Eccoci.

Consider the diophantine equation $\frac 1 a + \frac 1 b = \frac p {10^n}$ with $a, b, p, n$ positive integers and $a \le b$.
For $n=1$ this equation has $20$ solutions that are listed below: \begin{matrix} \frac 1 1 + \frac 1 1 = \frac{20}{10} & \frac 1 1 + \frac 1 2 = \frac{15}{10} & \frac 1 1 + \frac 1 5 = \frac{12}{10} & \frac 1 1 + \frac 1 {10} = \frac{11}{10} & \frac 1 2 + \frac 1 2 = \frac{10}{10}\\ \frac 1 2 + \frac 1 5 = \frac 7 {10} & \frac 1 2 + \frac 1 {10} = \frac 6 {10} & \frac 1 3 + \frac 1 6 = \frac 5 {10} & \frac 1 3 + \frac 1 {15} = \frac 4 {10} & \frac 1 4 + \frac 1 4 = \frac 5 {10}\\ \frac 1 4 + \frac 1 {20} = \frac 3 {10} & \frac 1 5 + \frac 1 5 = \frac 4 {10} & \frac 1 5 + \frac 1 {10} = \frac 3 {10} & \frac 1 6 + \frac 1 {30} = \frac 2 {10} & \frac 1 {10} + \frac 1 {10} = \frac 2 {10}\\ \frac 1 {11} + \frac 1 {110} = \frac 1 {10} & \frac 1 {12} + \frac 1 {60} = \frac 1 {10} & \frac 1 {14} + \frac 1 {35} = \frac 1 {10} & \frac 1 {15} + \frac 1 {30} = \frac 1 {10} & \frac 1 {20} + \frac 1 {20} = \frac 1 {10} \end{matrix}

How many solutions has this equation for $1 \le n \le 9$?