Chinese Remainder Theorem/General Result

Theorem
Let $n_1, n_2, \ldots, n_r$ be positive integers such that $n_i \perp n_j$ for all $i \ne j$ (that is, $\gcd \left\{{n_i, n_j}\right\} = 1$).

Then the system of linear congruences:
 * $x \equiv b_1 \pmod {n_1}$
 * $x \equiv b_2 \pmod {n_2}$
 * $\ldots$
 * $x \equiv b_r \pmod {n_r}$

has a simultaneous solution which is unique modulo $\displaystyle \prod_{i \mathop = 1}^r n_i$.

Proof
Let $\displaystyle N = \prod_{i \mathop = 1}^r n_i$ and $N_k = \dfrac N {n_k}$ for $k = 1, 2, \ldots, r$.

Because $n_i \perp n_k$ for each $i \ne k$, we have:
 * $\gcd \left\{{N_k, n_k}\right\} = \gcd \left\{{n_1 n_2 \cdots n_{k-1} n_{k+1} \cdots n_r, n_k}\right\} = 1$.

So by Solution of Linear Congruence, the linear congruence $N_k x \equiv 1 \pmod {n_k}$ has a unique solution modulo $n_k$.

Let this solution be $x_k$.

So $N_k x_k \equiv 1 \pmod {n_k}$

We are going to show that $x_0 = b_1 N_1 x_1 + b_2 N_2 x_2 + \cdots + b_r N_r x_r$ satisfies each of the congruences.

So, we evaluate $x_0$ modulo $n_k$ for each $k = 1, 2, 3, \ldots, r$.

We start by noting that $i \ne k \implies n_k \mathop \backslash N_i$ so $N_i \equiv 0 \pmod {n_k}$.

So for each $k = 1, 2, \ldots, r$ we have $x_0 \equiv b_k N_k x_k$ because each of the remaining terms in the sum is congruent modulo $n_k$ to $0$.

Finally, since $x_k$ was found such that $N_k x_k \equiv 1 \pmod {n_k}$ we have $x_0 \equiv b_k \pmod {n_k}$ as we claimed.

All we need to do now is show that this solution we have discovered is unique modulo $N$.

So, suppose that $x'$ is a second solution of the system.

That is:
 * $x_0 \equiv x' \equiv b_k \pmod {n_k}$ for each $k = 1, 2, \ldots, r$.

So each $n_k$ divides $x' - x_0$.

But because each of the moduli are pairwise coprime, we have:
 * $\displaystyle \prod_{i \mathop = 1}^r n_i \mathop \backslash x' - x_0$

That is:
 * $\displaystyle x' \equiv x_0 \pmod {\prod_{i \mathop = 1}^r n_i}$

as we wanted to show.

Historical Note
This theorem was found in the Sun Tzu Suan Ching of Sun Tzu, who was active sometime between 200 and 500 C.E (nobody knows exactly when).