Question

令 $N = 100^2 + 99^2 - 98^2 - 97^2$ $+ 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ ，其中加号和减号每两个一组交替出现。求 $N$ 除以 $1000$ 的余数。

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 $1000.$

答案

答案：100

知识点：平方差配对与分组求和

难度评级：1890

解答：

每四项分成一组。对 $k = 1, 2, \ldots, 25$ ，以 $(4k)^2$ 结尾的一组为 $\begin{aligned} &(4k)^2 + (4k-1)^2 \\ &\quad {}- (4k-2)^2 - (4k-3)^2 \\ &= 2(8k - 2) + 2(8k - 4) \\ &= 32k - 12, \end{aligned}$ 这里两次用了平方差公式 $a^2 - b^2 = (a + b)(a - b)$ ，且 $a - b = 2$ 。

对 $k = 1$ 到 $25$ ，求和， $\begin{aligned} N &= 32 \cdot \frac{25 \cdot 26}{2} - 12 \cdot 25 \\ &= 10400 - 300 = 10100, \end{aligned}$ 所以 $N$ 除以 $1000$ 的余数是 $100$ 。

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 第 1 题

答案：100

难度评级：1890

解答：

其他年份的第 1 题