1985 AJHSME Problems/Problem 15

Solution

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 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?

The reasoning behind this is, the number of numbers that have a 2 is the same as the total number of numbers subtract the number of numbers that DON'T have 2's. And since it's easy to count the number of numbers, and it's easy to count how many numbers don't have twos, this problem becomes extremely simple.

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. . Since there are 243 numbers that DON'T have any 2's, we can accurately assume that numbers WILL have at least one two.

137 is ... not a choice?