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

Rewriting this sequence with more terms, we have

\begin{align*} N &= 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + 95^2 - 94^2 - 93^2 + 92^2 + 91^2 + \ldots - 10^2 - 9^2 + 8^2 + 7^2 - 6^2 - 5^2 + 4^2 + 3^2 - 2^2 - 1^2 \mbox{, and reordering, we get}\\ N &= (100^2 - 98^2) + (99^2 - 97^2) + (96^2 - 94^2)  + (95^2 - 93^2) + (92^2 - 90^2) + \ldots +  (8^2 - 6^2) + (7^2 - 5^2) +(4^2 - 2^2) + (3^2 - 1^2) \mbox{.} \end{align*}

Factoring this expression yields

\begin{align*} N &= (100 - 98)(100 + 98) + (99 - 97)(99 + 97) + (96 - 94)(96 + 94) + (95 - 93)(95 + 93) + (92 - 90)(90 + 92) + \ldots + (8 - 6)(8 + 6) + (7 - 5)(7 + 5) + (4 - 2)(4 + 2) + (3 - 1)(3 + 1) \mbox{, leading to}\\ N &= 2(100 + 98) + 2(99 + 97) + 2(96 + 94) + 2(95 + 93) + 2(92 + 90) + \ldots + 2(8 + 6) + 2(7 + 5) + 2(4 + 2) + 2(3 + 1) \mbox{.} \end{align*}

Next, we get

\begin{align*} N &= 2(100 + 98 + 99 + 97 + 96 + 94 + 95 + 93 + 92 + 90 + \ldots + 8 + 6 + 7 + 5 + 4 + 2 + 3 + 1 \mbox{, and rearranging terms yields}\\ N &= 2(100 + 99 + 98 + 97 + 96 + \ldots + 5 + 4 + 3 + 2 + 1) \mbox{.} \end{align*}

Then,

\begin{align*} N &= 2\left(\frac{(100)(101)}{2}\right) \mbox{, and simplifying, we get}\\ N &= (100)(101) \mbox{, so}\\ N &= 10100 \mbox{.} \end{align*}

Dividing $10100$ by $1000$ yields a remainder of $\boxed{100}$.

Solution 2

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*}

See also

2008 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS