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Published byAlvin Jones Modified over 9 years ago
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Chinese Remainder Theorem Ying Ding Junru Chen
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Chinese Remainder Theorem Sun Zi suanjing ( 孫子算經 The Mathematical Classic by Sun Zi) Shushu Jiuzhang ( 數書九章 Mathematical Treatise in Nine Sections) Simultaneous Congruence
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DEF: Congruence Modulo n
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Example 1:
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Method 1:
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Method 2:
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Chinese Remainder Theorem Let m 1,m 2,…,m n be pairwise relatively prime positive integers and a 1, a 2, …, a n arbitrary integers. Then the system x ≡ a 1 (mod m 1 ) x ≡ a 2 (mod m 2 ) : x ≡ a n (mod m n ) has a unique solution modulo m = m 1 m 2 …m n.
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Proof: Let M k = m / m k 1 k n Since m 1, m 2,…, m n are pairwise relatively prime, gcd ( M k, m k ) = 1 (by the Definition of relatively prime. P274) integer y k s.t. M k y k ≡ 1 (mod m k ) (by the theorem on gcd(a,b). P273) a k M k y k ≡ a k (mod m k ), 1 k n since a k M k y k ≡ 0(mod m i ), i ≠ k Let x = a 1 M 1 y 1 +a 2 M 2 y 2 +…+a n M n y n x ≡ a i M i y i ≡ a i (mod m i ) 1 i n x is a solution. All other solution y satisfies y ≡ x (mod m k ). m = m 1 m 2 …m n x ≡ a 1 (mod m 1 ) x ≡ a 2 (mod m 2 ) : x ≡ a n (mod m n )
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