This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full 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.
 Spring '06
 CALLAHAN
 Remainder Theorem, Remainder

Click to edit the document details