Question

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 $1000.$

Answer

Answer: 100

Concepts:difference of squares pairing and grouping summation

Difficulty rating: 1890

Solution:

Group the terms four at a time. For $k = 1, 2, \ldots, 25,$ the block ending at $(4k)^2$ is $\begin{aligned} &(4k)^2 + (4k-1)^2 \\ &\quad {}- (4k-2)^2 - (4k-3)^2 \\ &= 2(8k - 2) + 2(8k - 4) \\ &= 32k - 12, \end{aligned}$ using the difference of squares $a^2 - b^2 = (a + b)(a - b)$ with $a - b = 2$ twice.

Summing over $k = 1$ to $25,$ $\begin{aligned} N &= 32 \cdot \frac{25 \cdot 26}{2} - 12 \cdot 25 \\ &= 10400 - 300 = 10100, \end{aligned}$ so the remainder when $N$ is divided by $1000$ is $100.$

2008 AIME II Problem 1

Answer: 100

Difficulty rating: 1890

Solution:

Problem 1 in Other Years