Summer math talks

I’ll be giving two talks at math camps this summer. The first is at the east coast Idea math camp at the beginning of July and the second is at a math camp at Williams college in the middle of July.

I’m super excited to be able to have the opportunity to give these talks and can’t wait for the chance to interact with the students at the two camps.

The topics I’ll cover – not surprisingly – will come from some of the projects I’ve done with the boys this year. The talk at Williams is about 45 minutes shorter than the Idea math talk, so one or two of the topics below will get cut out.

Here’s are the ideas I’d like to cover:

(1) Larry Guth’s “No Rectangles” Problem

Larry Guth’s “No Rectangles” problem

I covered the 3×3 and 4×4 cases with the 2nd and 3rd graders at my younger son’s school as part of Family Math night and the kids loved the problem. With high school students I’d like to try to explore some of the larger cases and also discuss why this is a difficult problem for computers to solve.

Patrick Honner also showed me this related problem which I’ll leave as a challenge for the students 🙂

(2) Ann-Marie Ison’s Math Art

Our projects with Ann-Marie Ison’s art

I’m still waiting to hear what sort of projection capability I’ll have at the two events, but oh do I hope I have the ability to share this program with the students:

The explorations you can do with this simple modular arithmetic idea are incredible.

 

(3) A problem from Po-Shen Loh’s MoMath talk

What I love about Loh’s talk is that he takes an extremely difficult problem – one from the 2010 International Mathematics Olympiad – and turns it into a talk that is accessible to the public.

His approach is so accessible that I talked through the first part of the problem with my 4th grade son:

I’m very excited to hear the different guesses that the students have for the answer to Loh’s two questions.

(4) Bjorn Poonen’s N-Dimensional Sphere problem

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Here’s the problem and our project on the problem:

A strange problem I overheard Bjorn Poonen discussing

Bjorn Poonen’s sphere problem

I’m guessing that not all of the kids will have seen geometry beyond 3 dimensions, so this problem will take a little bit of setting up.  Luckily the only complicated bit of math that they need to understand it is the Pythagorean theorem and I’m guessing that all of them will know that theorem.

I was blown away by the answers to Poonen’s questions when I finally worked through them.  This was also one of the most enjoyable projects that I’ve done with the boys this year.

I hope I have enough time to show the students the fun relationship between \pi and e hiding in this problem, too:

A fun surprise in Bjorn Poonen’s n-dimensional sphere problem

 

Can’t wait to talk about these problems with the kids!

Michael Fenton’s counting problem

[I hope this write up makes sense – I wrote it while my kids were watching Lord of the Rings “Two Towers” – so there was a bit of a distraction in the background]

Saw this problem posed on Twitter yesterday:

I like Fenton’s request to share solutions, and actually made a similar request previously with a problem from the Euorpean Girls’ Math Olympiad:

A Challenge / Plea to Math Folks

Having made a similar request, I’m sort of obligated to respond to this one!

There are a couple of different ways to approach Fenton’s problem, and I’d guess that the counting technique of “inclusion / exclusion” would probably be a common choice of how to attack the problem.

However, since the numbers were small enough I thought it would be fun to see if I could count the possibilities directly.

The main ideas I used to approach the problem were these:

(1) I’m going to try to count the arrangements with no songs by the artists next to each other. If I can find the probability of finding such an arrangement then 1 – that probability is the probability of having at least one pair.

(2) Next, I’ll ignore the different songs by different artists and just look for arrangements of the 8 numbers 1,1,2,2,3,3,4,4 with no numbers that are the same next to each other. If I can count these sets, the the arrangements of the number of songs with no songs by the same artist next to each other is just 2^4 times the number of these sets.

(2) For any arrangement of the 8 numbers – say 1,4,3,2,3,4,1,2 – a unique permutation of the numbers 1,2,3,4 will change the arrangement to one in which we encounter the numbers in increasing order. In my example, simply switching 2 and 4 changes the example to 1,2,3,4,3,2,1,4.

For a second example. Consider 3,2,4,3,2,1,4,1. If I map 1 to 3, 2 to 2, 3 to 4, and 4 to 1 I have the new sequence – 1,2,3,1,2,4,3,4 – in which we, again, encounter the numbers in the order 1,2,3,4. So, using this idea, if I can count the number of arrangements with no pairs in which the numbers appear in order, all I need to do is multiply by 4! to get the total number of arrangements of similar patterns.

So, let’s count:

(A) Arrangements of the form 1,2,3,4,_,_,_,_

There are 3 choices for the 5th numbers, and after that choice the remaining 3 numbers can be arranged in any of the 3! ways. So, there are 3*3! arrangements of this form, or 18.

(B) Arrangements of the form 1,2,3,1,_,_,_,_

There are two cases to consider:

Case 1: the 5th number is 4. There are then 2 choices for the 6th number (2 or 3) and 2 ways to arrange the remaining 2 numbers ( 4,2 or 2,4, for example, if the first choice was 3). Thus there are 2*2 = 4 arrangements of this form.

Case 2: The 5th number is 2 or 3. So, there are 2 choices for the 5th number. 4 must be the 6th number otherwise the arrangement would end 4,4 which we don’t want. The final two numbers are then forced. Thus, there are 2 arrangements of this form.

So, a total of 6 arrangements begin 1,2,3,1

(C) Arrangements of the form 1,2,3,2,_,_,_,_

This case is essentially the same as case (B) – so we get 6 more arrangements.

We’ve now covered all of the cases of the form 1,2,3,_,_,_,_,_ and the only other cases begin with 1,2,1 since the numbers must appear in order.

(D) Arrangements of the form 1,2,1,2,_,_,_,_

There is only 1 – which sort of surprised me (!) but the remaining slots are forced to be 3,4,3,4 by the increasing condition and the condition that no numbers appear next to the same number.

(E) Arrangements of the form 1,2,1,3,_,_,_,_

As with (B) above, we have two cases:

Case 1: The 5th number is 2.

Here the arrangement is forced to finish 4,3,4. So, there is only 1 arrangement in this case.

Case 2: The 5th number is 4.

Here we have 2 choices for the 6th number – 2 or 3. For either choice the remaining two numbers can be arranged in any order (4,2 or 2,4 if the choice for the 6th number was 3, for example). Thus there are 4 cases here.

So, part (E) has 5 total cases.

Putting it all together.

So, that’s all the cases, and we end up with 18 + 4 + 2 + 4 + 2 + 1 + 1 + 4 = 36 total arrangements in which (i) the numbers appear in increasing order, and (ii) there are no pairs.

As mentioned at the beginning, when we distinguish between the two 1’s, the two 2’s, the two 3’s, and the two 4’s, the number of arrangements is multiplied by 16. Also, when we no longer require the numbers to appear in order we multiply the number of arrangements by 24.

So, the total number of arrangements with no pairs is 36*16*24. The total number of arrangements is 8!, so the probably of having no pairs is:

(36 * 16 * 24) / 8! = (36 * 16 ) / ( 8 * 7 * 6 * 5) = 12 / 35.

Thus, the probability of having at least one pair is 23 / 35.

Talking through Ben Orlin’s exponent problem

Saw this problem last week via a Joshua Bowman retweet:

I thought it would be a good problem to talk through with the boys this morning, and it turned out to be an even better discussion that I was expecting.

Talking through this problem with kids gives you the opportunity to discuss lots of specific ideas in arithmetic. Three that come to mind are:

(i) exponent rules,
(ii) different ways to represent numbers,
(iii) the differences between adding and multiplying.

There’s also a great opportunity to discuss problem solving, and that discussion happens in the 2nd video when my older son says “we don’t really know anything about [the number] 2^{1/3}.”

Here’s the beginning of our discussion. The first idea that my older son has is writing both numbers in the form a^{1/10}.

Before we went down that path, though, I wanted to make sure that my younger son understood a little bit about the two different ways to write numbers as nth roots. So, the discussion here is half about exponent rules and half about our start at solving the problem:

In the last video we found that Orlin’s problem was equivalent to comparing 10 with 2^{10/3} but we sort of got stuck there. Getting stuck at this step was fascinating to me because [to me!] it was so obvious what the next step was. The ensuing discussion taught me a lot about how kids see numbers.

In the next part of the project we explored comparing 10 and 2^{10/3} a bit more. The first new idea in this part of the discussion was to remove a factor of 2. Now we were left comparing 5 and 2^{7/3}. What now?

My older son has a neat observation – if we knew if 2^{1/3} was larger or smaller than 1.25 we’d be able to answer the question. This particular observation comes from understanding a little bit about the relationship between addition and multiplication.

How, though, do we compare 2^{1/3} to 1.25?

So, having solved the problem at the end of the last video, I wanted to show the boys the actual values on the calculator. Unfortunately, I decided to zoom in on the calculator and then not zoom out for the rest of the discussion. Hopefully the words in that discussion are enough to know what we were talking about 🙂 Sorry about the camera goof up – at least the camera mistake was in the shortest video!

Definitely a fun project. I think Orlin’s problem is a great one to talk through with kids learning about exponents. There are so many different ways that conversation can go, and so many opportunities along the way for kids to think about and discussion ideas about numbers.

Playing “The Witness” with kids is an amazing experience

I learned about a new game called “The Witness” from this Vi Hart tweet last week:

https://twitter.com/vihartvihart/status/692128282876186626

After seeing a few reviews and the game trailer, I thought I’d give it a try:

 

The game is really fun if you like solving puzzles, but the pleasant surprise about this game is how fun in is to play with kids. So far none of the puzzles are so hard that kids can’t solve them, though figuring out how to solve to solve each particular one is far from obvious!

I think another Vi Hart tweet has the best description of the game:

https://twitter.com/vihartvihart/status/693927287343837184

Here’s a little pic from our first time playing

   

and a video of my son solving one of the puzzles – which probably doesn’t give you the exact right impression of the game, but it is better than nothing:

   

and when my son was sick this morning we sat on the couch together trying to draw the right solution to some puzzles we were seeing (essentially) in a mirror.


The kids love the open-endedness of the game – so if we can’t figure out how to solve all of the puzzles in one section we can just go somewhere else. My younger son called this a “sandbox game” which is a term I’ve not heard before – but he says the ability to go to other places is just like playing in a sandbox. The also love the challenge of solving the puzzles.

Anyway, with all of the hype around the release of the game last week, I just wanted to say that if you are looking for a mathy game to play with kids, this game is for you!

An interesting AMC 10 probability problem

Problem 20 from the 2006 AMC 10 gave my son some trouble last night:

Problem #20 from the 2006 AMC 10 A

Here’s the problem:

Six distinct positive integers are randomly chosen between 1 and 2006, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of 5?

I like this problem because it seems overwhelming at first, but a little exploration reveals the hidden structure. The lessons learned working through this problem seemed important enough to revisit again this morning.

Here’s the introduction to the problem and a few test cases:

 

At the end of the last video we started talking about the Pigeon Hole principle – a simple sounding, but really powerful idea in math.

 

In the last part of our talk today I introduced the idea of looking at the remainder when you divide by 5. This idea uses “pigeon holes” that are slightly easier to see (but only slightly), but it gave a chance to look at the problem one more time through a different lens:

 

So, a fun and really instructive problem. As I think about it, I’d love to see how other people would approach this problem with younger math students.

A challenging factorial problem

Yesterday my older son and I worked through a a challenging algebra problem. Today’s challenging problem involved factorials. The problem is #23 from the 2015 AMC 10b:

Problem 23 from the 2015 AMC 10b

Here’s the problem:

Let $n$ be a positive integer greater than 4 such that the decimal representation of n! ends in k zeros and the decimal representation of (2n)! ends in 3k zeros. Let s denote the sum of the four least possible values of n. What is the sum of the digits of s?

We started by just talking through the problem and coming up with a plan:

Next he implemented that plan and did a great job working through to the end of the problem:

Finally, we went to Wolfram alpha and double checked that the numbers he found were indeed solutions to the problem (and sorry for the stumbling around by me in this part!)

So, a challenging problem and a good solution from my son. We continue to work on ideas about problem solving. It is always nice when everything comes together like it did for today’s project!

An experiment with a difficult algebra problem

Instead of working through some medium level AMC 10 problems today, I decided it would be fun to walk through a single really challenging problem with my older son.

The problem we chose was #23 from the 2015 AMC 10b:

Problem 23 from the 2015 AMC 10a

Here’s the problem:

The roots of f(x) = x^2 - ax + 2a are integers.  Find the sum of all values of all possible values of a.

Though this problem is difficult, I choose it because my son is able to understand the solution.  Over the course of about 20 minutes we worked through that solution – slowly.  There were a few misconceptions along the way, but I hope this was a productive exercise.

We began by talking about the problem itself and getting a few initial thoughts from him:

In the second section of the talk he began to try to play around with the values of the roots, but that didn’t go anywhere. Eventually he thinks to try the quadratic formula which leads to an interesting breakthrough:

Now he’s found a new equation that he knows needs to be a perfect square to make the original equation in the problem have integer solutions. Here we start down the path to finding when our new equation is a perfect square, but we hit a strange problem about dividing by zero. This problem causes a little detour:

So, with the division by 0 problem out of the way we returned to trying to find when our new equation could be a perfect square. In this part of the solution he checks a few cases by hand.

Finally, we wrap up by trying to see if we’ve found all of the solutions. There a little bit of number theory / number sense that helps us out here:

So, a tough problem for sure, but nice to see that each individual step is something that my son was able to understand. Learning to pull together all these pieces is an important part of problem solving.

A walk through a challenging probability problem

Saw a great problem from the 2009 AMC 10 last night:

Here’s the problem:

Problem 24 from the 2009 AMC 10a

3 verticies of a cube are chosen at random.  What is the probability that the plane formed by those three verticies contains points inside of the cube?

This is a great problem (and a challenging one, too).   My older son took a few missteps walking through it, but our whole conversation about the problem this morning was great.

here’s the first part:

 

In the second part he takes a little misstep, but hopefully the mistake turned into a good learning opportunity:

Now we back up to where the misstep happened and try to get to the end of the problem:

A challenging algebra problem from the 2011 AMC 10b

This problem gave my son some trouble this morning:

Problem #19 from the 2011 AMC 10b

The problem asks you to find the product of the roots of the equation:

\sqrt{ 5|x| + 8} = \sqrt{x^2 - 16}

A lot of ideas from Algebra come into play on this problem and we had a good conversation about it tonight.

We started by talking through his ideas about the problem and found a couple of interesting misconceptions about algebra and equality right at the beginning:

Next we went back to look if the 4 solutions that we found to the equation were actually solutions:

Finally, I went back through the problem to show him how the 3 and -3 showed up even though they are not solutions.

The emergence of these spurious solutions makes this a really instructive problem.

“Simplify don’t complify”

My older son had trouble with a practice problem from his math club this morning. Part of the problem can from trying to slog through lots of arithmetic.

We went through the problem this morning using his approach – we were running a little bit low on time so we did some (though not all) of the calculations with a calculator. Part of the point of what I wanted to show him this morning was that all of this arithmetic did indeed make the problem a little confusing:

 

When he got home from school we talked through the problem using a slightly different approach. This approach involved trying to do as little calculating as possible:

 

I love my son’s old math motto – “simplify don’t complify”. He came up with this motto when he was 5 or 6 🙂 This is a nice example of a problem where getting caught up in the calculations makes the problem much more difficult and confusing than it needs to be.