Problem
NT-B1-M08-P009 Another Compatible System
#9
★★☆☆☆ Level 2 of 5
Solve \(x\equiv4\pmod6\), \(x\equiv10\pmod{15}\).
The two residues are congruent modulo \(3\).
Let \(x=6k+4\). Then \(6k+4\equiv10\pmod{15}\), so \(6k\equiv6\pmod{15}\). Divide by \(3\): \(2k\equiv2\pmod5\), hence \(k\equiv1\pmod5\). Then \(x=6(5t+1)+4=30t+10\). The answer is \(x\equiv10\pmod{30}\).
Useful practice dividing a congruence by a common divisor.