AIME 1994 · 第 1 题

Solution

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