\(7\le x\le12\)

We have: \(a+2016b⋮2017\)

\(\Rightarrow a+2016b=2017n\left(n\in N\right)\)

\(\Rightarrow a=2017n-2016b\)

We have: \(2a+2015b=2\left(2017n-2016b\right)+2015b=2.2017n-2017b⋮2017\)

Same as above. We hava: \(\left\{{}\begin{matrix}3a+2014b=3.2017n-2.2017b⋮2017\\4a+2013b=4.2017n-3.2017b⋮2017\\...\\2015a+2b=2015.2017n-2014.2017b\end{matrix}\right.\)

So that: \(A⋮2017.2017...2017\)

\(\Rightarrow A⋮2017^{2014}\)