Difference between revisions of "2009 AMC 8 Problems/Problem 22"
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Note that this is the same as finding how many numbers with up to three digits do not contain 1. | 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 <math>\boxed | + | 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 <math>\boxed{\textbf{(D)}\ 728}</math> numbers. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=21|num-a=23}} | {{AMC8 box|year=2009|num-b=21|num-a=23}} | ||
Revision as of 17:48, 5 November 2012
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 | ||
