The arithmetic mean of the nine numbers in the set \9, 99, 999, 9999, …, 999999999\ is a 9 -digit number M, all of whose digits are distinct. Which digit does the number M not contain?

2002 AMC 10B Problem 3 — Solution

2002 AMC 10B Problem 3

Attempt Problem 3 of the 2002 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

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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?

$0$

$2$

$4$

$6$

$8$

Answer: A

Concepts:mean digits place value

Difficulty rating: 960

Solution:

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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