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**Unformatted text preview: **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 mod49 x x ! ! ( ) 23 mod100 100 23 x x k ! " = + ( ) 100 23 31 mod49 k + ! , now reduce mod49 to get 2 K + 23 ! 31 mod49 ( ) " 2 K ! 8 mod49 ( ) Since ( ) gcd 2,49 1 = , we can divide both sides by 2 . ( ) 4 mod49 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 = . McCombs Math 381 The Chinese Remainder Theorem 2 (b) ( ) ( ) 13 mod27 7 mod16 x x !...

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