Problem
NT-B2-M07-P017 A Block of Consecutive Integers
Prove that for every integer \(N>2\), one can choose \(N\) consecutive positive integers whose product is divisible by every prime \(p\le 2N+1\).
Split into two cases: \(N+1\) composite and \(N+1\) prime. Recall that every prime \(p\le N\) divides the product of \(N\) consecutive integers.
If \(N+1\) is composite, take the block \(N+2,N+3,\ldots,2N+1\). Every prime \(p\le N\) divides the product of any \(N\) consecutive integers, because one of them is divisible by \(p\). Every prime \(p\) with \(N+2\le p\le 2N+1\) appears in the block as one of the factors. The skipped number \(N+1\) is not prime.
If \(N+1\) is prime, take the block \(N+3,N+4,\ldots,2N+2\). Primes \(p\le N\) are again covered by a block of length \(N\). The prime \(N+1\) divides the last term \(2N+2=2(N+1)\). The number \(N+2\) is not prime: if \(N+1\) is prime and greater than \(3\), then \(N+1\) is odd, so \(N+2\) is even and greater than \(2\); the case \(N=2\) is excluded. Primes between \(N+3\) and \(2N+1\) appear in the block. Thus the required block exists in both cases.
The source is used only for its method architecture: covering primes by a block and splitting according to \(N+1\).