Problem: Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if $ \frac{n}{810}=0 . d 25 d 25 d 25 \ldots $ Solution: There is a nice line where we can think of repeating decimal digital as an infinite geometric series. We can write $ 0.d 25 d 25 d 25 \ldots = \sum_{n=1}^{\infty} \frac{100d+25}{1000^{n}} = \frac{100d+25}{1-\frac{1}{1000^n}} $ and use a mod argument from here. There is an even cleaner way though that I found online that abuses the repeating nature of the decimal. Simply multiply both sides by $1000$ to get $\frac{1000n}{810} = d 2 5.d 2 5 d 2 5 \ldots$. Now, we may subtract the initial equation from this one to get the very clean $\frac{999n}{810} = 100d+25$. From here, we can simplify to get $\frac{37n}{30} = 100d+25$. From here, $\frac{37n}{30} \in \mathbb{Z}$ \since the RHS is, and so $n = 30 \cdot 5 \cdot (2k+1)$ because $\gcd(30,37)=1$ and five divides the RHS but not 10. From here, the only odd number that gives an individual digit $d$ abd 25 at the end is when $k = 2$ so we get $750$ as our answer.