Problem
NT-B1-M04-P023 Minus One and Zero
#23
★★★★☆ Level 4 of 5
Find the smallest positive \(n\) such that \(n\equiv-1\pmod2\), \(n\equiv-1\pmod3\), \(n\equiv-1\pmod5\), but \(n\equiv0\pmod7\).
The first three conditions mean \(n\equiv-1\pmod{30}\).
The first three conditions give \(n\equiv29\pmod{30}\). Let \(n=7k\). Then \(7k\equiv29\pmod{30}\). The inverse of \(7\) modulo \(30\) is \(13\), so \(k\equiv29\cdot13\equiv17\pmod{30}\). The smallest positive \(k\) is \(17\), hence \(n=119\).
A construction problem leading toward the future CRT module.