The Chinese Remainder Theorem provides a method to solve systems of simultaneous congruences. Given residual numbers r1, r2, ..., rk with mutually prime moduli m1, m2, ..., mk, it finds a number x such that x is congruent to ri modulo mi for each i. It does this by first computing values Si and Mi, then taking x to be the sum of MiSiri terms modulo the product of all the moduli. This allows solving congruences modulo large composite numbers more efficiently than directly working modulo the full modulus.

1. Chinese Remainder Theorem
Given a residual number (r1, r2, …, rk) with
moduli (m1, m2, …, mk), where all mi are
mutually prime, set M= m1×m2× …×mk, and
Mi=M/mi.
Let Si be the solution that (Mi×Si)%mi = 1
Then we have the corresponding number
x = (∑i=1,k(Mi Si ri))%M.
1

2. Example
Given (m1,m2,m3)=(2,3,7), M=2×3×7=42, we have
M1=m2×m3=3×7=21 (M1S1)%m1=(21S1)%2=1
M2=m1×m3=2×7=14 (M2S2)%m2=(14S2)%3=1
M3=m1×m2=2×3=6 (M3S3)%m3=(6S3)%7=1
Thus, (S1, S2, S3) = (1,2,6)
For a residual number (0,2,1):
x=(M1S1r1 + M2S2r2 + M3S3r3)%M
=(21×1×0 + 14×2×2 + 6×6×1 )%42
= ( 0 + 56 + 36 )%42 = 92%42 = 8
2

3. Example
For a residual number (1,2,5):
• x=(M1S1r1 + M2S2r2 + M3S3r3)%M
= (21×1×1 + 14×2×2 + 6×6×5)%42
= (21 + 56 + 180)%42
= 257%42 = 5
3

4. Proof of Chinese Remainder
Theorem
Let A = ∑i=1,k(Mi Si ri), we show that
1. A%mv = rv and 2. x=A%M is unique.
1. A%mv= (∑i=1,k(Mi Si ri) )% mv
= (Σ(MiSiri) % mv)%mv = (MvSvrv)%mv
= [(MvSv)%mv × rv%mv ]%mv = rv%mv = rv
2. Proof was shown in lecture 5.
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5. Chinese Remainder Theorem
 used to speed up modulo computations
 if working modulo a product of numbers
 e.g., mod M, where M = m1m2..mk
 Chinese Remainder theorem lets us work
in each modulus mi separately
 since computational cost is proportional to
size, this is faster than working in the full
modulus M

6. Chinese Remainder Theorem
 can implement CRT in several ways
 to compute A(mod M)
 first compute all ai = A mod mi separately
 determine constants ci below, where Mi = M/mi
 then combine results to get answer using: