assignment3_sol - MATH 135 Fall 2005 Assignment 3 Solutions...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 135 Fall 2005 Assignment 3 Solutions Hand-In Problems Exercise 2-11: Prove that gcd( ad,bd ) = | d | gcd( a,b ) . Solution: If d = 0, then both sides of the equation are equal to 0, so the result is true. Suppose d > 0 and let e = gcd( a,b ). Then the right side of the equation equals de . We must show that gcd( ad,bd ) = de . Since e | a , then a = qe for some integer q . This tells us that ad = qde and so de | ad by definition. Similarly, since e | b , then de | bd . Therefore, de is a common divisor of ad and bd . Since e = gcd( a,b ), by the Extended Euclidean Algorithm, there exist integers x and y so that ax + by = e . Multiplying through by d , we obtain ( ad ) x + ( bd ) y = de . But de is a common divisor of ad and bd , so by the GCD Characterization Theorem, de = gcd( ad,bd ). If d < 0, then set D =- d = | d | . Then gcd( ad,bd ) = gcd(- ad,- bd ) = gcd( aD,bD ) = D gcd( a,b ) = | d | gcd( a,b ) from the result for positive D above. This concludes the proof, as we have proven the result for d = 0, d > 0 and d < 0. Exercise 2-24: If a = 3953 and b = 1829, write gcd( a,b ) in the form ax + by where x,y Z . Solution: Using the Extended Euclidean Algorithm, 3953 x +1829 y = r q i 1 3953 1 1829 1- 2 295 2- 6 13 59 6 31- 67 5 Therefore 3953(- 6) + 1829(13) = 59 = gcd(3953 , 1829) so a (- 6) + b (13) = gcd( a,b ). Exercise 2-36: Find one integer solution, if possible, to the following Diophantine equation. 91 x + 126 y = 203 Solution: First, we use the Extended Euclidean Algorithm to find the gcd(91 , 126). 126 y +91 x = r q i 1 126 1 91 1- 1 35 1- 2 3 21 2 3- 4 14 1- 5 7 7 1 13- 18 2 Thus gcd(91 , 126) = 7 and 91(7) + 126(- 5) = 7. Since 203 = 7(29), then 7 | 203, so the Diophantine equation 91 x + 126 y = 203 has a solution. Since 91(7) + 126(- 5) = 7, then 91(29)(7) + 126(29)(- 5) = 203, or 91(203) + 126(- 145) = 203. Therefore, a solution is x = 203 and y =- 145. Check: 91 203- 126 145 = 18473- 18270 = 203. 1 Exercise 2-42: Find all the integer solutions of 169 x- 65 y = 91. Solution: Set w =- y . We use the Extended Euclidean Algorithm to find a solution to the linear Diophantine equation 169 x + 65 w = 91, and at the end, set y =- w : 169 x +65 w = r q i 1 169 1 65 1- 2 39 2- 1 3 26 1 2- 5 13 1- 5 13 2 Therefore, gcd(169 , 65) = 13 and 169(2) + 65(- 5) = 13. Since gcd(169 , 65) = 13, then gcd(169 ,- 65) = 13. Also, we have 169(2)- 65(5) = 13. Notice that 91 = 13 7, so the Diophantine equation 169 x- 65 y = 91 does have a solution. Therefore, a particular solution to 169 x- 65 y = 91 is x = 2 7 = 14, y = 5 7 = 35. This tells us that the complete integer solution to 169 x- 65 y = 91 is x = 14 + (- 65 13 ) n = 14- 5 n y = 35- ( 169 13 ) n = 35- 13 n for all n Z ....
View Full Document

Page1 / 8

assignment3_sol - MATH 135 Fall 2005 Assignment 3 Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online