Problem
NT-B2-M09-P016 A Long Composite Block
#16
★★★★☆ Level 4 of 5
Prove that for every \(k\), there exist \(k\) consecutive positive composite integers.
Use \((k+1)!+2,\ldots,(k+1)!+(k+1)\).
Let \(A=(k+1)!\). The numbers \(A+2,A+3,\ldots,A+(k+1)\) form \(k\) consecutive integers. For each \(j=2,\ldots,k+1\), \(A+j\) is divisible by \(j\), because \(A\) is divisible by \(j\). Also \(A+j>j\), so \(A+j\) is composite. This gives the required block.
The factorial construction is a special CRT-style construction where all conditions are built in automatically.