[tex]\it a)\ (2-x+1)|6 \Rightarrow(3-x)|6 \Rightarrow (x-3)|6 \Rightarrow x-3\in D_6 \Rightarrow \\ \\ \Rightarrow x-3\in\{\pm1,\ \pm2,\ \pm3,\ \pm6\}\Rightarrow x-3\in\{-1,\ -2,\ -3,\ -6,\ 1,\ 2,\ 3,\ 6\}\bigg|_{+3} \Rightarrow \\ \\ \Rightarrow x\in\{2,\ 1,\ 0, \ -3,\ 4,\ 5,\ 6,\ 9\}[/tex]
[tex]\it b)\ \overline{xy}+7y=10x+y+7y=10x+8y=2(5x+4y)=\ num\breve ar\ \ par[/tex]