Final Sample 515

Final Sample 515 - INFS 501 - Sample Final Exam 5/10/2010 -...

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1. Suppose the sequence a 0 ,a 1 ,a 2 ,... =− 3, 7, 37, . .. satisfies the recurrence relation a k = 4a k 1 3a k 2 k ∈ ℤ where k 2. Calculate a 17 . a 0 ,a 1 ,a 2 ,... =− 3, 7, 37, . .. a k = 4a k 1 3a k 2 t 2 = 4t - 3 0 = t 2 - 4t + 3 = (t-1)(t-3) a k = A*3 k + B* 1 k -3 = A + B 7 = A* 3 1 + B* 1 1 7= 3A + B so, B = 7 - 3A -3 = A + 7 – 3A -3 = 7 – 2A so A = 5, B = -8 a k = 5*3 k - 8*1 k (check: a 0 = 5 – 8 = –3 a 1 = 5*3 – 8*1 = 7 a 2 = 5*3 2 – 8 = 37) a 17 = 5*3 17 - 8*1 17 = 645700807 2. Find gcd(67966, 34255). 67966 = 34255 + 33711 34255 = 33711 + 544 33711 = 544 *61 + 527 544 = 527 + 17 527 = 17 * 314 67966 = 17 * 3998 34255 = 17 * 2015 gcd(67966, 34255) = 17 3. f: 383 1151  ℤ 383 x 1151 is defined by f x = mod 383 x , mod 1151 x  . f is an isomorphism. a) Give a similar definition for f 1 . 1151 = 383 * 3 + 2 383 = 2 * 191 + 1 1 = 383 – 2*191 = 383 – 191(1151-383*3) = (-191)* 1151 + 574 * 383 f 1 ((x,y)) = (-191)* 1151*x + 574 * 383*y b) Find an integer x that satisfies the following 3 conditions: 1 x 440832 x 100 mod 383 x 200 mod 1,151 = x (-191)* 1151 * 100 + 574 * 383 * 200
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This note was uploaded on 11/01/2010 for the course IFNS 501 taught by Professor Ellis during the Fall '09 term at George Mason.

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Final Sample 515 - INFS 501 - Sample Final Exam 5/10/2010 -...

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