2002 AMC 10B Problem 3

Question

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

Each term is

9

times a repunit, so divide the sum by

9

first

\dfrac19(9 + 99 + \cdots)

= 1 + 11 + 111 + \cdots

+ 111111111

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.

Problem 3 in other years