Difference between revisions of "2002 AMC 10B Problems/Problem 3"
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We wish to find <math>\frac{9+99+\cdots +999999999}{9}</math>, or <math>\frac{9(1+11+111+\cdots +111111111)}{9}=123456789</math>. This does not have the digit 0, so <math>\mathrm{ (A) \ }</math> | We wish to find <math>\frac{9+99+\cdots +999999999}{9}</math>, or <math>\frac{9(1+11+111+\cdots +111111111)}{9}=123456789</math>. This does not have the digit 0, so <math>\mathrm{ (A) \ }</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2002|ab=B|num-b=2|num-a=4}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 13:57, 27 December 2008
Problem
The arithmetic mean of the nine numbers in the set 
is a 
-digit number 
, all of whose digits are distinct. The number does not contain the digit
Solution
We wish to find 
, or 
. This does not have the digit 0, so 
See Also
| 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 | ||
