The increasing [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) $3, 15, 24, 48, \ldots\,$ consists of those [positive](https://artofproblemsolving.com/wiki/index.php/Positive) multiples of 3 that are one less than a [perfect square](https://artofproblemsolving.com/wiki/index.php/Perfect_square).  What is the [remainder](https://artofproblemsolving.com/wiki/index.php/Remainder) when the 1994th term of the sequence is divided by 1000?
One less than a perfect square can be represented by $n^2 - 1 = (n+1)(n-1)$. Either $n+1$ or $n-1$ must be divisible by 3. This is true when $n \equiv -1,\ 1 \equiv 2,\ 1 \pmod{3}$. Since 1994 is even, $n$ must be congruent to $1 \pmod{3}$. It will be the $\frac{1994}{2} = 997$th such term, so $n = 4 + (997-1) \cdot 3 = 2992$. The value of $n^2 - 1 = 2992^2 - 1 \pmod{1000}$ is $\boxed{063}$.
~minor edit by [Yiyj1](https://artofproblemsolving.com/wiki/index.php/User:Yiyj1)