MATH 216T Homework 1 Solutions

Size: px

Start display at page:

Download "MATH 216T Homework 1 Solutions"

Mervin Simmons
5 years ago
Views:

1 MATH 216T Homew 1 Solutions 1. Find the greatest ommon divis of 5321 and 1235 and write it as a linear omination of 5321 and Solution : Using our implementation of the Eulidean Algithm, we easily get Hene = R 3 = R 1 R = R = R 2 2R = R 5 = R 3 2R = R 6 = R 16R = so we stop gd(1235,5321)= 3 = ( 822) Reall the Division Algithm : Let a, Z with 0. Then there exist q,r Z suh that a = q +r and 0 r <. Use this to prove the following : Let a, Z with 0. Then there exist q,r Z suh that a = q +r and 0 r 2. Proof : Let q (resp. r ) e the quotient (resp. remainder) of a divided y. So q,r Z, a = q +r and 0 r <. Suppose first that r 2. Put q = q and r = r. Then q,r Z, a = q +r and 0 r 2. Suppose next that r > 2. Put r = r and q = q q +sgn() = +1 if > 0 q 1 if < 0 Then q,r Z. Sine 2 < r <, we get that 2 < r <. So 2 Finally, a = q +r = (q + )+(r ) = q +r. < r < 0. Hene 0 < r < Consider the diophantine equation 1x+17y = 1000 (a) Find all (x,y) Z 2 with 1x+17y = () How many ouples (x,y) N 2 are there with 1x+17y = 1000? Solution : (a) We start y alulating gd(1,17) and writing it as a linear omination of 1 and = R 3 = R 1 2R = R = R 2 2R = R 5 = R 3 2R = so we stop So gd(1,17) = 1. Sine , the equation 1x + 17y = 1000 will have integral solutions. To find a partiular solution, note that 5 1+( 12) 17 = 1. Multiplying oth sides y 1000, we get that ( 12000) 17 = 1000 So (5000, 12000) is a partiular solution of 1x+17y = Hene all the integral solutions of 1x+17y = 1000 are given y 1

2 x = t y = t where t Z () We are interested in finding the solutions where x and y are positive. So t 0 and t 0 Hene we get t and t Sine t is an integer, we get that t 293,29}. There are only two ouples (x,y) N 2 with 1x+17y = Nadir Airways offers three types of tiets on their Boston-new Y flights. First-lass tiets are $10, seondlass tiets are $110 and stand-y tiets are $78. If 69 passengerspay a total of $658 f their tiets on a partiular flight, how many of eah type of tiet were sold? Show your w! Going over all ominations will not reeive redit! Solution : Let x (resp. y and z) e the numer first-lass (resp. seond-lass and stand-y) tiets sold. Then we have x+y +z = 69 (1) 10x+110y+78z = 658 (2) x,y,z N First, we find all integral solutions to (1) and (2). Solving (1) f z and sustituting the result in (2), we get z = 69 x y 10x+110y+78(69 x y) = 658 z = 69 x y 62x+32y = 1166 We now how to find all the integral solutions of 62x+32y = Dividing oth sides y 2, we get 31x+16y = 583 We start y alulating gd(31,16) and writing it as a linear omination of 1 and = 16 2+( 1) R 3 = R 1 2R 2 and hange signs = so we stop So gd(31,16) = 1. Sine 1 583, the equation 31x + 16y = 583 will have integral solutions. To find a partiular solution, note that ( 1) = 1. Multiplying oth sides y 583, we get that ( 583) = 583 So ( 583,1166) is a partiular solution of 31x+16y = 583. Hene all the integral solutions of 31x+16y = 583 are given y x = t where t Z y = t x+y +z = 69 Sustituting this result in the expression f z, we find that all the integral solutions of 10x+110y+78z = 658 are given y x = t y = t where t Z z = 51+15t 2

3 Sine we are interested in natural solutions, we must have that x 0, y 0 and z 0. Hene t 0 and t 0 and 51+15t 0 So t = and t Sine t Z, we must have that t = 37. Hene 37.6 and t x = = 9, y = = 19 and z = = 1 9 first-lass tiets, 19 seond-lass tiets and 1 stand-y tiets were sold. 5. Let a, N 0 with a 3 2. Use unique prime fatization to prove that a. Proof : Put a = p α1 1 pα2 2 pα and = p β1 1 pβ2 2 pβ where is a positive integer, p 1 < p 2 < < p are primes and α i,β i are nonnegative integers f i = 1,2,...,. Then Sine a 3 divides 2, we get that So a 3 = p 3α1 1 p 3α2 2 p 3α sine β i 0 f i = 1,2,...,. But a = p α1 1 pα2 2 pα Sine α i β i f i = 1,2,...,, we get that a divides. and 2 = p 2β1 1 p 2β2 2 p 2β 3α i 2β i f i = 1,2..., α i 2 3 β i β i f i = 1,2,..., and = p β1 1 pβ2 2 pβ 6. Let a,, N 0 with gd(a,) = 1. (a) Let r > 0. Prove that the line ax+y = intersets the irle x 2 +y 2 = r 2 in two different points if and only if r > a () Prove that the equation ax+y = has a solution (x 0,y 0 ) Z 2 suh that x 2 0 +y0 2 2 a a2 + 2 Proof : (a) There will e one point of intersetion if the line is tangent to the irle. We have the following piture : Y x 2 +y 2 = r 2 ax+y = X 3

4 ( ) ( We easily get that the X-interept of ax+y = is a,0 while the Y-interept is 0, ). Hene P(0, ) R r O(0, 0) Q( a,0) Using similar triangles, we get that OR OQ = OP PQ. Hene r = OR = OQ OP PQ = a ( a) 2 + ( ) 2 = = a2 + 2 Hene the line ax+y = will interset the irle x 2 +y 2 = r 2 in two different points if and only if r > a Note that we an get this result pure algeraially. There are two points of intersetion if and only if the system ax+y = (1) x 2 +y 2 = r 2 (2) has two solutions. It follows from (1) that y = ax. Sustituting this into (2), we find ( ) 2 ax x 2 + = r 2 (a )x 2 2ax+ 2 2 r 2 = 0 This quadrati equation is x has two solutions if and only if it s disriminant is stritly positive : ( 2a) 2 (a )( 2 2 r 2 ) > 0 Solving f r 2, we get r 2 > 2 a () Put r = a a Sine r > x 2 +y 2 = r 2 in two different points, say P 1 (x 1,y 1 ) and P 2 (x 2,y 2 ). Suppose we an prove that the distane d etween P 1 and P 2 is at least a Then it follows from what we have seen in lass that there exists a ouple (x 0,y 0 ) Z 2 suh that ax 0 + y 0 = and the point P(x 0,y 0 ) elongs to the line segment P 1 P 2. Hene P lies inside the irle x 2 +y 2 = r 2. So it follows from (a) that the ax+y = will interset the irle a2 +2, x 2 0 +y0 2 r 2 = 2 a a2 + 2 It remains to prove that d a This is equivalent to showing that d 2 a We have the following piture :

5 ax+y = P 1 Y x 2 +y 2 = r 2 R P 2 X O P 1 r R r O It follows from (a) that OR = Hene a2 +2. ( d 2 = RP 1 2 = (r 2 OR 2 2 ) = a a2 + 2 ) 2 a = a P 2 Again, we an derive this result algeraially. Using the Distane Fmula, we get that d 2 = (x 2 x 1 ) 2 +(y 2 y 1 ) 2 Note that (x 1,y 1 ) and (x 2,y 2 ) are the solutions of ax+y = (1) x 2 +y 2 = r 2 (2) It follows from (1) that y = ax. Sustituting this into (2), we find Hene So ( ) 2 ax x 2 + = r 2 (a )x 2 2ax+ 2 2 r 2 = 0 x 1,x 2 = a± (a) 2 (a )( 2 2 r 2 ) a = = a± 1 2 (a2 + 2 ) a (x 2 x 1 ) 2 = ( a+ 1 2 (a2 + 2 ) a a 1 2 (a2 + 2 )2 ) a = 2 5

6 Similarly, we get that (y 2 y 1 ) 2 = a 2 (so we would solve (1) f x and sustitute into (2); this omes down to interhanging a and ). Hene d 2 = (x 2 x 1 ) 2 +(y 2 y 1 ) 2 = 2 +a 2 6

Intermediate Math Circles November 18, 2009 Solving Linear Diophantine Equations

Intermediate Math Circles November 18, 2009 Solving Linear Diophantine Equations 1 University of Waterloo Faulty of Mathematis Centre for Eduation in Mathematis and Computing Intermediate Math Cirles November 18, 2009 Solving Linear Diophantine Equations Diophantine equations are equations