Problem
NT-B1-M08-P015 Arbitrarily Many Composite Numbers
#15
★★★☆☆ Level 3 of 5
Prove that for every \(k\ge1\), there exist \(k\) consecutive composite positive integers.
Use the number \((k+1)!\).
Consider \((k+1)!+2,(k+1)!+3,\ldots,(k+1)!+(k+1)\). For each \(r=2,3,\ldots,k+1\), the number \((k+1)!+r\) is divisible by \(r\), because \((k+1)!\) is divisible by \(r\). It is larger than \(r\), so it is composite. There are \(k\) such numbers.
Classical construction, close in spirit to CRT.