Problem
NT-B1-M08-P011 Residues \(2,4,6\)
#11
★★☆☆☆ Level 2 of 5
Find the smallest positive \(x\) such that \(x\equiv2\pmod3\), \(x\equiv4\pmod5\), \(x\equiv6\pmod7\).
Notice that this is \(x\equiv-1\) modulo each modulus.
The conditions are equivalent to \(x\equiv-1\pmod3\), \(x\equiv-1\pmod5\), \(x\equiv-1\pmod7\). Thus \(x+1\) is divisible by \(3\cdot5\cdot7=105\). The smallest positive \(x\) is \(104\). All solutions are \(x\equiv104\pmod{105}\).
A nice case of equal negative residues.