Let $a,b,c$ be three pairwise coprime positive integers, and $\Gamma=\langle a,b,c\rangle$ be the corresponding numerical semigroup. Consider the linear equations:


$n\_1a=m\_{12}b+m\_{13}c$


$n\_2b=m\_{21}a+m\_{23}c$


$n\_3c=m\_{31}a+m\_{32}b$


Where $n\_i,m\_{ij}$ are positive integers. Also, let $n\_1=\min\{x\in\mathbb{Z}\_{>0}|xa\in\langle b,c\rangle\}$, i.e, we choose the smallest $n\_1$ that satisfies the equation, and $n\_2$ and $n\_3$ are chosen similarly.


Then it's not hard to prove that $n\_1=m\_{21}+m\_{31}$, $n\_2=m\_{12}+m\_{32}$, and $n\_3=m\_{13}+m\_{23}$. For the prove see <http://hera.ugr.es/doi/15773139.pdf>.


My question is, does there exist $a,b,c$ such that $n\_1a<F(\langle b,c\rangle)=bc-b-c$, $n\_2b<F(\langle a,c\rangle)=ac-a-c$, $n\_3c<F(\langle a,b\rangle)=ab-a-b$where $F$ is the Frobenius number? 


Edit: thanks for the example, the answer is positive. Furthermore, does there exist $a,b,c$ such that $\max\{n\_1a,n\_2b,n\_3c\}<F(\Gamma)?$



Positive integer solutions of linear equations under the constraint of Frobenius number
Let $a=30$, $b=31$ and $c=37$. Then


\begin{array}{ll}
7a=2b+4c=210 \\
7b=6a+c=217 \\
5c=a+5b=185 \\
\end{array} 


and the Frobenius number of $\langle a,b,c \rangle$ is 267.


