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Unformatted text preview: MATH 135 Fall 2005 Assignment 3 Solutions HandIn Problems Exercise 211: Prove that gcd( ad,bd ) =  d  Â· 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 Â· gcd( a,b ) =  d  Â· 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 224: If a = 3953 and b = 1829, write gcd( a,b ) in the form ax + by where x,y âˆˆ 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 236: 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 Â· 203 126 Â· 145 = 18473 18270 = 203. 1 Exercise 242: 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 Â· 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 Â· 7 = 14, y = 5 Â· 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 âˆˆ Z ....
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 Spring '08
 Vanderburgh
 Greatest common divisor, Euclidean algorithm, gcd, Diophantine

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