Problem
NT-B1-M04-P024 Mixed System
#24
★★★★★ Level 5 of 5
Find all \(x\pmod{420}\) satisfying \(x\equiv1\pmod4\), \(x\equiv2\pmod5\), \(x\equiv3\pmod7\), \(6x\equiv12\pmod9\).
First solve the last linear congruence.
Reduce \(6x\equiv12\pmod9\) by \(3\): \(2x\equiv4\pmod3\), so \(-x\equiv1\pmod3\), hence \(x\equiv2\pmod3\). Now solve the system with coprime moduli \(3,4,5,7\): \(x\equiv2\pmod3\), \(x\equiv1\pmod4\), \(x\equiv2\pmod5\), \(x\equiv3\pmod7\).
Combine \(x\equiv1\pmod4\) and \(x\equiv2\pmod5\): this gives \(x\equiv17\pmod{20}\). The number \(17\) also satisfies \(17\equiv2\pmod3\) and \(17\equiv3\pmod7\). Since \(3,4,5,7\) are pairwise coprime, the answer is unique modulo \(420\): \(x\equiv17\pmod{420}\).
Level 5 for Book 1: a linear congruence plus a four-modulus system.