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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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
Small hint:
The mean is 19(9+99+⋯+999,999,999)\dfrac{1}{9}(9+99+\cdots+999{,}999{,}999)
Big hint:
Dividing each term by 99 turns the sum into 1+11+111+⋯1+11+111+\cdots; add these and see which digit never appears
Solution:
Each of the nine numbers is 10k−1,10^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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