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Unformatted text preview: MATH 135 Fall 2007 Assignment #7 Due: Wednesday 07 November 2007, 8:20 a.m. HandIn Problems 1. In each part, determine if the congruence has solutions. If it does, determine the complete solution. (a) 1653 x 77 (mod 2000) (b) 1492 x 77 (mod 2000) (c) x 2 4 x (mod 12) (d) x 11 + 3 x 10 + 5 x 2 (mod 11) 2. (a) Determine the inverse of [41] in Z 64 . (b) Determine the integer a with 0 a < 64 such that [ a ] = [13] 1 [5] + [41] 1 [9] in Z 64 . (c) Suppose that a is an odd integer and k is a positive integer. Explain why [ a ] has an inverse in Z 2 k . 3. Let p be an odd prime number. (a) Prove that x 2 + ab ( a + b ) x (mod p ) has exactly two solutions modulo p . (b) Solve the linear congruence 2 x 1 (mod p ). (c) Solve the congruence 4 x 3 x (mod p ). 4. Solve the following system of simultaneous equations in Z 13 . [3] [ x ] + [5] [ y ] = [7] [9] [ x ] + [6] [ y ] = [2] 5. Solve the simultaneous congruences 4 x 11 (mod 61) 7 x 21 (mod 30) 6. Solve the simultaneous congruences x 3 (mod 7) x 9 (mod 13) x 6 (mod 11) ...continued 7. (a) Consider the system of simultaneous congruences x 14 (mod 27) x 2 (mod 51) Determine the unique solution of this system modulo lcm(27 , 51). (b) Consider the system of simultaneous congruences x 14 (mod 27) x 1 (mod 51) Prove that this system does not have a solution. 8. Strings of length n are to composed using only the digits 0, 1 and 2 such that the first 0 may not be included until after a 2 is included. (For example, 1112010 is an allowable string of length 7, but 1110210 is not.) For each positive integer n , a n is the number of such strings. (a) Write down all such strings of length at most 3 and hence determine a 1 , a 2 and a 3 . (b) Prove that a n = 3 a n 1 1 for n P , n 2. (c) Suppose that we are told that, for every positive integer n , a n = x n + y z for some positive integers x , y and z . We can thus use the values of a 1 , a 2 and a 3 to determine the values of x , y and z . Solve the system of equations a 1 = x 1 + y z a 2 = x 2 + y z a 3 = x 3 + y z for x , y and z . (This problem is not directly related to the course material, but is included to keep your problem solving skills sharp.) Recommended Problems 1. Text, page 83, #35 2. Text, page 83, #39 3. Text, page 83, #48 4. Text, page 83, #55 5. Text, page 86, #81 6. Text, page 87, #95 MATH 135 Fall 2007 Assignment #7 Solutions HandIn Problems 1. (a) We convert to the linear Diophantine equation 1653 x + 2000 y = 77 and solve using the EEA: 1 2000 1 1653 1 1 347 4 5 4 265 4 6 5 82 1 23 19 19 3 98 81 6 4 317 262 1 3 2000 1653 6 Therefore, gcd(1653 , 2000) = 1, so there is a unique solution mod 2000....
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 Fall '08
 ANDREWCHILDS
 Math, Congruence

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