The Chinese Remainder Theorem gives a unique solution to a set of linear congruences if their moduli are coprime.

This means, that given a set of arbitrary integers $a^{i}$, and pairwise coprime integers $n^{i}$, such that the following linear congruences hold:

Note "pairwise coprime integers" means that if we have a set of integers $\{n^{1}, n^{2}, ..., n^{i}\}$, all pairs of integers selected from the set are coprime: $\gcd(n^{i}, n^{j}) = 1$.

$x \equiv a^{1} \mod n^{1}$
$x \equiv a^{2} \mod n^{2}$
$\ldots$
$x \equiv a^{n} \mod n^{n}$

There is a unique solution $x \equiv a \mod N$ where $N = n^{1} \cdot n^{2} \cdot ... \cdot n^{n}$.

In cryptography, we commonly use the Chinese Remainder Theorem to help us reduce a problem of very large integers into a set of several, easier problems.

Given the following set of linear congruences:

$x \equiv 2 \mod 5$
$x \equiv 3 \mod 11$
$x \equiv 5 \mod 17$

Find the integer $a$ such that $x \equiv a \mod 935$

Starting with the congruence with the largest modulus, use that for $x \equiv a \mod p$ we can write $x = a + k \cdot p$ for arbitrary integer $k$.