Problem
NT-B1-M01-P024 Arbitrarily Many Composite Numbers in a Row
#24
★★★★★ Level 5 of 5
Prove that for every positive integer \(k\), there exist \(k\) consecutive positive integers, each of which is composite.
Try the numbers \((k+1)!+2,(k+1)!+3,\ldots,(k+1)!+(k+1)\).
Consider the \(k\) consecutive numbers \((k+1)!+2,(k+1)!+3,\ldots,(k+1)!+(k+1)\). For each \(j=2,3,\ldots,k+1\), the number \((k+1)!\) is divisible by \(j\). Hence \((k+1)!+j\) is also divisible by \(j\). Also \((k+1)!+j>j\), so \(j\) is a nontrivial divisor of this number. Therefore all these \(k\) consecutive numbers are composite.
Level 5 for Book 1: a parameter construction with the hidden factorial idea.