Call a $3$-digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. So, consider the number $931$. Let us note $931-792=139$ which means reverse order of $931$. In over words $9·100+10·3+1-792=100·1+10·3+9$.
In addition to that note that $(9-4)·100+3·10+1=531$, the numbers $1$, $3$, $5$ form arithmetic progression.
The standard way to find the $3$-digit number is to solve the system of equations: $$\begin{cases} b^2 &= ac &\,\,\,(a)\\ 2b &= a-4 &\,\,\,(b)\\ 100a+10b+c-792 &= 100c+10b+a &\,\,\,(c) \\ \end{cases}$$, here a,b,c - the members of the geometric progression
I wonder if the numbers with similar properties exist amongst $n$-digit number and how to find the numbers avoiding brute force? I think more clarification about $n$-digit geometric numbers is required and how they are organized. For eg. the $4$-digit number is $8421$, the $5$-digit number are $10010$, $11111$, $16421$,$81931$. So, for $4$-digit numbers: system of equations: $$\begin{cases} b^2 &= ac &\,\,\,(a)\\ c^2 &= bd &\,\,\,(b)\\ 2b &= a-4+c &\,\,\,(c)\\ 2c &= b+d &\,\,\,(e)\\ 1000a+100b+10c+d-792 &= 1000d+100c+10b+a &\,\,\,(f) \\ \end{cases}$$