Define $n \in \mathbb{N}^* $ and $\overline{abcd}$ 4 digit number to satisfy the following equation: $$ \overline{abcd} + \frac{\overline{abcd}}{6} + \frac{\overline{abcd}}{6^2} + ... + \frac{\overline{abcd}}{6^n} = \frac{6^{n+1} -1}{5} $$
I've been trying to write up the sum of the geometric progression with $a_1=\overline{abcd}$ and $q=\frac{1}{6}$, and then find an $n$ to get a 4 digit number, but no luck, for $n=4$ for example I don't get a whole number.
So we have a $4$-digit number $x$ and the equation $$ x(1+\tfrac16+\tfrac1{6^2}+...+\tfrac1{6^n})=\frac{6^{n+1}-1}{6-1} $$ where we know that partial sums for the geometric series with powers of $q=\frac16$ have the form $$ 1+\tfrac16+...+\tfrac1{6^n}=\frac{1-\frac1{6^{n+1}}}{1-\frac1{6}}=\frac{6^{n+1}-1}{6^{n+1}-6^n} $$ Dividing by this expression on both sides in the original equation then yields $$ x=\frac{6^{n+1}-6^n}{6-1}=\frac{6^n(6-1)}{6-1}=6^n $$ and to have a $4$-digit number we want $3\leq \log_{10}(x)\leq4$ meaning $$ 3\leq \log_{10}(6^n)=n\cdot\log_{10}(6)\leq 4 $$ and we see that $$ 3.855\approx\frac3{\log_{10}(6)}\leq n\leq\frac{4}{\log_{10}(6)}\leq5.140 $$ showing that $n=4,x=6^4=1296$ or $n=5,x=6^5=7776$ will work.