### Instruction: If $n$ is an integer, $1 \leq n \leq 2010,$ how many fractions $\frac{n^2}{2010}$ yield repeating decimals?### Response: A fraction yields a repeating decimal if and only if the denominator, when fully factored, has a prime factor other than 2 or 5.
Since 2010 = 2 x 3 x 5 x 67, the denominator 2010 has one such prime factor, namely 3 or 67.
Therefore, to get a repeating decimal, the numerator must be divisible by either 3 or 67, or both.
If the numerator is divisible by 3, then $n$ must be a multiple of 3, since $n^2$ is a perfect square.
There are $\lfloor \frac{2010}{3} \rfloor = 670$ multiples of 3 between 1 and 2010, inclusive.
If the numerator is divisible by 67, then $n$ must be a multiple of 67, for the same reason.
There are $\lfloor \frac{2010}{67} \rfloor = 30$ multiples of 67 between 1 and 2010, inclusive.
If the numerator is divisible by both 3 and 67, then $n$ must be a multiple of 201, which is the least common multiple of 3 and 67.
There are $\lfloor \frac{2010}{201} \rfloor = 10$ multiples of 201 between 1 and 2010, inclusive.
By the inclusion-exclusion principle, the number of fractions that yield repeating decimals is the sum of the multiples of 3 and the multiples of 67, minus the multiples of both 3 and 67, to avoid double counting.
That is, 670 + 30 - 10 = 690.