2010 AIME I 第 2 题

Question

求

9 \cdot 99 \cdot 999 \cdot \cdots \cdot \underbrace{99\ldots9}_{\text{999 个 9}}

除以

1000

的余数。

Find the remainder when

9 \cdot 99 \cdot 999 \cdot \cdots \cdot \underbrace{99\ldots9}_{\text{999 nines}}

is divided by

1000.

答案

从第三个因子开始，每个因子末尾至少有三个

9

，所以它

\equiv -1 \pmod{1000}

Every factor from the third one on ends in at least three

9

s, so it is

\equiv -1 \pmod{1000}

数一数有多少个因子

\equiv -1 \pmod{1000}

，并判断这个个数的奇偶；剩下只需计算

9 \cdot 99

Count how many factors are

\equiv -1 \pmod{1000}

and check whether that count is odd or even; only

9 \cdot 99

remains to compute

在模

1000

下计算。从第三个因子开始，每个因子末尾至少有三个

9

，所以每个都

\equiv -1 \pmod{1000}

。一共有

999

个因子，因此其中

997

个

\equiv -1

。

所以乘积满足

\begin{aligned} &\equiv 9 \cdot 99 \cdot (-1)^{997} \\ &\equiv -891 \equiv 109 \pmod{1000} \end{aligned}\text{，}

因此余数是

109

。

Work modulo

1000.

Every factor from the third one on ends in at least three

9

s, so each is

\equiv -1 \pmod{1000}.

There are

999

factors in all, hence

997

of them are

\equiv -1.

The product is therefore

\begin{aligned} &\equiv 9 \cdot 99 \cdot (-1)^{997} \\ &\equiv -891 \equiv 109 \pmod{1000}, \end{aligned}

so the remainder is

109.

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