1985 AJHSME Problems/Problem 15
Contents
Problem
How many whole numbers between 
and 
contain the digit 
?
Solution 1
This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.
If you've ever learned about complementary counting, this would be the best time to utilize it. Instead of counting how many DO have 2's, why don't we count how many that DON'T?
So let's find the number of numbers. Obviously, we'd start by subtracting 100 from 400, getting us 300, but we're not done. Since just subtracting includes the number 400, we must subtract one (because 400 isn't allowed - it says between), getting us 299.
So how many numbers are there that DON'T have a 2? Well, we have 2 possibilities for the hundreds digit (1, 3, note that 2 is not allowed), 9 possibilities for the tens digit (1, 3, 4, 5, ... , 9, 0), and 9 possibilities for the ones digit. 
. However, one of the numbers we counted is , which isn't allowed, so there are 
numbers without a 2.
Since there are 299 numbers in total and 161 that DON'T have any 2's, 
numbers WILL have at least one two.
Video Solution
~savannahsolver
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
