Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 1E (Numbers and Sets)

Since $4^{-1}\equiv 16(\mathrm{mod}21)$ and $2^{-1}\equiv 23(\mathrm{mod}45)$, the two linear congruences reduce toWrite $x=16+21k$. The second congruence becomesor $7k\equiv 3(\mathrm{mod}15)$ after division by three. Since $7^{-1}\equiv 13(\mathrm{mod}15)$, $k\equiv 9(\mathrm{mod}15)$. The Chinese remainder theorem therefore gives the complete familyThe modulus is $\mathrm{lcm}(21,45)=315$ because the compatibility condition modulo their greatest common divisor is satisfied.