2008 AIME II Problems/Problem 1

Problem

Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$ .

Solution

Since we want the remainder when $N$ is divided by $1000$ , we may ignore the $100^2$ term. Then, applying the difference of squares factorization to consecutive terms, $\begin{align*} N &= (99-98)(99+98) - (97-96)(97+96) + (95-94)(95 + 94) + \cdots + (3-2)(3+2) - 1 \\ &= \underbrace{197 - 193}_4 + \underbrace{189 - 185}_4 + \cdots + \underbrace{5 - 1}_4 \\ &= 4 \cdot \left(\frac{197-5}{8}+1\right) = \boxed{100}. \end{align*}$

2008 AIME II (Problems • Answer Key • Resources)
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2008 AIME II Problems/Problem 1

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