Problem
NT-B1-M12-P012 Set 2. System of Residues
#12
★★☆☆☆ Level 2 of 5
Solve the system \(n\equiv2\pmod5\), \(n\equiv3\pmod7\).
Let \(n=5k+2\).
Substitute: \(5k+2\equiv3\pmod7\), so \(5k\equiv1\pmod7\). The inverse of \(5\) modulo \(7\) is \(3\), hence \(k\equiv3\pmod7\). Thus \(n=5(7t+3)+2=35t+17\).
Short CRT problem.