We have \(\overline{a0b}⋮9\Rightarrow a+0+b\in\left\{9;18;27;36;...\right\}\)

But \(a+0+b\le18\)

So \(a+b=\left\{9;18\right\}\)

We have pair (a;b) = (1;8) , (8;1) , (2;7) , (7;2) , (6;3) , (3;6) , (5;4) , (4;5) , (9;0).

But there is only the pair : (4;5) match the condition \(\overline{a0b}=\overline{ab}\cdot9\).

So the answer of the two digit number \(\overline{ab}\) is 45.

\(\overline{a0b}=\overline{ab}.9\Leftrightarrow100a+b=9\left(10a+b\right)\Leftrightarrow100a+b=90a+9b\)

\(\Leftrightarrow10a=8b\Leftrightarrow5a=4b\Leftrightarrow\dfrac{a}{4}=\dfrac{b}{5}\)

\(b< 10\Rightarrow\dfrac{b}{5}< 2\). Moreover, \(a>0\Rightarrow\dfrac{a}{4}>0\)

So, \(\dfrac{a}{4}=\dfrac{b}{5}=1\Rightarrow a=4;b=5;\overline{ab}=45\)