Question

Find all numbers $r$ for which the system of congruences
\begin{align*}
x &\equiv r \pmod{6}, \\
x &\equiv 9 \pmod{20}, \\
x &\equiv 4 \pmod{45}
\end{align*}
has a solution.

Answer 1

we'll solve using the Chinese remainder theorem first $$x\equiv 9\pmod{20}\\x\equiv4\pmod{45}$$... which gives \(x=180n+49\). now observe \(\pmod6\): $$180n+49\equiv49\equiv1\pmod6$$ so all \(r\equiv1\pmod6\) give a solution