Problem
NT-B2-M09-P008 A Block of Four Composites
#8
★★★☆☆ Level 3 of 5
Construct \(N\) such that \(N+1,N+2,N+3,N+4\) are composite.
Assign divisors \(2,3,5,7\), but choose a sufficiently large solution.
Impose \(N+1\equiv 0\pmod 2\), \(N+2\equiv 0\pmod 3\), \(N+3\equiv 0\pmod 5\), \(N+4\equiv 0\pmod 7\). Thus \(N\equiv 1\pmod 2\), \(N\equiv 1\pmod 3\), \(N\equiv 2\pmod 5\), \(N\equiv 3\pmod 7\). By CRT, a solution exists modulo \(210\). One solution is \(N=157\). Then \(158,159,160,161\) are divisible by \(2,3,5,7\), respectively, and are larger than those divisors; hence they are composite.
Give as a construction problem without requiring minimality.