MATH 216T Homework 1 Solutions 1. Find the greatest common ...

{x = 5000−17ty = −12000+41twhere t ∈ Z(b) We are interested in finding the solutions where x and y are positive. So5000−17t ≥ 0 and −12000+41t ≥ 0Hence we gett ≤ 5000 12000≈ 294.12 and t ≥ ≈ 292.6817 41Since t is an integer, we get that t ∈ {293,294}.There are only two couples (x,y) ∈ N 2 with 41x+17y = 10004. Nadir Airways offers three types of tickets on their Boston-new York flights. First-class tickets are $140, secondclasstickets are $110 and stand-by tickets are $78. If 69 passengerspay a total of $6548 for their tickets on a particularflight, how many of each type of ticket were sold? Show your work! Going over all combinations will not receive credit!Solution : Let x (resp. y and z) be the number first-class (resp. second-class and stand-by) tickets sold. Then wehave⎧⎨ x+y +z = 69 (1)140x+110y+78z = 6548 (2)⎩x,y,z ∈ NFirst, we find all integral solutions to (1) and (2). Solving (1) for z and substituting the result in (2), we get{z = 69−x−y140x+110y+78(69−x−y) = 6548or{ z = 69−x−y62x+32y = 1166We know how to find all the integral solutions of 62x+32y = 1166. Dividing both sides by 2, we get31x+16y = 583We start by calculating gcd(31,16) and writing it as a linear combination of 41 and 17.1 0 310 1 16 31 = 16·2+(−1) R 3 = R 1 −2R 2 and change signs−1 2 1 16 = 1·16+0 so we stopSo gcd(31,16) = 1. Since 1|583, the equation 31x + 16y = 583 will have integral solutions. To find a particularsolution, note that (−1)·31+2·16 = 1. Multiplying both sides by 583, we get that(−583)·31+1166·16 = 583So (−583,1166) is a particular solution of 31x+16y = 583. Hence all the integral solutions of 31x+16y = 583 aregiven by{ x = −583+16twhere t ∈ Zy = 1166−31t{ x+y +z = 69Substituting this result in the expression for z, we find that all the integral solutions of140x+110y+78z = 6548are given by⎧⎨ x = −583+16ty = 1166−31t where t ∈ Z⎩z = −514+15t2