Possible last digits

Since $p>q$ , and these terms cannot contain factors of both $2$ and $5$ , hence $p$ is the power of $5$ and $q$ is the power of $2$ .

So, $q$ has to be a power of $2$ , i.e. $q=2^n$ and $p$ has to be the same power, but of $5$ instead of $2$ , i.e. $p=5^n$ .

This means that $p$ will always be odd and $q$ will always be even. So $p-q$ will always be odd, and will possibly be, at first, ending with $1,3,5,7,9$ .

But $5$ is not possible, since $p$ will always end in $5$ and, for $p-q$ to end in $5$ , $q$ would have to end in $0$ which is not possible, since it is a power of $2$ .

The other vales, $1,3,7,9$ , are attainable for $q$ ending in $4,2,8,6$ respectively, which are all possible last digits for a power of $2$ .

So the possible values are $1,3,7,9$ , which sum up to $\color{#3D99F6} \fbox{20}$ . In fact, the values appear periodically in order $3, 1,7, 9,3,1,7,9...$ for $n = 1,2,3,4,5,6,7,8...$