Let $m\_1,\ldots, m\_n$ be pairwise coprime natural numbers $\geq 1$. We consider the product $$G(m\_1,\ldots,m\_n) := \prod\_{i=1}^{n} \mathbb{Z} / m\_i \mathbb{Z}.$$ We define $M(n)$ as the set $n$-tuple of natural numbers $\geq 1$ with the property that the entrys are pairwise coprime. We define $l : M(n) \rightarrow \mathbb{N}\_0$ by $$l(m\_1,\ldots,m\_n) = \max\_{(x\_1,\ldots,x\_n) \in G(m\_1,\ldots,m\_n)} \min \{ k \in \mathbb{N}\_0 | x\_i+k \neq 0 \forall 1 \leq i \leq n \},$$
hence $l(m\_1,\ldots,m\_n)$ may be regarded as the maximal distance of the elements of $G$ from the elements which have no identity in their entrys. It is clear, that $l(m\_1,\ldots,m\_n)$ is always a natural number.


The question is now: Does there exist a real number $r$, such that $$r \cdot n \cdot \ln (n) \geq \sup\_{(m\_1,\ldots,m\_n) \in M(n)} l(m\_1,\ldots,m\_n)$$ holds for all $n >> 0$ and is it possible to take $r=2$ ?



A question concerning products of finite cyclic groups
We have $l(m\_1,\cdots,m\_n)$ is ~~at least~~ equal to the maximum length of a sequence of consecutive integers each of which is divisible by some $m\_i$. Let's suppose for a moment that $m\_i$ is the $i$th prime. In "On the problem of Jacobsthal", Iwaniec proves that this number is $\ll (n\log n)^2$, while your bound would be much stronger, and if you see the lower bound I mention in [this equvalent question](https://mathoverflow.net/questions/17526/smallest-integer-not-divisible-by-integers-in-a-finite-set) ($\sim\frac{n(\log n)^2 \log \log \log n}{(\log \log n)^2}$), it is actually false. In any case the question reduces to finding the best upper bound on [Jacobsthal's function](http://oeis.org/A058989) (In the comments of the question I linked to above there is an argument for why it suffices to reduce to the $m\_i$ prime case).


