# A little screw up by me . . .

This morning I wanted my son to have an easy morning and just start in on the next chapter in his book. I don’t remember the exact words – it was 6:30 in the morning – but they were something like “just start in on the next page.”

What my son did was sit alone and work on the problem that was on the back of the page he was on . . . here it is:

Let A and B be positive integers. If A! / B! is a multiple of 4 but not a multiple of 8, then what is the largest possible value for A – B?

I’d say that this problem is a little bit above 4th grade level . . . . whoops.

But we talked about it and how you might begin to think about it. First, though, I let him share some thoughts. You’ll see that it was tough for him to even know where to begin:

Because he was struggling to see where to go, I just went back to the beginning. We started by looking at a chart with some small numbers. The largest difference we could find was 3.

Finally we looked a bit more carefully at the situation when the difference was 3 and when the difference was 4. He was then able to see the reason why the difference could never be 4. I was pretty excited and happy to see him make this last connection – at least a bad mistake by we ended on an up note!

# Justin Solonynka’s Train counting project

I few weeks (months?) ago I learned about Justin Solonynka’s train counting project from a Tracy Johnson Zager tweet:

It looked like a super fun project and we finally got around to trying it ourselves last night. This is one that I’d like to have a 2nd crack at because the first time around didn’t go quite as well as I was hoping – but that’s life I suppose.

Still . . . I was happy about the counting ideas we did get to discuss in the project, and I was happy that we got to the end of a pretty difficult counting problem. Hopefully Justin’s incredible video all by itself is enough proof of how great a project this is, but here’s how ours went:

I started by having them open the puzzle box and ask some mathematical questions:

So, the boys decided to count the total number of trains you could make, and we started counting. The first path we went down didn’t quite capture all of the arrangements, but the mistake in counting was actually pretty instructive.

At the end of the last video the kids thought there would be only two trains with one train car, but at the beginning of this video they notice that there are actually 20. This observation means that we’ll have to correct the counting from the last video and they work through this revised counting problem here.

There’s a little bit of confusion about how to revise our counting procedure, but working through that confusion is a really important part of learning to count.

Finally, I wanted to show them a slightly different way to count the trains by keeping the factorials from the second video around. I didn’t do nearly as good a job here as I’d have liked to do, but eventually they found the idea of choosing numbers.

So, we got a nice contrast from the last video where we just looked at permutations. For three train cars, for example, the contrast is that 10*9*8 arrangements is the same number of arrangements as choosing 3 cars out of 10 and looking at the 3! ways you can arrange those cars.

So, a nice counting project even if our walk through it wasn’t so great. I’m happy that Tracy shared Justin’s video – it is a great way for kids to learn about counting.

# 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!

# Introduction to counting with combinations

This morning we started looking at combinations and “choosing” numbers. I think this is one of those topics that seems much easier after you’ve learned it than it seemed when you were learning it. After a quick introduction to the idea of choosing number, the boys worked on an example problem that involved counting the number of games in a round robin tournament:

The were able to work through that problem, but I thought that thinking through it again would still be helpful, so we reviewed the problem from start to finish.

After that quick review we began to talk about choosing numbers. I let the boys explain their ideas about these numbers. One fun thing that happened in this part of our project is the boys discovered that there is a little bit of symmetry in these numbers:

Now we looked at patterns that arise in the choosing numbers. There’s a little trick that I mistakenly thought we’d already talked about – that 0! = 1. After telling them that was simply a definition, we moved on to finding a fun pattern that comes up in the choosing numbers – Pascal’s triangle!

Finally, as a way of confirming the connection to Pascal’s triangle, we looked to see if the addition relationship between two rows of the triangle also shows up in the choosing numbers. This is one of the first examples of a combinatorial proof that the boys have seen!

So, a really fun project showing that counting has some surprising connections to other topics in math. It was fun to hear their ideas (and their surprise) when they found the connection to Pascal’s triangle. Showing them a basic combinatorial proof at the end was fun, too – those proofs can be absolutely amazing!

# Fibonacci Factorials

As part of the publicity around the Fields Medal announcement, the American Mathematical Monthly’s Facebook page pointed out this paper written by Manjul Bhargava in 2000:

The Factorial Function and Generalizations

The reason that this paper caught my attention is that I actually felt like I had a prayer of understanding the ideas that the paper was explaining.  I’m sure, of course, that the work of all of the 2014 Fields Medal winners is off the charts brilliant, but when I searched for lectures or papers by them everything but this paper was miles over my head.  So many miles, actually, that the factorial function might come in handy if I needed to describe that distance accurately!

One particularly helpful example that Bhargava gives in the paper is on top of the 6th page (page number 788) where he calculates the first six values of the generalized factorial function over the primes.  Since he said that it was “an easy matter to compute” these six values, I thought that replicating this calculation would be a fun way to see if I had understood some bits of the paper.    After a few times through the paper I finally had understood the ideas well enough to replicate the calculation, so yay!  I was also surprised to see that those six numbers (1,1,2,24,48, and 5760)  appear in the On-line Encyclopedia of Integer Sequences only once (and without reference to Bhargava’s result):

Is this the full generalized factorial function over the primes?

I don’t know if the full sequences would match each other (or, obviously, why that sequence would arise from the Taylor series of $log(1 + x)^2 / \sqrt{1 + x}$) but at least my calculation of the next term in Bhargava’s sequence does match the next term in the OEIS example.

So, having (hopefully) understood how to calculate this generalized factorial function over the primes, I wanted to try it for another sequence of integers.  The Fibonacci numbers seemed like as good a place as any to start, so I gave that a shot this weekend.  Unluckily I was traveling this weekend, but still found a little time early this morning at the Lone Wolf diner in Amherst, MA to get things going while enjoying their Santa Fe omelette.    According to my calculations, Bhargava’s factorial function over the Fibonacci numbers would evaluate as follows:

0! = 1

1! = 1

2! = 2

3! = 6

4! = 24

5! = 240

6! = 720

7! = 443,520  ($2^7 * 3^2 *5 * 7 * 11$)

8! = 443,520 (yes, the same as 7!.  That’s a surprise, but I haven’t been able to see the mistake)

9! = $2^8 * 3^4 * 5 * 7 * 11* 13$ = 103,783,680

10! = $2^9 * 3^4 * 5^2 * 7 * 11* 13$ = 1,037,836,800

[edit note:  after publishing earlier today, I noticed that I left of the 13 on both 9! and 10! ]

Fingers crossed that these are the correct calculations but since I slept at a farm last and was woken up early by roosters, it wouldn’t be super surprising if there was an error 🙂  In any case, one interesting thing that I learned playing around with this is that the Fibonacci numbers have an interesting pattern modulo 11.   I’m hoping to play around with this and other sequences in the next month.  I think there is a really fun project for kids hiding in here somewhere, too.