Does additive quasigroup generated by arbitrary subset $K$ of natural numbers contain each sufficiently big $\gcd (K)$ multiplication? I'm wondering if for every set of positive integers $\{a_0, .., a_{n-1}\}\subset\mathbb{N}_+$ with $g:=\gcd\{a_i:i<n\}$ there is $M_0$ such, that for each $M\geq M_0$ there exists $\{b_i:i<n\}\subset\mathbb{N}$, satisfying $\sum_{i<n}a_ib_i=Mg$.
I'm stuck twofold. First of all, I can't even prove/disprove it for $n=2$. Secondly, easiest idea to reduce $n$ to $n-1$ would be to replace $a_0, a_1$ with something like $c\gcd\{a_0,a_1\}$, where $c$ is coprime with each $a_i,i>1$. But again, is there guaranteed triple $b_0', b_1', c$ with $a_0b_0'+a_1b_1'=c\gcd\{a_0,a_1\}$? This is instant over integers, but I can't see this over naturals. Some elementary tools like Bézout theorem, or Chinese remainder theorem specify existence of solutions, but not precise "shape" of these.
Side note: this is reformulation of more linguistic question: is there $M_L\in\mathbb{N}$ for each language $L$ over single letter (say A), such that $\{A^{M_L}\}\#\{A^g\}^*\subset L^*$ where $^*$ is Kleene star, $\#$ is concatenation, and $g=\gcd\{|w|:w\in L\}$. Previous considerations are sufficient, because it is known, that for each such $L$ there exists finite sublanguage $L'$ satisfying $L^*=L'^*$.
For yet another idea, it could be seen as subset sum problem with unlimited supply of elements. This would probably require termination proof.
 A: First, I'll show how to prove the property for $n$ given the property for $2$ and the property for $n-1$, which reduces the problem to the $n=2$ case.
Given a subset $K$ of the natural numbers, let $\overline K$ denote its closure under addition. Note that, if $K=\{a,b\}\sqcup K'$, then
$$\overline K = \overline{K'\cup \overline{\{a,b\}}}\supset \overline{K'\cup\{xa+yb\}}$$
for any integers $x,y\geq 0$. If we assume the property for all finite sets of size $2$, then $xa+yb$ can represent any sufficiently large multiple of $\gcd(a,b)$.
So, for any finite $K$ with $|K|=n$, we can choose $xa+yb=c\gcd(a,b)$ with $c$ coprime to every element of $K$. Then, by the property for $K'\cup\{c\gcd(a,b)\}$, we have that
$$\overline K\supset \overline{K'\cup\{c\gcd(a,b)\}}$$
contains every sufficiently large multiple of
$$\gcd(\gcd(K'),c\gcd(a,b))=\gcd(K'\cup\{a,b\})=\gcd(K),$$
as desired.
Now, to show the $n=2$ case, it suffices to (a) only prove the $\gcd(a,b)=1$ case for simplicity and (b) only show that $\overline{\{a,b\}}$ contains some number in each residue class modulo $b$, as then you can just take each class's representative and add multiples of $b$. How can you get different residue classes modulo $b$?
