Suppose $n$ is a [positive integer](https://artofproblemsolving.com/wiki/index.php/Positive_integer) and $d$ is a single [digit](https://artofproblemsolving.com/wiki/index.php/Digit) in [base 10](https://artofproblemsolving.com/wiki/index.php/Base_10). Find $n$ if

$\frac{n}{810}=0.d25d25d25\ldots$
Repeating decimals represent [rational numbers](https://artofproblemsolving.com/wiki/index.php/Rational_number).  To figure out which rational number, we sum an [infinite](https://artofproblemsolving.com/wiki/index.php/Infinite) [geometric series](https://artofproblemsolving.com/wiki/index.php/Geometric_series), $0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$.  Thus $\frac{n}{810} = \frac{100d + 25}{999}$ so $n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}$.  Since 750 and 37 are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime), $4d + 1$ must be [divisible](https://artofproblemsolving.com/wiki/index.php/Divisible) by 37, and the only digit for which this is possible is $d = 9$.  Thus $4d + 1 = 37$ and $n = \boxed{750}$.

(Note: Any repeating sequence of $n$ digits that looks like $0.a_1a_2a_3...a_{n-1}a_na_1a_2...$ can be written as $\frac{a_1a_2...a_n}{10^n-1}$, where $a_1a_2...a_n$ represents an $n$ digit number.)