Math 5410 Chinese Remainder Theorem

Chinese Remainder Theorem

Theorem: Suppose that m₁, m₂, ..., m_r are pairwise relatively prime positive integers, and let a₁, a₂, ..., a_r be integers. Then the system of congruences, x = a_i (mod m_i) for 1 <= i <= r, has a unique solution modulo M = m₁ x m₂ x ... x m_r, which is given by:
x = a₁M₁y₁ + a₂M₂y₂ + ... + a_rM_ry_r (mod M), where M_i = M/m_i and y_i = (M_i)^-1 (mod m_i) for 1 <= i <= r.

Pf: Notice that gcd(M_i, m_i) = 1 for 1 <= i <= r. Therefore, the y_i all exist (and can be determined easily from the extended Euclidean Algorithm). Now, notice that since M_iy_i = 1 (mod m_i), we have a_iM_iy_i = a_i (mod m_i) for 1 <= i <= r. On the other hand, a_iM_iy_i = 0 (mod m_j) if j is not i (since m_j | M_i in this case). Thus, we see that x = a_i (mod m_i) for 1 <= i <= r.

If there were two solutions, say x₀, and x₁, then we would have x₀ - x₁ = 0 (mod m_i) for all i, so x₀ - x₁ = 0 (mod M), i.e., they are the same modulo M.

Example

Find the smallest multiple of 10 which has remainder 2 when divided by 3, and remainder 3 when divided by 7.

We are looking for a number which satisfies the congruences, x = 2 mod 3, x = 3 mod 7, x = 0 mod 2 and x = 0 mod 5. Since, 2, 3, 5 and 7 are all relatively prime in pairs, the Chinese Remainder Theorem tells us that there is a unique solution modulo 210 ( = 2x3x5x7). We calculate the M_i's and y_i's as follows:
M₂ = 210/2 = 105; y₂ = (105)^-1 (mod 2) = 1
M₃ = 210/3 = 70; y₃ = (70)^-1 (mod 3) = 1
M₅ = 210/5 = 42; y₅ = (42)^-1 (mod 5) = 3 and
M₇ = 210/7 = 30; y₇ = (30)^-1 (mod 7) = 4.

So, x = 0(M₂y₂) + 2(M₃y₃) + 0(M₅y₅) + 3(M₇y₇) = 0 + 2(70)(1) + 0 + 3(30)(4) = 140 + 360 = 500 mod 210 = 80.

Remark 1: The theorem is valid in much more general situations than we have presented here.

Remark 2: The condition given is sufficient, but not necessary for a solution. Necessary and sufficient conditions exist but we are not presenting them.

Remark 3: It is purported that Sun Tsu was aware of this result in the first century A.D.