We have: 

$\Rightarrow a=2017n-2016b$

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

Same as above. We hava: $\{\begin{array}{c}3a+2014b=3.2017n-2.2017b⋮2017\\ 4a+2013b=4.2017n-3.2017b⋮2017\\ ...\\ 2015a+2b=2015.2017n-2014.2017b\end{array}$

So that: $A⋮\mathrm{2017.2017...2017}$

$\Rightarrow A⋮{2017}^{2014}$

Trịnh Đức Phát 22/03/2017 at 12:43

We have: 

⇒a=2017n−2016b

We have: 2a+2015b=2(2017n−2016b)+2015b=2.2017n−2017b⋮2017

Same as above. We hava: ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩3a+2014b=3.2017n−2.2017b⋮20174a+2013b=4.2017n−3.2017b⋮2017...2015a+2b=2015.2017n−2014.2017b

So that: A⋮2017.2017...2017

⇒A⋮20172014