Homework 7 Solution

Homework 7 Solution - a h4 ig6 e k# 7 i b iC a aha 2.6...

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: a h4 ig6 e k# 7 i b iC a aha 2.6 Problem 18: We want to solve x 2 (mod 3) x 1 (mod 4) x 3 (mod 5) . We have a 1 = 2, a 2 = 1 and a 3 = 3. The moduli m 1 = 3, m 2 = 4 and m 3 = 5 are pairwise relatively prime, so we can use the Chinese Remainder Theorem. We see that a 1 = 2, a 2 = 1 and a 3 = 3. We let M 1 = m 2 m 3 = 20 , M 2 = m 1 m 3 = 15 , M 3 = m 1 m 2 = 12 . Then we find the inverses of each M i modulo m i . To find the inverse of M 1 modulo m 1 we first note that M 1 = 20 2 (mod 3). It is easy to see that the inverse of 2 is 2 itself (modulo 3). We let y 1 = 2. For M 2 we have M 2 = 15 3 (mod 4), and the inverse of 3 is 3 itself (because 3 3 = 9 1 (mod 4)). We let y 2 = 3. For M 3 we have M 3 = 12 2 (mod 5), and the inverse of 2 is 3 (because 3 2 1 (mod 5)). We let y 3 = 3. Now we let x = a 1 M 1 y 1 + a 2 M 2 y 2 + a 3 M 3 y 3 = 2 20 2 + 1 15 3 + 3 12 3 = 233 . Of course, we can add any multiple of m = m 1 m 2 m 3 = 3 4 5 = 60 to this and still get a solution. In fact, we note that 233 mod 60 = 53, so we might as well let x = 53. Actually, the problem asks for all solutions, so these would be x = 60 n + 53 for any integer n ....
View Full Document

This note was uploaded on 07/12/2008 for the course MAT 243 taught by Professor Callahan during the Spring '06 term at ASU.

Page1 / 4

Homework 7 Solution - a h4 ig6 e k# 7 i b iC a aha 2.6...

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