Let $\mathcal{N}(n)$ be the next prime greater than $n$.
Which is the smallest natural number $n>0\;$ such that:
$\mathcal N(2\cdot 3\cdot 5\cdot 7\cdot 11\cdot n)−2\cdot 3\cdot 5\cdot 7\cdot 11\cdot n\;\;\;$ is neither a prime nor $1$?
I'm sorry! I thought I had good reasons to believe that $n$ should be very big, but I should have tested it, because I was wrong about everything and $n=100284$. I might delete the question.
It's clear that the difference must be divisible by no prime less than $11$ (otherwise $\mathcal N(2\cdot 3\cdot 5\cdot 7\cdot 11\cdot n)$ would not be prime), so the smallest possible difference is $169$. The first occurrence of a prime gap of that magnitude is the pair $\{17051707, 17051887\}$. That doesn't fully explain why the smallest solution ($n=100284$ with product $231656040$) is as large as it is, but it comes close. Even though the prime gap is larger than $169$, you still need a product with high divisibility to fall in just the right place in that prime gap in order to get a solution.