Difference between revisions of "2006 AMC 10A Problems/Problem 21"
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== Problem == | == Problem == | ||
| − | How many four-[[digit]] [[positive integer]]s have at least one digit that is a 2 or a 3? | + | How many four-[[digit]] [[positive integer]]s have at least one digit that is a <math>2</math> or a <math>3</math>? |
| − | <math>\ | + | <math>\textbf{(A) } 2439\qquad\textbf{(B) } 4096\qquad\textbf{(C) } 4903\qquad\textbf{(D) } 4904\qquad\textbf{(E) } 5416\qquad</math> |
== Video Solution == | == Video Solution == | ||
Revision as of 08:39, 23 September 2021
Problem
How many four-digit positive integers have at least one digit that is a 
or a 
?
Video Solution
https://youtu.be/0W3VmFp55cM?t=3291
~ pi_is_3.14
Solution (Complementary Counting)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
