2009 AMC 8 Problems/Problem 22: Difference between revisions
No edit summary |
|||
| Line 13: | Line 13: | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=21|num-a=23}} | {{AMC8 box|year=2009|num-b=21|num-a=23}} | ||
{{MAA Notice}} | |||
Revision as of 00:49, 5 July 2013
Problem
How many whole numbers between 1 and 1000 do not contain the digit 1?
Solution
Note that this is the same as finding how many numbers with up to three digits do not contain 1.
Since there are 10 total possible digits, and only one of them is not allowed (1), each digit has its choice of 9 digits, for a total of 9*9*9=729 such numbers. However, we over counted by one; 000=0 is not between 1 and 1000, so there are
numbers.
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing