A short conversation a frisbee tournament with my friend Gwen got me thinking about my math projects and what I might be able to do to improve them. After month of pondering I’ve decided that I want to take both the videos I’m making and the blog writing a little more seriously than I have in the past. Not totally sure what that’s going to mean – maybe things will start to look as though I spent 10 minutes on them rather than 5 – but my old camera breaking this morning got the ball rolling. At minimum, the videos should look like they’ve been shot with a newer camera – ha!
This morning my son got a little stuck on a neat counting problem from an old AMC 8. The problem is #22 from the 2009 exan – how many integers from 1 to 1000 do not contain 1 as a digit? You can see the entire 2009 AMC 8 exam here (with the link going to problem 22, specifically):
I spent the summer working through a little of Art of Problem Solving’s “Introduction to Counting and Probability” book with both kids, so I thought working through this problem would be a fun for them as well as a great way to test out the new camera. We started with a quick statement of the problem and talked about a few different techniques we could use to solve it. The boys thought of two ways to approach the problem – (i) case by case counting, and (ii) complimentary counting. We started with the case by case approach:
The second approach the boys came up with was complimentary counting. In this approach you count the “things you don’t want” and then use that information to get the answer you are looking for. A typical complimentary counting problem would be something like this:
I flip a fair coin 4 times. What is the probability that I get at least one tail in the sequence of flips? The compliment of getting “at least one tail” is getting no tails, and that situation turns out to be a little easier to think through than all the cases where you get at least one tail. Getting no tails means that you get heads on every flip, and that will occur with probability (1/2) x (1/2) x (1/2) x (1/2) = 1/16. Thus the probability of getting at least one tail is 15 / 16.
In the situation we are looking at in this problem, the compliment of the set we are counting is the set of integers between 1 and 999 that have at least one 8. We go through how to count that set in this video – it is a little harder to count than the example I gave above.
In the last video I wanted to show the kids a completely new way to approach the problem – partly because this solution is pretty cool and partly because this solution really hits on the topic of place value. There aren’t many situations in which thinking about the number 25 as 025 is useful, but in this problem allowing leading zeros leads to a pretty clever solution. The hint that there’s a clever solution in the background to this problem is that the answer – 728 – is pretty close to 729 which is .
So, an old AMC8 problem leads to a fun review of both counting and place value. It also gave us a nice opportunity to look at a problem from several different angles. Luckily all of the different approaches led to the same solution!
I’d love to hear any ideas of how to improve these blogs entries and/or videos. With about 2500 math and frisbee videos made with that old camera I am a little sad that it broke, but hopefully the videos with the 2014 version are a least a tiny step up in quality 🙂