[tex]\it \overline{0,1(a)}=\dfrac{\overline{1a}-1}{90} =\dfrac{10+a-1}{90}\\ \\ \\ 0,(a) =\dfrac{^{10)}a}{\ \ 9}=\dfrac{10a}{90} \\ \\ \\ \overline{0,a(1)}=\dfrac{\overline{a1}-a}{90} =\dfrac{10a+1-a}{90}=\dfrac{9a+1}{90}\\ \\ \\[/tex]
Acum, se poate scrie:
[tex]\it \dfrac{9+a+10a+9a+1}{90}=\dfrac{20a+10}{90}=\dfrac{10(2a+1)^{{(10}}}{90}=\dfrac{2a+1}{9}\ \ \ (*)[/tex]
[tex]\it (*) \Rightarrow \begin{cases}\it 2a+1=impar\\ \\ 2a+1\in M_9\end{cases}\ \ (**)\\ \\ \\ (**) \Rightarrow 2a+1=9 \Rightarrow 2a=8 \Rightarrow a=4[/tex]
