1994 AIME Problems/Problem 1

Problem

The increasing sequence $3, 15, 24, 48, \ldots\,$ consists of those positive multiples of 3 that are one less than a perfect square. What is the remainder when the 1994th term of the sequence is divided by 1000?

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 $\equiv 1 \pmod{3}$. 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 $063$.

See also

1994 AIME (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
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All AIME Problems and Solutions

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