Difference between revisions of "2006 AMC 10A Problems/Problem 21"
m (+box) |
|||
| Line 13: | Line 13: | ||
Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their [[decimal representation]] is <math>9000-3584=5416 \Longrightarrow \mathrm{(E)} </math> | Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their [[decimal representation]] is <math>9000-3584=5416 \Longrightarrow \mathrm{(E)} </math> | ||
| − | == See | + | == See also == |
| − | + | {{AMC10 box|year=2006|ab=A|num-b=20|num-a=22}} | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 16:22, 26 February 2007
Problem
How many four-digit positive integers have at least one digit that is a 2 or a 3?
Solution
Since we are asked for the number of positive 4-digit integers with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits.
The total number of 4-digit integers is 
, since we have 10 choices for each digit except the first (which can't be 0).
Similarly, the total number of 4-digit integers without any 2 or 3 is 
.
Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is 
See also
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 | ||
