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 $M$ doesn't 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 doesn't 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.

Solution 3

The arithmetic mean is $\frac{(10^1-1)+(10^2-1)+\ldots+(10^9-1)}{9}=\frac{1111111101}{9}=123456789$ . So select $\boxed{\mathrm{A}}$ . ~hastapasta

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 10 Problems and Solutions

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2002 AMC 12B Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

Solution 3

See also