2012 AIME I Problem 10

Let \mathcal{S} be the set of all perfect squares whose rightmost three digits in base 10 are 256. Let \mathcal{T} be the set of all numbers of the form \frac{x-256}{1000}, where x is in \mathcal{S}. In other words, \mathcal{T} is the set of numbers that result when the last three digits of each number in \mathcal{S} are truncated. Find the remainder when the tenth smallest element of \mathcal{T} is divided by 1000.

Leading zeroes must be inputted, so if your answer is 34, then input 034

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.