4.2 question 1 letters d,e and f 160CongruencesHence, 1=5-2.2=5- (7-5- 1) 2=6-3-2-7=(12 – 7 - 1) · 3 - 2-7= 12 .3-5-7. Therefore, a particular solution to the linear diophantine equation is xo =-20and yo-12. Hence, all solutions of the linear congruences are given by x =-20 =4 (mod 12).%3DLater e wili want to know which integers are their own inverses modulopis prime. The following theorem tells us which integers have this property.P,whereTheorem 4.12. Let p be prime. The positive integer a is its own inverse modulo p ifand only if a = i (mod p) or a =- (mod p).Proof. lf a =1 (mod p) or a = -1 (mod p), then a =1 (mod p), so that a is its owninverse modulo p.Conversely, if a is its own inverse modulo p, then a² = a · a = 1(mod p). Hence,P| (a² – 1). Because a - 1= (a – 1)(a + 1), this implies that p| (a – 1) or p | (a + 1).Therefore, a = 1 (mod p) or a =-1 (mod p).%3D4.2 EXERCISES1. Find all solutions of each of the following linear congruences.c) 19x = 30 (mod 40)a) 2x = 5 (mod 7)e) 103x = 444 (mod 999)b) 3x = 6 (mod 9)d) 9x = 5 (mod 25)f) 980x = 1500 (mod 1600)2. Find all solutions of each of the following linear congruences.a) 3x = 2 (mod 7)c) 17x = 14 (mod 21)e) 128x = 833 (mod 1001)b) 6x = 3 (mod 9)d) 15x =9 (mod 25) f) 987.x 610 (mod 1597)3. Find all solutions to the congruence 6,789,783x = 2,474,010 (mod 28,927,591).4. Suppose that p is prime and that a and b are positive integers with (p, a) = 1. The followingmethod can be used to solve the linear congruence ax = b (mod p).a) Show that if the integer x is a solution of axr b (mod p), then x is also a solution of thelinear congruenceax = -b(m/a] (mod p),where a, is the least positive residue of p modulo a. Note that this congruence is ofthe same type as the original congruence, with a positive integer smaller than a as thecoefficient of x.b) When the procedure of part (a) is iterated, one obtains a sequence of linear congruenceswith coefficients of x equal to ao =aa,az .Show that there is a positive integern with anI, so that at the nth stage, one obtains a linear congruence x = B (mod pl-%3Dc) Use the method described in part (b) to solve the linear congruence 6.x =7(mod 25.5. An astronomer knows that a satellite orbits the Earth in a period that is an exact multipieI hour that is less than I day. If the astronomer notes that the satellite completes 11 orbits inan interval that starts when a 24-hour clock reads 0 hours and ends when the clock reads 17bours, how long is the orbital period of the satellite?

Question

4.2 question 1 letters d,e and f 160CongruencesHence, 1=5-2.2=5- (7-5- 1) 2=6-3-2-7=(12 – 7 - 1) · 3 - 2-7= 12 .3-5-7. Therefore, a particular solution to the linear diophantine equation is xo =-20and yo-12. Hence, all solutions of the linear congruences are given by x =-20 =4 (mod 12).%3DLater e wili want to know which integers are their own inverses modulopis prime. The following theorem tells us which integers have this property.P,whereTheorem 4.12. Let p be prime. The positive integer a is its own inverse modulo p ifand only if a = i (mod p) or a =- (mod p).Proof. lf a =1 (mod p) or a = -1 (mod p), then a =1 (mod p), so that a is its owninverse modulo p.Conversely, if a is its own inverse modulo p, then a² = a · a = 1(mod p). Hence,P| (a² – 1). Because a - 1= (a – 1)(a + 1), this implies that p| (a – 1) or p | (a + 1).Therefore, a = 1 (mod p) or a =-1 (mod p).%3D4.2 EXERCISES1. Find all solutions of each of the following linear congruences.c) 19x = 30 (mod 40)a) 2x = 5 (mod 7)e) 103x = 444 (mod 999)b) 3x = 6 (mod 9)d) 9x = 5 (mod 25)f) 980x = 1500 (mod 1600)2. Find all solutions of each of the following linear congruences.a) 3x = 2 (mod 7)c) 17x = 14 (mod 21)e) 128x = 833 (mod 1001)b) 6x = 3 (mod 9)d) 15x =9 (mod 25) f) 987.x 610 (mod 1597)3. Find all solutions to the congruence 6,789,783x = 2,474,010 (mod 28,927,591).4. Suppose that p is prime and that a and b are positive integers with (p, a) = 1. The followingmethod can be used to solve the linear congruence ax = b (mod p).a) Show that if the integer x is a solution of axr b (mod p), then x is also a solution of thelinear congruenceax = -b(m/a] (mod p),where a, is the least positive residue of p modulo a. Note that this congruence is ofthe same type as the original congruence, with a positive integer smaller than a as thecoefficient of x.b) When the procedure of part (a) is iterated, one obtains a sequence of linear congruenceswith coefficients of x equal to ao =a>a,>az > .Show that there is a positive integern with anI, so that at the nth stage, one obtains a linear congruence x = B (mod pl-%3Dc) Use the method described in part (b) to solve the linear congruence 6.x =7(mod 25.5. An astronomer knows that a satellite orbits the Earth in a period that is an exact multipieI hour that is less than I day. If the astronomer notes that the satellite completes 11 orbits inan interval that starts when a 24-hour clock reads 0 hours and ends when the clock reads 17bours, how long is the orbital period of the satellite?

Accepted Answer

To find the solution of the linear congruence: 9x≡5mod 25.
A linear congruence in one variable x is…

1. Find all solutions of each of the following linear congr a) 2x = 5 (mod 7) c) 19x = 30 (mod 40) e) 103x = 444 (mod 999) b) 3x = 6 (mod 9) d) 9x = 5 (mod 25) f) 980x = 1500 (mod 1600)