{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MATH135 - Assignment4 Solutions

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 )).
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 | 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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}