x 2 mod 6 (1) (b) x 7 mod 15 (2) x 12 mod 25 (3) k 3 2 k mod 5 x = 7k + 5 = 7(5 l + 4) + 5 = 35 l l mod 3 2 l 2 mod 3

Size: px

Start display at page:

Download "x 2 mod 6 (1) (b) x 7 mod 15 (2) x 12 mod 25 (3) k 3 2 k mod 5 x = 7k + 5 = 7(5 l + 4) + 5 = 35 l l mod 3 2 l 2 mod 3"

Scarlett Newton
5 years ago
Views:

1 MATH 116 Homework 6 lutions 1. lve the following systems of congruences. x 1 2 (1) x 2 3 (2) (a) x 3 5 (3) x 5 7 (4) x 2 6 (1) (b) x 7 15 (2) x (3) (c) { 11x 2 15 (1) 13x 1 21 (2) lution : a) From (4), we get that x = 7k + 5 where k Z Substituting this into (3), we find 7k , we get 2k 3 5 Since 3 5 is the inverse of 2 5, we multiply both sides by 3 : k 3 2 k k = 5 l + 4 where l Z Substituting this into (2), we get 3, we have Dividing both sides by 2, we get x = 7k + 5 = 7(5 l + 4) + 5 = 35 l l l 1 l = 3m + 1 Substituting this into (1), we get 2, we have 2 l (3, 2) = 3 where m Z x = 35 l + 33 = 35(3m + 1) + 33 = 105m m m 1 2 m = 2n + 1 where n Z x = 105 m + 68 = 105(2n + 1) + 68 = 210n x = 210n where n Z 1

2 Note that we can rewrite this answer in ular form x b) From (3), we get that Substituting this into (2), we find 15, we get Dividing both sides by 10, we get x = 25k + 12 where k Z 25k k k k 1 k = 3 l + 1 Substituting this into (1), we get 6, we have Since (3, 6) = 3 and 3 1, there are no solutions 15 (10, 5) = 3 where l Z x = 25k + 12 = 25(3 l + 1) + 12 = 75 l l l l 1 6 No solutions c) To solve (2) for x, we need to find an inverse of = ( 5) R 3 = R 1 2R 2 and change signs = ( 2) R 4 = R 2 3R 3 and change signs = R 5 = R 3 2R = so we stop ( 8) 13 = 1. ( 8) and 8 is an inverse of Multiplying both sides of 13 x 1 21 by 8, we get x ( 8) 13 x ( 8) Substituting this into (1), we find x = 21k + 13 where k Z 11 (21k + 13) k

3 15, we get Dividing both sides by 6, we find 6k k 1 4 6k 6 15 k = 5 l + 4 Substituting this into our expression for x, we get Note that we can rewrite this answer in ular form 15 (15, 6) = 5 where l Z x = 21k + 13 = 21(5 l + 4) + 13 = 105 l + 97 x = 105 l + 97 where l Z x A band of 17 pirates has stolen a number of gold coins. When they divide the coins into equal piles, 3 coins are left over. When they fight over who should get the extra coins, one of the pirates is slain. The remaining pirates divide the coins again into equal piles and 10 coins are left over. Another round of fighting leaves 1 pirate dead. Now they can finally divide the coins into equal piles with no coins remaining. What is the least number of gold coins that the pirates stole? lution : Let x be the number of gold coins. Then we have that x 3 17 (1) x (2) x 0 15 (3) x 0 From (1), we get that Substituting this in (2), we get 16, we find Substituting this in our expression for x, we get x = 17k + 3 where k Z 17k k k 7 16 k = 16 l + 7 where l Z x = 17k + 3 = 17(16 l + 7) + 3 = 272 l Substituting this in (3), we find 15, we get 272 l l

4 Dividing by 2, we have l l 2 15 l = 15m + 14 Substituting this into the expression for x, we get 15 (15, 2) = 15 where m Z x = 272 l = 272(15m + 14) = 4080m we get the smallest number of gold coins when m = 0. The smallest number of gold coins is In this exercise, you will solve the equation x 2 + x (a) By trial and error, find all the solutions of x 2 + x 44 8 (b) By trial and error, find all the solutions of x 2 + x 44 7 (c) Using the Chinese remainder Theorem, find all the solutions of x 2 + x lution : a) x 3 8 or x 4 8 b) x 1 7 or x 5 7 c) Since 56 = 8 7 and (8, 7) = 1, we have that x 2 + x x 2 + x 44 8 and x 2 + x 44 7 (x 3 8 or x 4 8) and (x 1 7 or x 5 7) { { { { x 3 8 x 3 8 x 4 8 x 4 8 or or or x 1 7 x 5 7 x 1 7 x 5 7 We solve the first system. The other systems are similar. From the first equation, we get that Substituting this into the second equation, we get x = 8k + 3 where k Z 8k , we find k

5 k k = 7 l + 5 where l Z Substituting this in our expression for x, we get x = 8k + 3 = 8(7 l + 5) + 3 = 56 l + 43 x x or x or x or x Use Hensel s Method to solve the following equations: (a) x 2 + x (b) x 2 2x lution : a) We put f(x) = x 2 + x Then f (x) = 2x + 1. x 2 + x By trial and error, we get x 1 3 x 2 + x f (1) = and f(1) = t is a solution for all t. x 1 9 or x 4 9 or x 7 9 x 2 + x f (1) = and f(1) = so no solutions of the form 1 + 9t. f (4) = and f(4) = t is a solution for all t. f (7) = and f(7) = so no solutions of the form 7 + 9t. x 4 27 or x or x x 2 + x f (4) = and f(4) = so no solutions of the form t. f (13) = and f(13) = so no solutions of the form t. f (22) = and f(22) = so no solutions of the form t b) We put f(x) = x 2 2x 35. Then f (x) = 2x 2. x 2 2x By trial and error, we get No solutions x 0 5 or x 2 5 5

6 x 2 2x f (0) = and f(0) = t is a solution where t ( 7) f (2) = 2 5 and f(2) = t is a solution where t ( 7) x 7 25 or x x 2 2x f (7) = and f(7) = 0 so t is a solution where t f (20) = and f(20) = t is a solution where t x or x x 2 2x f (7) = and f(7) = 0 so t is a solution where t f (120) = and f(20) = t is a solution where t x or x Remark: Note that x 2 2x 35 = 0 x = 7 or x = 5. x = 7 and x = 5 are always solutions of x 2 2x k for k = 1, 2,.... It turns out that they are the only solutions for any k. 6

Solutions to Section 2.1 Homework Problems S. F. Ellermeyer

Solutions to Section 2.1 Homework Problems S. F. Ellermeyer Solutions to Section 21 Homework Problems S F Ellermeyer 1 [13] 9 = f13; 22; 31; 40; : : :g [ f4; 5; 14; : : :g [3] 10 = f3; 13; 23; 33; : : :g [ f 7; 17; 27; : : :g [4] 11 = f4; 15; 26; : : :g [ f 7;