{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# chinesekey - McCombs Math 381 The Chinese Remainder Theorem...

This preview shows pages 1–3. Sign up to view the full content.

McCombs Math 381 The Chinese Remainder Theorem 1 Basic Idea: We want to devise a strategy for finding simultaneous solutions to a system of linear congruences. Chinese Remainder Theorem: Given pairwise relatively prime positive integers 1 2 3 , , ..., k n n n n , and integers 1 2 3 , , ..., k a a a a . The system of congruences ( ) ( ) ( ) 1 1 2 2 mod mod . . . mod k k x a n x a n x a n ! ! ! has a unique simultaneous solution ( ) mod x n , where 1 2 3 k n n n n n = ! ! !!! . Moreover, ( ) mod x n is the smallest positive integer that solves the system. Examples: 1. Solve each system. (a) ( ) ( ) 23 mod100 31 mod 49 x x ! ! ( ) 23 mod100 100 23 x x k ! " = + ( ) 100 23 31 mod 49 k + ! , now reduce mod 49 to get 2 k + 23 ! 31 mod49 ( ) " 2 k ! 8 mod49 ( ) Since ( ) gcd 2,49 1 = , we can divide both sides by 2 . ( ) 4 mod 49 49 4 k k m ! " = + x = 100 k + 23 = 100 49 m + 4 ( ) + 23 = 4900 + 423 Thus, the smallest positive integer solution is 423 x = .

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

View Full Document