An introductory stars and bars problem

Yesterday a counting problem from my son’s math team homework gave him a little trouble. The problem went something like this:

There are 5 different types of fasteners and you need to buy 10 total. If you need to buy at least one of each, how many different ways can you do it?

First we talked about the problem and got their initial thoughts. Then I introduced the stars and bars counting idea:

Next I tried to go through a few more examples by changing the numbers a little. The main ideas seemed a little confusing to the boys and I’d hoped a few extra examples would help. Unfortunately things weren’t going so well.

The last example in the prior video confused my younger son, so I moved on to the next video to talk about that example in more detail. By the end of this example I hoped that the general idea had sunk in, but there was still a little confusion.

So, we talked through the problem a few more times. Now the ideas seemed to be sinking in. IF you have N groups of objects (in the original problem 5 fasteners) and you have to pick M total objects (in the original problem we were trying to pick 5 fasteners) you can represent the problem with M stars and N – 1 bars. So the total number of different ways to make the selections are (M + N – 1) “choose” M or, alternately, (M + N – 1) choose (N – 1).

Not all of our projects go super well. Here my mistake was thinking that I could introduce an advanced concept and the boys would immediately understand it. I feel like the ideas here are definitely within their grasp and will probably spend a bit more time this weekend covering the concept. Hopefully a few more examples will do the trick.

Working through two old contest problems

I’ve been sort of on pause doing new math with the boys for the last couple of weeks. I want them to find their stride with the new school year before seeing what additional enrichment math we can do at home.

So, while on pause they’ve just been working through problems from old AMC tests in the morning. When they finish we talk through some of the problems that gave them trouble. Both problems were pretty interesting lessons (for them and me, I mean) today.

Here’s what my older son had to say about problem 18 from the 2013 AMC 10b. It was fascinating to me how he counted the numbers in this problem.

and here’s what my younger son had to say about problem #17 from the 1985 AMC 8 (which then was the American Junior High School Math Exam). It was fascinating to me to see both how he played with the averages and how he found his arithmetic mistake.

I love using these old AMC problems to keep the kids engaged with math. It is always fun to see what sorts of ideas give them problems and just as fun to see their problem solving strategies.

A great counting problem for kids from the AMC 8

This problem from the 2015 AMC 8 gave my son some trouble today. Actually quite a bit of trouble:

I’m not 100% sure what caused the difficulty. It might be that once you start thinking about this problem one way that it is hard to switch. Whatever the cause, though, we had a really good conversation about the problem.

Here’s his original approach that is incorrect:

So, after finding out that the answer in the last video was incorrect, we went to try to find the error. He found it pretty quickly.

After that I tried to explain an alternate approach to the problem. Unfortunately my explanation ended up causing quite a bit of confusion:

In the last part of our discussion I tried to dig my way out of the hole I created in the last video.

Even watching this video after the fact, I’m not sure what was the original source of his confusion. There was definitely some difficulty going from 4 parallel edges to 6 pairs of parallel edges.

By the end of our conversation he was able to walk through the argument, but I think that I’ll revisit some similar problems with him just to be sure the main ideas have sunk in.

I think this short project is a nice example of how old contest problem can help kids learn math. For me anyway, it is really challenging to come up with good problems and the fact that all of the old AMC problems are available for kids to work through is an incredibly helpful resource. Hopefully I can find some similar counting problems on other old AMC contests.

Going through three AMC 8 problems

My younger son has been doing a little practice for the AMC 8. Yesterday three problems from the 2013 exam gave him a little trouble. We went over them together.

The first was a tricky geometry problem. Both the words and ideas needed to solve this problem were new to him.

Next up was a challenging counting problem – we broke this into two pieces. This is a great counting problem for kids. In the first part we found out how to calculate the answer, but didn’t finish the calculation:

In the second part we talked about strategies to finish the calculation:

Finally – a fantastic geometry problem. It has a few little traps in it, but my son found a nice solution.

I love using the old AMC contest problems to help the boys see math that is both fun and challenging. These problems were really fun to talk through.

Two AMC8 problems that gave the boys a bit of trouble

I had a couple of things going on today and just asked the kids to work through an AMC 8 rather than doing a longer project. Each had one problem that gave them some trouble, so we turned those problems into a short discussion.

Here’s the first problem – this one gave my younger son some trouble – it is #21 from the 1992 AMC 8:

Here’s our discussion of the problem:

Here’s the 2nd problem – it is problem #24 from the 1999 AMC 8.

There’s some questionable advice from me and also some terrible camera work, but it was a nice discussion!

I like using the AMC problem to help the kids see a wide variety of accessible mathematical ideas. Despite being in a bit of a rush today, this was a fun project.

Going through an IMO problem with kids

Last week I saw this problem on the IMO and thought that the solution was accessible to kids:

The problem is problem #1 from the 2017 IMO, just to be clear.

My kids were away at camp during the week, but today we had a chance to talk through the problem. We started by reading it and thinking about some simple ideas for approaching it:

The boys thought we should begin by looking at what happens when you start with 2. Turns out to be a good way to get going – here’s what we found:

In the last video we landed on the idea that looking at the starting integer in mod 3 was a good idea. The case we happened to be looking at was the 2 mod 3 case and we found that there would never be any repetition in this case. Now we moved on to the 0 mod 3 case. One neat thing about this problem is that kids can see what is going on in this case even though the precise formulation of the idea is probably just out of reach:

Finally, we looked at the 1 mod 3 case. Unfortunately I got a little careless at the end and my attempt to simply the solution for kids got a little to simple. I corrected the error when I noticed the mistake while writing up the video.

The error was not being clear that when you have a perfect square that is congruent to 1 mod 3, the square root can be either 1 or 2 mod 3. The argument we go through in the video is essentially the correct argument with this clarification.

It is pretty unusual for an IMO problem to be accessible to kids. It was fun to show them that this problem that looks very complicated (and was designed to challenge some of the top math students in the world!) is actually a problem they can understand.

Working through an AMC 8 geometry problem

My younger son was working through the 1989 American Junior High Mathematics Exam this morning and got stuck on this problem:

Here’s a link to the entire exam on Art of Problem Solving’s website:

The 1989 AJHME on Art of Problem Solving’s website

I thought this problem would make for a nice project since there are a couple of good mathematical ideas in it, so we sat down to talk about it. My younger son talked through his approach first:

My older son went next and had a different approach:

To wrap up we talked about how the answer would change if the problem was set up with a slightly different arrangement of the cubes and the boys found their way to an important idea in geometry:

I’m really happy that the old AMC problems are available – they are a wonderful resource to use to find challenging but accessible problems for kids.

A challenging counting problem for kids learning algebra

My son is in a weekend enrichment math program and that program has been great for him. It comes to an end this week. The last problem on this week’s homework assignment gave him some trouble, so I thought it would be fun to see if we could work through it together.

I was a little worried because I’d not seen the problem until just before the project, but luckily things went ok.

Here’s the problem:

a, b, c, and d are positive integers less than 10. How many solutions are there to the equation a + bcd = ab + cd?

[post publication note: Originally the text presented the problem incorrectly. It is correct in the videos. Karen Carlson pointed out the typo to me. Sorry about that.]

Here’s how we got started – my son had found several cases, but not quite all of them:

After the introduction to the problem and my son’s work so far, we moved on to try to find more solutions. The main idea I gave my son involved writing the equation in a slightly different form:

Now that we had a plan, we moved on to counting the rest of the cases that we found in the last video:

Finally, we went to Mathematica to write a little program to count the solutions for us. This part of our project turned out to be more interesting than I was expecting. It was interesting to compare the brute force solution of the computer to the case by case counting technique that we’d just gone through.

So, a fun problem that definitely made my son think this week. It is

Christopher Long’s fun generalization of an Expii problem

Twitter is really great place to see fun math. Before showing the fun generalization, though, just to avoid spoilers I want to show the original problem. Here’s the direct link:

https://www.expii.com/solve/69/5

and here’s the problem itself:

So, I’ll pause here to not ruin the problem for anyone who wants to work on it.

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Ok . . . here’s the really cool set of tweets I saw from Christopher Long this morning:

which continued as follows:

To see how delightful Long’s general solution is, maybe walking through my chicken scratch solution which happened to be still sitting on my desk will be helpful:

Here’s a sketch of my approach.

In order to maximize your chance of winning a bet like the one in the problem (one that you expect to lose) you should bet as much as you can at each step (subject to a maximum of the amount you need to win) at each stage.

So,

(i) at step 1 the probability of getting the tree to grow to 40 feet is 1/5 and probability of losing is 4/5.

(ii) Assuming you win, you now have a 1/5 probability of getting the tree to grow to 80 feet and a 4/5 probability of losing.

(iii) Assuming you win, you now have a 1/5 probability of getting the tree to grow to 100 feet and a 4/5 probability of having it shrink to 60 feet.

(iv) If you win on stage (iii) you win (1 out of 125 times). If you lose, you now have a 1/5 probability of having the tree now grow to 100 feet and a 4/5 probability of having the tree shrink to 20 feet.

So, after the 4th branch in my picture you’ve either won (probability 1/125 + 4/625), returned to 20 feet (probability 16/625) or lost (the only other case).

Thus, your probability of winning the game from the start, $x$, satisfies the equation:

$x = 1/125 + 4/625 + (16/625)*x$

We can solve this pretty easily to see that $x = 3/203$.

The really fun – and honestly, amazing – thing about Long’s solution is that he notices that the pattern in the branches of the binary tree corresponds exactly to the pattern in the digits of the binary expansion of 1/5. For clarity, the 1/5 here comes from the growth multiple – 20 feet growing to 100 feet – and not from the probability which, by coincidence, also has a 1/5 in it.

Anyway, Long’s solution also allows you to immediately see how to solve any problem like the Expii one, and, for extra fun, problems where the growth multiple is irrational:

The answer is in Long’s timeline, but it is a good challenge to see if you can work out the answer just from the tweets I’ve included here. Since he skips a bit of algebra in his tweets, working through his tweets is also an important way to make sure that you really understand his work.

I think the sequence of tweets from Long are a great thing to show kids who are learning math – especially kids learning probability and stats. Those tweets really show how a mathematician thinks about a problem.

A neat expected value problem from Expii

[sorry for the quick write up – I got asked to help out with my son’s archery class today, so I just decided to publish this one as it was when I get asked to help . . . ]

I saw a neat expected value problem from Expii yesterday. In case you’ve not see their site, here’s the link to their main site:

Expii’s front page

and here’s a direct link to the problem:

A neat expected value problem from Expii

The problem goes like this:

“You are planting some trees as environmental action for Earth Day. At each of 200 spots around a circle, you place a seed. Each seed will sprout into a small tree with probability 1/2. Sadly, some of these small trees will die. In particular, a small tree dies if it has another small tree as its neighbor, because they will be fighting for sunlight.

What is the expected value of the number of trees that are still alive at the end of the year?”

I thought this would be a great problem to discuss with the boys. We just got back from a vacation in San Diego and my younger son was still on west coast time, though, so I just talked through this one with my older son.

First I introduced the problem and we double checked that he understood it:

Next we discussed some simple cases to see if we could get our arms around the problem:

Now we moved on to the general case. My son understood some of the main ideas about the problem, but made a small mistake at the end that led to a very small expected value.

Finally, we wrapped up by looking at the error at the end of the last video and trying to calculate the expected value slightly more carefully: