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).
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