[tex]\it \overline{0,1(a)} +\overline{0,(a)}+\overline{0,a(1)}=\dfrac{10+a-1}{90}+\dfrac{\ a^{(10}}{9}+\dfrac{10a+1-a}{90}=\dfrac{9+a+10a+9a+1}{90}=\\ \\ \\ =\dfrac{20a+10}{90}=\dfrac{10(2a+1)^{(10}}{90}=\dfrac{2a+1}{9}\in \mathbb{N} \Rightarrow a=4[/tex]
