This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 135 Midterm Exam, Winter 2010 Page 1 UNIVERSITY OF WATERLOO MIDTERM EXAMINATION WINTER TERM 2010 Last Name: First Name: Signature: Id.#: Indicate your instructor: D. Leung (Section 1) J. McKinnon (Section 2) J. Franklin (Section 3) P. Haxell (Section 4) J. Cheriyan (Section 5) Course Number MATH 135 Course Title Algebra Date of Exam February 8 2010 Time Period 7:009:00pm Number of Exam Pages 7 (including this cover sheet) Exam Type Closed Book Additional Materials Allowed Math Faculty approved calculators only Additional Instructions Write your answers in the space provided. If the space is insufficient, use the back of the page and indicate clearly where your solution continues. Show all your work. You may use the result of any theorem proved in class. State clearly which theorems you are using. Problem Value Mark Awarded Problem Value Mark Awarded 1 12 4 9 2 10 5 9 3 10 6 10 total 60 MATH 135 Midterm Exam, Winter 2010 Page 2 1. (a) Find all solutions to the linear Diophantine equation [7] 697 x + 561 y = 27472 . (Check your particular solution to make sure it is correct.) We begin by applying the Extended Euclidean Algorithm to find a solution to 697 x + 561 y =gcd(697,561). 1 697 1 561 11 1364 5 17 3341 From this we find that gcd(697,561)= 17, and one solution to 697 x + 561 y = 17 is given by x 1 = 4, y 1 = 5. Since 27472 = (17)(1616), we see that 17 divides 27472 and so the original equation does have solutions. Since 697( 4) + 561(5) = 17, by multiplying by 1616 we obtain 697( 6464) + 561(8080) = (17)(1616) = 27472. Therefore a particular solution to the original equation is given by x = 6464 and y = 8080. Then by our theorem on complete solutions to linear Diophantine equations, the complete solution is { ( x = 6464 + (561 / 17) n, y = 8080 (697 / 17) n ) : n Z } which equals { ( x = 6464 + 33 n, y = 8080 41 n ) : n Z } . (b) Find all positive solutions to 697 x + 561 y = 27472 . [3] To get both x and y to be positive, we need 6464 + 33 n > 0 and 8080 41 n > 0. Solving these gives 6464 33 < n < 8080 41 . Since 6464 33 195 . 88 and 8080 41 197 . 07, and since n is an integer, the only two values of n that yield positive solutions are n = 196 and n = 197. Therefore the only two positive solutions are ( x 2 , y 2 ) where x 2 = 6464 + (196)(33) = 4 and y 2 = 8080 (196)(41) = 44, and ( x 3 , y 3 ) where x 3 = 6464 + (197)(33) = 37 and y 3 = 8080 (197)(41) = 3. So we get (4 , 44) and (37 , 3). (c) Prove that there are no integers x and y that satisfy 697 x 2 + 561 y 2 = 27472 . [2] Suppose that a solution ( x, y ) exists. Since x 2 0 and y 2 0 for every integer, we find in particular that ( x 2 , y 2 ) is a nonnegative solution to the LDE in (a). We can quickly see) is a nonnegative solution to the LDE in (a)....
View Full
Document
 Spring '08
 ANDREWCHILDS
 Math

Click to edit the document details