Problem
NT-B1-M08-P018 A System Modulo \(900\)
#18
★★★☆☆ Level 3 of 5
Solve \(x\equiv3\pmod4\), \(x\equiv7\pmod9\), \(x\equiv12\pmod{25}\).
The moduli are pairwise coprime, so the answer will be modulo \(900\).
First solve \(x\equiv3\pmod4\), \(x\equiv7\pmod9\). The sequence \(7,16,25,34,43,\ldots\) modulo \(9\) first has residue \(3\) modulo \(4\) at \(7\). Thus \(x\equiv7\pmod{36}\). Now solve \(x=36k+7\equiv12\pmod{25}\). We get \(36k\equiv5\pmod{25}\), i.e. \(11k\equiv5\pmod{25}\). The inverse of \(11\) is \(16\), so \(k\equiv80\equiv5\pmod{25}\). Hence \(x=36(25t+5)+7=900t+187\). The answer is \(x\equiv187\pmod{900}\).
A more computational problem, but all steps are transparent.