2008 AIME II 第 1 题

Question

令

\begin{aligned} N &= 100^2 + 99^2 - 98^2 - 97^2 \\ &\quad {}+ 96^2 + \cdots + 4^2 \\ &\quad {}+ 3^2 - 2^2 - 1^2 \end{aligned}\text{，}

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

\begin{aligned} N &= 100^2 + 99^2 - 98^2 - 97^2 \\ &\quad {}+ 96^2 + \cdots + 4^2 \\ &\quad {}+ 3^2 - 2^2 - 1^2, \end{aligned}

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

小提示：

将每个带正号的平方项与比它小两个单位的带负号平方项配对，并用

a^2 - b^2 = (a+b)(a-b)

，其中

a - b = 2

Pair each positive square with the negative square two below it, and factor

a^2 - b^2 = (a+b)(a-b)

with

a - b = 2

大提示：

每四个连续项分为一组；以

(4k)^2

结尾的一组可化简为

32k - 12

Group four consecutive terms at a time; the block ending in

(4k)^2

simplifies to

32k - 12

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}\text{，}

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}\text{，}

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