MATH135 - Assignment4 Solutions

MATH135 - Assignment4 Solutions - MATH 135 Assignment #4...

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MATH 135 Fall 2007 Assignment #4 Due: Thursday 11 October 2007, 8:20 a.m. Hand-In Problems 1. Which of the following linear Diophantine equations have solutions? In each case, explain briefly why or why not. If there is a solution, determine the complete solution. (a) 28 x + 91 y = 40 (b) 2007 x - 897 y = 15 2. (a) Determine all non-negative integer solutions to the linear Diophantine equation 133 x + 315 y = 98000. (b) Determine all non-negative integer solutions to the linear Diophantine equation 133 x + 315 y = 98000 with the additional property that x 640 - 2 y . 3. Find the smallest positive integer x so that 141 x leaves a remainder of 21 when divided by 31. 4. Suppose a, b, c Z . Prove that gcd( a, c ) = gcd( b, c ) = 1 if and only if gcd( ab, c ) = 1. 5. Suppose a, b, n Z . Prove that if n 0, then gcd( an, bn ) = n · gcd( a, b ). 6. Suppose that f ( x ) is a polynomial of degree n with coefficients from Q and suppose that g ( x ) = x - c for some c Q . (a) When f ( x ) is divided by g ( x ), we obtain a quotient q ( x ) and a remainder r ( x ). (See Assignment #3.) Explain why, in this case, r ( x ) is a constant r ( x ) = r Q . (b) Prove that r = f ( c ). 7. Consider the system of equations a + b = 2 m 2 b + c = 6 m a + c = 2 Determine all real values of m for which a b c . (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 50, #42 2. Text, page 51, #44 3. Text, page 51, #48 4. Text, page 52, #75 5. Text, page 52, #79 ...continued
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6. Let a , b and c be non-zero integers. Their greatest common divisor gcd( a, b, c ) is the largest positive integer that divides all of them. (a) If d = gcd( a, b, c ), prove that d is a common divisor of a and gcd( b, c ). (b) If f is a common divisor of a and gcd( b, c ), prove that f is a common divisor of a , b and c . (c) Prove that gcd( a, b, c ) = gcd( a, gcd( b, c )).
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MATH 135 Fall 2007 Assignment #4 Solutions Hand-In Problems 1. (a) By inspection, gcd(28 , 91) = 7. (We could calculate this using the Euclidean Algorithm instead.) Since 7 6 | 40, there are no solutions. (b) Set z = - y . We solve 2007 x + 897 z = 15 first. We find gcd(2007 , 897) using the Extended Euclidean Algorithm: 1 0 2007 0 1 897 1 - 2 213 2 - 4 9 45 4 17 - 38 33 4 - 21 47 12 1 59 - 132 9 2 - 80 179 3 1 299 - 669 0 3 So gcd(2007 , 897) = 3 | 15 so there are solutions. Since 2007( - 80) + 897(179) = 3, then, multiplying both sides by 5, we get 2007( - 400) + 897(895) = 15. Therefore, ( - 400 , 895) is a particular solution to 2007 x + 897 z = 15. Therefore, the complete solution to 2007
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This note was uploaded on 09/21/2008 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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MATH135 - Assignment4 Solutions - MATH 135 Assignment #4...

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