Problem
NT-B2-M09-P012 Divisors for Five Shifts
#12
★★★★☆ Level 4 of 5
Prove that there are infinitely many \(N\) such that \(N+1,N+2,N+3,N+4,N+5\) have divisors \(3,5,7,11,13\), respectively.
Write five congruences for \(N\).
We need to solve \(N\equiv -1\pmod 3\), \(N\equiv -2\pmod 5\), \(N\equiv -3\pmod 7\), \(N\equiv -4\pmod {11}\), \(N\equiv -5\pmod {13}\). The moduli are pairwise coprime, so by CRT a solution exists modulo \(M=3\cdot 5\cdot 7\cdot 11\cdot 13\). If \(N_0\) is one solution, then all \(N=N_0+tM\) are also solutions. There are infinitely many.
No explicit \(N\) is required; the existence idea is the focus.