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: $$\begin{array}{ccccc}\frac{1}{1}+\frac{1}{1}=\frac{20}{10}& \frac{1}{1}+\frac{1}{2}=\frac{15}{10}& \frac{1}{1}+\frac{1}{5}=\frac{12}{10}& \frac{1}{1}+\frac{1}{10}=\frac{11}{10}& \frac{1}{2}+\frac{1}{2}=\frac{10}{10}\\ \frac{1}{2}+\frac{1}{5}=\frac{7}{10}& \frac{1}{2}+\frac{1}{10}=\frac{6}{10}& \frac{1}{3}+\frac{1}{6}=\frac{5}{10}& \frac{1}{3}+\frac{1}{15}=\frac{4}{10}& \frac{1}{4}+\frac{1}{4}=\frac{5}{10}\\ \frac{1}{4}+\frac{1}{20}=\frac{3}{10}& \frac{1}{5}+\frac{1}{5}=\frac{4}{10}& \frac{1}{5}+\frac{1}{10}=\frac{3}{10}& \frac{1}{6}+\frac{1}{30}=\frac{2}{10}& \frac{1}{10}+\frac{1}{10}=\frac{2}{10}\\ \frac{1}{11}+\frac{1}{110}=\frac{1}{10}& \frac{1}{12}+\frac{1}{60}=\frac{1}{10}& \frac{1}{14}+\frac{1}{35}=\frac{1}{10}& \frac{1}{15}+\frac{1}{30}=\frac{1}{10}& \frac{1}{20}+\frac{1}{20}=\frac{1}{10}\end{array}$$

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