This is the second short (in words) post for today.
Problem #7 from the 2008 AMC 10 gave my son a little trouble this morning, but talking about it led to a fun conversation about geometry and algebra. Here’s the problem:
Problem #7 from the 2008 AMC 10 A
After talking about the solution to this problem, I scrapped the actual lesson for today and talked a bit about some special Pythagorean triangles. Then I gave my son this problem as a challenge:
He was able to solve the problem in the last video by finding a pattern in the side lengths. Next I challenged him to find a different pattern. This part was a little bit of a struggle, but he did eventually find a different pattern that connects the side lengths:
Finally, I wanted to use the pattern that we found in the second video to find some new triangles. We found the next couple of triangles in the pattern and that showed him, I think, that this new pattern could be pretty useful.
This was an interesting little project for me. I guess there’s no way to know what patterns that people will find easy to see and what ones that they will find hard to see, but I do think looking for patterns that you don’t see initially is an important skill in problem solving. I’ll be on the lookout for similar project connecting geometry, algebra, and patterns in the future.
This is the first of two short blog posts today.
My younger son and I started the section on divisibility rules in our Introduction to Number Theory book. I don’t remember the context, but we have talked a little bit about divisibility rules before. He knows some of the rules, but now we are going to learn how to understand these rules through the lens of modular arithmetic.
Last night we talked about divisibility rules without even looking at the book. I just wanted to hear what he had to say:
Today we talked a little more in depth about some of the basic rules – namely divisibility by 2, 5, and 10. He seemed to be able to understand the ideas and gave a really nice explanation of why the divisibility rules for these numbers work. It was fun to hear his explanations (despite my stumbling explanation of the problem that we were working on . . . .):
I remember being fascinated by these divisibility rules as a kid, though I’m sure that I just learned the rules without really understanding why they worked. Learning the ideas behind these rules isn’t too complicated, though, and hopefully helps build up number sense and a little bit of sense about place value, too. Definitely a fun little project.
We’ve traveling for a few days and I finally got some down time driving home from Boston during the day today. I have Jordan Ellenberg’s “How not to be Wrong” on audiobook and was listening to chapter 10 – “Are you there God? It’s me, Bayesian Inference.”
I really like one of the examples that he gives and want to figure out a way to turn it into a project with my kids. Here’s roughly what he goes through:
You have a coin that is drawn from a large population of coins. In the overall population 10% of the coins flip heads 60% of the time and tails 40% of the time. 80% of the coins flip heads and tails 50% of the time (a “fair” coin), and the remaining 10% flip heads 40% of the time and tails 60% of the time.
The example he works through is this one: You flip the coin you’ve selected 5 times and get heads every time. What is the probability that you have selected one of the fair coins?
It turns out the answer is about 86.5%.
Here’s what I wondered, though. Suppose that you get heads every flip. How many flips do you need to conclude that is more likely that you’ve selected one of the 60% heads coins rather than one of the fair coins?
I haven’t worked through the numbers yet, but I never seem to guess very well on problems like this. I’m trying to formulate a good guess before doing the number crunching.