2002 AMC 10B 第 3 题

Question

集合

\{9, 99, 999, 9999, \ldots, 999999999\}

中九个数的算术平均数是一个

9

位数

M

，且它的所有数字互不相同。数字

M

不含哪个数字？

The arithmetic mean of the nine numbers in the set

\{9, 99, 999, 9999, \ldots, 999999999\}

is a

9

-digit number

M,

all of whose digits are distinct. Which digit does the number

M

not contain?

0

2

4

6

8

每一项都是

9

乘以一个各位均为一的数，所以先把和除以

9

。

Each term is

9

times a repunit, so divide the sum by

9

first

\dfrac19(9 + 99 + \cdots)

= 1 + 11 + 111 + \cdots

+ 111111111

。

\dfrac19(9 + 99 + \cdots)

= 1 + 11 + 111 + \cdots

+ 111111111

平均数可以写成

\begin{aligned} &\dfrac{1}{9} \\ &\quad {}\cdot \small \left(9 + 99 + 999 + \cdots + 999999999\right) \\ &= 1 + 11 + 111 + \cdots \\ &\quad {}+ 111111111 \end{aligned}\text{。}

把这九个各位均为一的数按列相加，得到

M = 123456789

。

M

唯一缺少的数字是

0

。

所以正确答案是 A。

The mean is

\begin{aligned} &\dfrac{1}{9} \\ &\quad {}\cdot \small \left(9 + 99 + 999 + \cdots + 999999999\right) \\ &= 1 + 11 + 111 + \cdots \\ &\quad {}+ 111111111. \end{aligned}

Adding these nine repunits column by column gives

M = 123456789.

The only digit missing from

M

is

0.

Thus, the correct answer is A.

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