Fraction

Suppose $x$ has $n$ number of digits. Then: $\dfrac{x}{y}=y+x*10^{-n}\Longleftrightarrow 10^n(x-y^2)=xy$ Since $x$ and $y$ are coprime, $x$ and $x-y^2$ are coprime, and $y$ and $x-y^2$ are coprime too. So $x-y^2=1$ and $10^n=xy$ . It's obvious that $x>y>1$ , so $x=5^n$ and $y=2^n$ . From that $\begin{aligned} 5^n-4^n & = 1 \\ \ \\ \left (\dfrac{5}{4}\right )^n & = 1+\left (\dfrac{1}{4}\right )^n\end{aligned}$ If $n=1$ , then the equation above is true. Since $\left (\dfrac{5}{4}\right )^n$ is strictly monotonous growing, and $1+\left (\dfrac{1}{4} \right )^n$ is strictly monotonous decreasing, there aren't more solutions. From that $x=5, y=2$ , $\text{answer}=5+2=\boxed{7}$