Problem
NT-B1-M04-P011 Compatible Moduli 6 and 9
#11
★★☆☆☆ Level 2 of 5
Solve \(x\equiv4\pmod6\), \(x\equiv1\pmod9\).
Let \(x=6k+4\).
Substitute: \(6k+4\equiv1\pmod9\), so \(6k\equiv6\pmod9\). Divide by \(3\): \(2k\equiv2\pmod3\), hence \(k\equiv1\pmod3\). Then \(x=6(3t+1)+4=18t+10\). The answer is \(x\equiv10\pmod{18}\).
The answer is modulo the LCM of the moduli.