TITLE: Occurence of numbers not divisible by primes
QUESTION [6 upvotes]: Consider a prime number $q$.  Is it true that every sequence of q consecutive integers contains at least one integer not divisible by any prime $p, p < q$?  It seems to be so in any specific instance I look at, but I don't know how to prove it as a general statement.

REPLY [2 votes]: No. Here's a disproof that works for all $q\ge 13$.
Label the primes $p_1<p_2<\cdots <p_{n+1}=q$.
First some motivation. A fairly long sequence of consecutive numbers, each divisible by some prime $<q$ is $q!+2,\ldots, q!+q-1$, but this isn't quite long enough. We also have the sequence $q!-q+1,\ldots, q!-2$ on the other side. If there was only a way to connect up these two!
Sure, if we knocked some primes off of $q!$.
Let $P=cp_1\cdots p_{n-2}$. Choose $c$ such that $p_{n-1}\mid P-1$ and $p_{n}\mid P+1$, possible by the Chinese remainder theorem. Now $P-p_{n-1}+1,\ldots, P-2$ and $P+2,\ldots, P-p_{n-1}-1$ are each divisible by some prime in $p_1,\ldots, p_{n-2}$. We obtain a sequence of $2p_{n-1}-1$ consecutive integers each divisible by some prime in $p_1,\ldots, p_n$. It suffices that $\boxed{2p_{n-1}-1\ge p_{n+1}=q}$, which happens for $q\ge 13$.