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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