Chinese Remainder Theorem

Ying DingJunru Chen

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

DEF: Congruence Modulo n

•For integers x & y and positive integer n,

Example 1:

•Solve X:

Method 1:

•Enumeration

X: 5, 8, 11, 14, 17, 20, 23…

X: 3, 10, 17, 24, 31…

Method 2:

•Chinese Remainder Theorem

M= a + b

a = 14 b = 24

Chinese Remainder Theorem

•Let m1,m2,…,mn be pairwise relatively prime positive integers and a1, a2, …, an arbitrary integers. Then the system

x ≡ a1 (mod m1)x ≡ a2 (mod m2)

:x ≡ an (mod mn)

has a unique solution modulo m = m1m2…mn.

Proof: Let Mk = m / mk 1 k n Since m1, m2,…, mn are pairwise relatively

prime, gcd (Mk , mk) = 1

(by the Definition of relatively prime. P274) integer yk s.t. Mk yk ≡ 1 (mod mk)

(by the theorem on gcd(a,b). P273) ak Mk yk ≡ ak (mod mk) , 1 k n

since ak Mk yk ≡ 0(mod mi), i ≠ k

Let x = a1 M1 y1+a2 M2 y2+…+an Mn yn

x ≡ ai Mi yi ≡ ai (mod mi) 1 i nx is a solution. All other solution y satisfies y ≡ x (mod mk).

m = m1m2…mn

x ≡ a1 (mod m1)x ≡ a2 (mod m2):x ≡ an (mod mn)

Han Xin Count Solders

•M= a + b + c

a = 35 b = 84 c = 90

Since

Thus n = 8 and X = 1049