7 - Math 135 Assignment 7 Solutions Winter 2009 1. (a)...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 135 Assignment 7 Solutions Winter 2009 1. (a) Using the Extended Euclidean Algorithm, we can find the gcd(1653,2000) and the required inte- gers x and y such that 1653 x + 2000 y = 77. x column y column r column 1 2000 1 1653- 1 1 347 5- 4 265- 6 5 82 23- 19 19- 98 81 6 317- 262 1 Thus, gcd(1653 , 2000) = 1 and since 1 | 77 the congruence has solutions. A particular solution of 1653 x 1(mod 2000) is x = 317. Thus, a particular solution of 1653 x 77(mod 2000) is x = 317 77 = 24409. Since 24409 409(mod 2000), the complete solution is x 409(mod 2000). (b) Since 2 | 1492 and 2 | 2000, 2 | gcd(1492 , 2000). But 2- 77, and thus, the congruence has no solution. (c) Using the table approach and working modulo 12: x 1 2 3 4 5 6 7 8 9 10 11 x 2- 4 x 9 8 9 5 9 8 9 5 Thus, the complete solution is x , 4 , 6 , 10(mod 12). (d) Checking x 0, we have 5 6 2(mod 11). Thus, 11- x and using Fermats Little Theorem we have x 10 1(mod 11) and x 11 x (mod 11)....
View Full Document

Page1 / 2

7 - Math 135 Assignment 7 Solutions Winter 2009 1. (a)...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online