Problem
NT-B2-M09-P002 Three Residues
#2
★★☆☆☆ Level 2 of 5
Find the least positive \(x\) such that \(x\equiv 1\pmod 4\), \(x\equiv 2\pmod 5\), \(x\equiv 3\pmod 7\).
First combine moduli \(4\) and \(5\), then add modulus \(7\).
From the first two conditions: \(x=4a+1\), \(4a+1\equiv 2\pmod 5\), so \(4a\equiv 1\), \(a\equiv 4\pmod 5\). Hence \(x\equiv 17\pmod {20}\). Now \(20b+17\equiv 3\pmod 7\), or \(6b\equiv 0\pmod 7\), so \(b\equiv 0\pmod 7\). The least positive solution is \(x=17\).
Shows that the answer may be much smaller than the product of the moduli.