Problem
NT-B1-M08-P003 System Modulo \(8\) and \(9\)
#3
★☆☆☆☆ Level 1 of 5
Solve \(x\equiv5\pmod8\), \(x\equiv7\pmod9\).
Let \(x=8k+5\).
Substitute: \(8k+5\equiv7\pmod9\), so \(8k\equiv2\pmod9\). Since \(8\equiv-1\), \(-k\equiv2\), hence \(k\equiv7\pmod9\). Then \(x=8(9t+7)+5=72t+61\). The answer is \(x\equiv61\pmod{72}\).
First solution by substitution.