This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: a Øh4Ô Ö iÒg6ÓÑ eÛ Ó Ök# 7 Ó ÐÙ Ø iÓÒ ×bÝ iÑC a ÐÐahaÒ § 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
 Spring '06
 CALLAHAN
 Remainder Theorem, Remainder, Prime number, M3

Click to edit the document details