 2 ) 2 8019 2 64304361 10672 (mod 12193) 2762 8 (2762 4 ) 2 10672 2 113891584 8964 (mod 12193) 2762 16 (2762 8 ) 2 8964 2 80353296 1426 (mod 12193) Therefore, 2762 21 2762 16 2762 4 2762 1 (mod 12193) 1426 10672 2762 (mod 12193) 15218272 2762 (mod 12193) 1408 2762 (mod 12193) 3888896 (mod 12193) 11522 (mod 12193) Thus, the encrypted message is 11522. (b) We must calculate 765 871 (mod 1147). Since 1147 = 31(37), we can appeal to Proposition 3.64 to split this into calculations mod 31 and mod 37 and then recombine. So 765 871 21 871 (mod 31) (21 30 ) 29 21 (mod 31) 1 29 21 (mod 31) (by F ` T, since 31 6 | 21) 21 (mod 31) and 765 871 25 871 (mod 37) (25 36 )25 7 (mod 37) 1 24 25 7 (mod 37) (by F ` T, since 37 6 | 25) 25 4 25 3 (mod 37) 390625 15625 (mod 37) 16 11 (mod 37) 176 (mod 37) 28 (mod 37) Therefore, if x = 765 871 , then x 21 (mod 31) x 28 (mod 37) so x = 28 + 37 y for some y Z , so 28 + 37 y 21 (mod 31) 6 y - 7 (mod 31) 6 y 24 (mod 31) y 4 (mod 31) (since gcd(6 , 31) = 1) Thus, y = 4 + 31 z for some z Z , so x = 28 + 37(4 + 31 z ) = 176 + 1147 z so x 176 (mod 1187). Therefore, the decrypted message is 176. Solution to Problem 2 Here is a sample solution to #2. Your solution will be different, because it will use different random numbers and your own UWID. O O (4) (4) O O (5) (5) O O (1) (1) (8) (8) O O O O O O (12) (12) O O (7) (7) (10) (10) (6) (6) O O O O O O O O (9) (9) (2) (2) O O (14) (14) (13) (13) O O (11) (11) O O (3) (3) O O restart ; R1 d rand 10 50 ; R1 := 7630549802283981984830860818461680342012649711141 R2 d rand 10 50 ; R2 := 99657431905029310669125664294977241480637976155876 p d nextprime R1 ; p := 7630549802283981984830860818461680342012649711299 q d nextprime R2 ; q := 99657431905029310669125664294977241480637976155879 isprime p ; true isprime q ; true n d p $ q ; n := 76044099731905080464602990385952956188893605225350463668721531360975470313\ 9361539727325515171576821 phi d p K 1 $ q K 1 ; f := 76044099731905080464602990385952956188893605225339734870550800031710074661\ 4248100805502864545709644 e d 20068761; e := 20068761 gcd e , phi ; 3 e d e C 2; e := 20068763 gcd e , phi ; 1 msolve d $ e =1, phi ; d =117925068526220104195418780691463356445061500786105841489113716212945794\ 121903043957534889789183967 d d 1179250685262201041954187806914633564450615007861058414891137162129457\ 94121903043957534889789183967; d := 11792506852622010419541878069146335644506150078610584148911371621294579412\ 1903043957534889789183967 Solution to Problem 2 (continued) O O (15) (15) O O (16) (16)...
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M135F08A9S - MATH 135 Fall 2008 Assignment #9 Solutions...

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