Problem
NT-B1-M04-P006 Congruence with Two Solutions
#6
★★☆☆☆ Level 2 of 5
Solve \(4x\equiv6\pmod{10}\).
Divide \(4,6,10\) by their common divisor \(2\).
\(\gcd(4,10)=2\), and \(2\mid6\). Divide by \(2\): \(2x\equiv3\pmod5\). The inverse of \(2\) modulo \(5\) is \(3\), so \(x\equiv9\equiv4\pmod5\). Modulo \(10\): \(x\equiv4\) or \(9\).
Shows how one solution modulo the reduced modulus gives several modulo the original modulus.