assignment6_sol

assignment6_sol - MATH 135 Assignment 6 Solutions Hand-In...

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MATH 135 Fall 2005 Assignment 6 Solutions Hand-In Problems Exercise 3-34: Solve the congruence 5 x 7 (mod 15) . Solution: Here, gcd(5 , 15) = 5, but 5 6 | 7, so there is no solution by Theorem 3.54. Exercise 3-44: Find the inverse of [23] in Z 41 . Solution: We must solve 23 x 1 (mod 41). Apply the Extended Euclidean Algorithm to 41 and 23. 41 y + 23 x = r q i 1 0 41 0 1 23 1 - 1 18 1 - 1 2 5 1 4 - 7 3 1 - 5 9 2 3 9 - 16 1 1 - 23 41 0 1 Hence 23( - 16) + 41(9) = 1. Since gcd(23 , 41) = 1 the solution to the linear congruence is unique. Therefore 23( - 16) 1 (mod 41) and so [23] - 1 = [ - 16] = [25] Z 41 . Check: 23 · 25 = 575 = 14 · 41 + 1. Exercise 3-51: Solve the following simultaneous congruences. x 1 (mod 2) x 2 (mod 3) x 3 (mod 7) Solution: By the Generalized Chinese Remainder Theorem, this system of linear congruences will have a unique solution modulo 2(3)(7) = 42, since the moduli (2, 3 and 7) are pairwise coprime (that is, gcd(2 , 3) = gcd(3 , 7) = gcd(2 , 7) = 1). Now, x 1 (mod 2) and x 2 (mod 3) are equivalent to x ≡ - 1 (mod 2) and x ≡ - 1 (mod 3) respectively. Hence, by Proposition 3.64, the solution to the ﬁrst two is x ≡ - 1 (mod 6). (We have could have also obtained x ≡ - 1 (mod 6) (or x 5 (mod 6)) using the usual Chinese Remainder Theorem technique). We have reduced the three simultaneous congruences to two simultaneous congruences. x ≡ - 1 (mod 6) x 3 (mod 7) .

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An integer x satisﬁes the second congruence if and only if x = 3 + 7 y for y Z . Substitute this into the ﬁrst congruence. 3 + 7 y ≡ - 1 (mod 6) 7 y ≡ - 4 (mod 6) y ≡ - 4 (mod 6) y 2 (mod 6) This is equivalent to y = 2 + 6 z for z Z . Since x = 3 + 7 y , then x = 3 + 7(2 + 6 z ) = 17 + 42 z . The solution to the original system of congruences is therefore x 17 (mod 42). Check: 17 + 42 z 1 (mod 2), 17 + 42 z 2 (mod 3), 17 + 42 z 3 (mod 7). Exercise 3-54: Solve the following simultaneous congruences. 161 x 49 (mod 200) 74 x 1 (mod 53) Solution: The ﬁrst congruence is equivalent to the Diophantine Equation 161 x + 200 y = 49 for some y Z . Apply the Extended Euclidean Algorithm to 200 and 161: 200 y + 161 x = r q i 1 0 200 0 1 161 1 - 1 39 1 - 4 5 5 4 29 - 36 4 7 - 33 41 1 1 200 - 161 0 7 The gcd(200 , 161) = 1, so the congruence has a solution. From the next to last row we see that 161(41)+200( - 33) = 1. Multiplying by 49 gives 161(2009) + 200( - 1617) = 49 . Therefore x = 2009 and y = - 1617 is a particular solution, and by Theorem 3.54, the complete solution to the congruence is x 2009 (mod 200) (in other words, x 9 (mod 200)). An integer x satisﬁes the ﬁrst congruence if and only if x = 9 + 200 y for some y Z . Substitute this into the second congruence and determine if there is a solution.
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assignment6_sol - MATH 135 Assignment 6 Solutions Hand-In...

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