Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 4 / 1E / c / Solution

Write $x=11+100t$. Reduction of the other integer congruence modulo $79$ gives $21t\equiv -3(\mathrm{mod}79)$. The Bézout's identity $19\cdot 79-15\cdot 100=1$ gives $-15\cdot 21\equiv 1(\mathrm{mod}79)$, so $t\equiv 45(\mathrm{mod}79)$. Consequently,Indeed, $4511=8+79\cdot 57=11+100\cdot 45$. The Chinese remainder theorem guarantees one class modulo $79\cdot 100$, since the two moduli are coprime.