Find all $x$ such that
\begin{align} x&\equiv 1 \pmod {12}\\ x&\equiv 4 \pmod {21}\\ x&\equiv 18 \pmod {35} \end{align}
Im not quite sure if this system of linear congruence is solvable. Since $\gcd(12,21) =3$, $\gcd (12,35)=1$ and $\gcd(21,35) = 7$, and the CRT states that "If(m1, m2) = 1, then the system has its complete solution a single resident class (mod m1.....mr).