Difference between revisions of "2002 AMC 12B Problems/Problem 1"

m
Line 1: Line 1:
{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #1]] and [[2002 AMC 10B Problems|2002 AMC 10B #3]]}}== Problem ==
+
{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #1]] and [[2002 AMC 10B Problems|2002 AMC 10B #3]]}}
 +
== Problem ==
 
The [[arithmetic mean]] of the nine numbers in the set <math>\{9, 99, 999, 9999, \ldots, 999999999\}</math> is a <math>9</math>-digit number <math>M</math>, all of whose digits are distinct. The number <math>M</math> does not contain the digit
 
The [[arithmetic mean]] of the nine numbers in the set <math>\{9, 99, 999, 9999, \ldots, 999999999\}</math> is a <math>9</math>-digit number <math>M</math>, all of whose digits are distinct. The number <math>M</math> does not contain the digit
  

Revision as of 16:19, 28 July 2011

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

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 $\boxed{\mathrm{(A)}\ 0}$

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