Difference between revisions of "2002 AMC 12B Problems/Problem 1"
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== Solution 1 == | == Solution 1 == | ||
| − | 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 | + | 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 doesn't have the digit 0, so the answer is <math>\boxed{\mathrm{(A)}\ 0}</math> |
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== Solution 2 == | == Solution 2 == | ||
Notice that the final number is guaranteed to have the digits <math>\{1, 3, 5, 7, 9\}</math> and that each of these digits can be paired with an even number adding up to 9. <math>\boxed{\mathrm{(A)}\ 0}</math> can be taken out, with the other digits fulfilling divisibility by 9. | Notice that the final number is guaranteed to have the digits <math>\{1, 3, 5, 7, 9\}</math> and that each of these digits can be paired with an even number adding up to 9. <math>\boxed{\mathrm{(A)}\ 0}</math> can be taken out, with the other digits fulfilling divisibility by 9. | ||
Revision as of 16:39, 23 October 2021
- The following problem is from both the 2002 AMC 12B #1 and 2002 AMC 10B #3, so both problems redirect to this page.
Contents
Problem
The arithmetic mean of the nine numbers in the set 
is a 
-digit number 
, all of whose digits are distinct. The number doesn't contain the digit
Solution 1
We wish to find 
, or 
. This doesn't have the digit 0, so the answer is 
Solution 2
Notice that the final number is guaranteed to have the digits 
and that each of these digits can be paired with an even number adding up to 9. can be taken out, with the other digits fulfilling divisibility by 9.
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 | ||
| 2002 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
