= Solution
Since $4^{-1}\equiv16\pmod{21}$ and $2^{-1}\equiv23\pmod{45}$, the two <linear congruences> reduce to
$$
x\equiv16\pmod{21},
\qquad
x\equiv25\pmod{45}.
$$
Write $x=16+21k$. The second congruence becomes
$$
21k\equiv9\pmod{45},
$$
or $7k\equiv3\pmod{15}$ after division by three. Since $7^{-1}\equiv13\pmod{15}$, $k\equiv9\pmod{15}$. The <Chinese remainder theorem> therefore gives the complete family
$$
\boxed{x\equiv205\pmod{315}}.
$$
The modulus is $\operatorname{lcm}(21,45)=315$ because the compatibility condition modulo their greatest common divisor is satisfied.
Back to article page
