Problem
NT-B1-M08-P004 Parity Conflict
#4
★☆☆☆☆ Level 1 of 5
Prove that \(x\equiv1\pmod2\), \(x\equiv0\pmod4\) has no solutions.
The second congruence already determines the parity of \(x\).
If \(x\equiv0\pmod4\), then \(x\) is even, so \(x\equiv0\pmod2\). This contradicts \(x\equiv1\pmod2\). No solutions exist.
Minimal compatibility problem.