2002 AMC 12B Problems/Problem 1

The following problem is from both the 2002 AMC 12B #1 and 2002 AMC 10B #3, so both problems redirect to this page.

Problem

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. The number $M$ does not contain the digit

$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$

Solution 1

We wish to find $\frac{9+99+\cdots +999999999}{9}$, or $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$. This does not have the digit 0, so the answer is $\boxed{\mathrm{(A)}\ 0}$

Solution 2

Notice that the final number is guaranteed to have the digits $\{1, 3, 5, 7, 9\}$ and that each of these digits can be paired with an even number adding up to 9. $\boxed{\mathrm{(A)}\ 0}$ can be taken out, with the other digits fulfilling divisibility by 9.

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2002 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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