Problem
NT-B1-M08-P007 Shifts \(n+1,n+2,n+3\)
#7
★★☆☆☆ Level 2 of 5
Find the smallest positive \(n\) such that \(2\mid n+1\), \(3\mid n+2\), \(5\mid n+3\).
Rewrite the conditions as congruences for \(n\).
We get \(n\equiv1\pmod2\), \(n\equiv1\pmod3\), \(n\equiv2\pmod5\). The number \(7\) works: \(8\) is divisible by \(2\), \(9\) by \(3\), \(10\) by \(5\). Since the moduli are pairwise coprime, all solutions are \(n\equiv7\pmod{30}\), and the smallest positive is \(7\).
Watch the signs when moving from \(n+r\).