11 and 13

$\begin{aligned} A & = \frac {11}{13} + \frac {1111}{1313} + \frac {111111}{131313} + \cdots + \frac {\overbrace{1111...1111}^{\text{\# of 11 = 13}}}{\underbrace{1313...1313}_{\text{\# of 13 = 13}}} \\ & = \frac {11}{13} \left(1 + \frac {101}{101} + \frac {10101}{10101} + \cdots + \frac {\overbrace{1010...1010}^{\text{\# of 10 = 13}}}{\underbrace{1010...1010}_{\text{\# of 10 = 13}}} \right) \\ & = \frac {11}{13} \left(\underbrace{1+1+1+\cdots + 1}_{\text{\# of 1 = 13}} \right) \\ & = \frac {11}{13}\times 13 = \boxed{11} \end{aligned}$

Note that $\begin{aligned} \dfrac{1111}{1313}=\dfrac{11*101}{13*101} & = \dfrac{11}{13} \\ \ \\ \dfrac{111111}{131313}=\dfrac{11*10101}{13*10101} & = \dfrac{11}{13} \\ \ \\ \dfrac{11111111}{13131313}=\dfrac{11*1010101}{13*1010101} & = \dfrac{11}{13} \\ \dots \end{aligned}$ Therefore $A=13*\dfrac{11}{13}=\boxed{11}$

This series can be written as:

$A = \large \Sigma_{p=0}^{12} \frac{11 \cdot \Sigma_{q=0}^{p} 100^q}{ 13 \cdot \Sigma_{q=0}^{p} 100^q} = \frac{11}{13} \Sigma_{p=0}^{12} \frac{\Sigma_{q=0}^{p} 100^q}{ \Sigma_{q=0}^{p} 100^q} = \frac{11}{13} \Sigma_{p=0}^{12} 1 = \frac{11}{13} \cdot 13 = \boxed{11}.$