2002 AMC 12B Problem 1

Attempt Problem 1 of the 2002 AMC 12B 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 12B solutions, or check the answer key.

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1.

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

00

22

44

66

88

Answer: A
Concepts:place valuedigits
Difficulty rating: 950
Solution:

Each of the nine numbers is 10k1,10^k-1, so their sum is 9+99++999,999,999.9+99+\cdots+999{,}999{,}999. Dividing by 9,9, M=1+11+111++111,111,111=123,456,789. \begin{aligned} M &= 1+11+111+\cdots \\ &\quad {}+111{,}111{,}111 \\ &= 123{,}456{,}789. \end{aligned} Its digits are 11 through 9,9, so the missing digit is 0.0.

Thus, the correct answer is A.

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