2002 AMC 10B Problems/Problem 3

Problem

The arithmetic mean of the nine numbers in the set $\{9,99,999,9999,...,999999999\}$ 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 $\mathrm{ (A) \ }$

2002 AMC 10B (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2002 AMC 10B Problems/Problem 3

Problem

Solution

See Also