2008 AIME II Problems/Problem 1

From AoPSWiki

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,

$\begin{align*}N &= (99-98)(99+98) - (97-96)(97+96) + (95-94)(95 + 94) + \cdots + (3-2)(3+2) - 1 \\&= \underbrace{197 ...$

2008 AIME II (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Personal tools

Navigation

Search

Toolbox

2008 AIME II Problems/Problem 1

From AoPSWiki

Problem

Solution

See also