MATH 187 sample_quiz_weekof0522_solution

# MATH 187 sample_quiz_weekof0522_solution - & 6) 2 = 33...

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Math 187 Prof. Garsia prepared by Alex Brik 05-19-2009 Sample Quiz 6 Solution 1. A=111111111111111 B=111111111 k r -2 111111111111111 -1 111111111 0 111111 1 111 2 0 Using r k = r k 2 a k r k 1 . Thus GCD ( A;B ) = 111 2. a) 131 b) 253 = 11 ± 23 So (253) = 253 ± 1 1 11 1 1 23 ± = 220 : 3. (17) = 17 1 1 17 ± = 16 Recall: if a;m;x;y are integers with gcd ( a;m ) = 1 then x ² y (mod ( m )) ) a x = a y (mod m ) : Let m = 17 , x = 1023 , a = 2 . Then 1023 = y (mod (17)) ) 2 1023 = 2 y (mod 17) 1023 = 15 (mod 16) ) 2 1023 = 2 15 (mod 17) 2 1023 = 9 mod 17 : 4. Let C ² 904187 : 1

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Note 9386 2 = aC + 390857 1138 2 = bC + 390857 : Thus 9386 2 1138 2 ± = ( a b ) C = 96 ± C Thus (9386 1138) (9386 + 1138) = 96 C ) 8248 ± 10524 = 96 C ) 8248 8 ± 10524 12 = C ) 1031 ± 877 = C: Thus 904187 = 1031 ± 877 : 5. a. f 1 ; 3 ; 4 ; 7 ; 9 ; 10 ; 11 ; 12 ; 16 ; 21 ; 25 ; 26 ; 27 ; 28 ; 30 ; 33 ; 34 ; 36 g b. x 2 12 x = 40 (mod 37) x 2 12 x + 36 = 33 (mod 37) ( x
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Unformatted text preview: & 6) 2 = 33 (mod 37) Thus ² ( x & 6) 2 = 25 2 (mod 37) ( x & 6) 2 = 12 2 (mod 37) Thus x = 31 (mod 37) and x = 18 (mod 37) : 6. K- key C- ciphertext M- message (plaintext) Recall H ( K j C ) = H ( K ) & H ( C ) + H ( M ) Let N be the length of the message. Then using the value in the lecture notes: H ( M ) = 2 N H ( C ) = log 2 (26) N ² 4 : 7 N 2 Since the key is 8-letter word we have that H ( K ) = log 2 & 26 8 ± & 37 : 6 Want H ( K j C ) = 0 : Thus 37 : 6 ± 4 : 7 N + 2 N = 0 ) N & 13 : 9 3...
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## This note was uploaded on 05/22/2010 for the course MATH 187 taught by Professor Staff during the Spring '08 term at UCSD.

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MATH 187 sample_quiz_weekof0522_solution - & 6) 2 = 33...

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