Find the set of integers $a$ such that 35 divides $a^{3}-1$.
We could look at the 35 congruence classes modulo 35 one by one. However, there is a faster way using the Chinese Remainder Theorem.

Indeed, since $35=5 \times 7$, $a^{3} \equiv 1(\bmod 35)$ if and only if $a^{3} \equiv 1(\bmod 5)$ and $a^{3} \equiv 1(\bmod 7)$. We then check by hand that $0^{3} \equiv 0(\bmod 7)$, that $1^{3} \equiv 2^{3} \equiv 4^{3} \equiv 1(\bmod 7)$, and that $3^{3} \equiv 5^{3} \equiv 6^{3} \equiv 6(\bmod 7)$; we also check that $0^{3} \equiv 0(\bmod 5)$, that $1^{3} \equiv 1(\bmod 5)$, that $2^{3} \equiv 3(\bmod 5)$, that $3^{3} \equiv 2(\bmod 5)$, and that $4^{3} \equiv 4(\bmod 5)$.

Thus, $a$ is a solution to the problem if and only if $a \equiv 1, 2$, or 4 (mod 7), and $a \equiv 1$ (mod 5). According to the Chinese Remainder Theorem, these conditions are equivalent to saying that $a \equiv 1$, 11, or 16 (mod 35). This means that 35 divides $a^{3}-1$ if and only if $a$ is of the form $35 \ell+1$, $35 \ell+11$, or $35 \ell+16$.