## Tuesday, October 27, 2015

The most exciting part of this week was I think I've found another father who is willing to help assist me. Today was his first time coming and helping out. This is great on two fronts. First the optimal ratio of kids:adults for me 8:1 and a second person lets two activities occur at the same. Secondly if this works out I'll be able to open up the wait list. At this point I think I have at least 5-6 kids whom I did not have room for this quarter in the math club. I'm really excited to be able to plan more flexible and complex transitions in the upcoming weeks.

So for today we started with candy in the lunchroom. I also had 5 sheets turned in from last week's take home problem. As I've mentioned before its fairly crude, but the naked bribery experiment is producing results. What I'm most excited about from today is the kids produced the three main solutions I was hoping they would find. As  a result I was able to I have them demo all of them on the whiteboard.

There is a curious 5 digit number A which when you add a 1 to the end of it is three times larger than when you add a one to the front. What is A?

The first way is to setup the algebraic equation 10A + 1 = 3(A + 100000) which I wasn't sure was within their skill set. (In full disclosure I think some of the parents are working with their children)

Secondly you can treat this as decoding problem:

1ABCDE
x          3
ABCDE1

and working right to left use a little number theory to find each digit. For example E must be 7 since
only 3 x 7 ends in a 1 and so on.

Finally its also possible to do a brute force search through the number space especially with a bit of intelligent bounding.

I'm super proud of how well they all did this week.Which leads to my other observation. This years kids are really good at working during club as well. They almost always form productive groups and grind a way at the problems. I can definitely see the fourth grade vs fifth grade difference in terms of algebraic awareness. But I'm hoping I'll be able to move them forward quite a bit over the year if this keeps up.

For the main activity I had everyone do a practice Olympiad since the first real one is coming in 3 weeks. This went well. I will report that it contained the STAR * 4 = RATS problem in it yet again. This must be the third or fourth place this problem has showed up. As expected this was the hardest problem for most of fourth graders. In followup when we worked through the problems as a group it looked like the decoding procedures were understandable. Also for the followup I was able to split the club into two sections and had the smaller size groups each with an adult which is really great for focusing on listening to each other.

Finally with the extra time after we finished going over the solutions we did some match stick problems (well actually tooth pick problems in my case since I don't risk burning down the school):

Example:

Form a grid like on the left and then moving only 12 sticks form two big squares: (ignore the right hand picture)

These worked really well I presented 4 challenges which was more than enough for the end of the club. Finally I left the kids with following take home challenge.

In triangle ABC, three lines are drawn parallel to side AC dividing the altitude of the triangle into four equal parts. If the area of the second largest part is 35 (the blue region), what is the area of the whole triangle ABC ?

## Tuesday, October 20, 2015

### 10/20 Grids and Graphs

This week I planned a series of more playful activities. We started with a review of the take home problem of the week: (September from www.moems.org/zinger.htm).  I already knew this was easier than October since 3 kids had told me the answer almost immediately when I handed it out. I had 8 kids complete the problem so I will be bringing M&M's to math club next week.  To wrap it up, we did another white board exercise with 2 kids showing their work. Listening worked better this week. My only concern was one student took fairly long to copy her work onto the board and I fretted unnecessarily that I would lose the club's attention.

After that I decided to do a quick demonstration with the distributive law why a negative times a negative is a positive.

For example:
-5 * -6 + -5 * 6 = -5 (6 + -6) = 5 * 0 = 0
Then go back to the original sum -5 * -6 + -5 * 6 = 0.
We know the second term is -30 already since its a negative times a positive.
So -5 * -6 + -30 = 0   which implies  -5 * -6 =  30.

I had each kid pick their own product and work along. This strategy worked OK but I ended needing to go over several more examples from the kids on the board. Overall this was a stretch and next time I should start with a review of factoring with the distributive law even if we did it the week before.

This entire process took about 20 minutes and we then switched to over to a 3x3 logic problem from http://www.logic-puzzles.org. I went much simpler than last year based on my previous experience and even so this was about 30+ minutes of work for most of the kids. Moral of the story I estimate a 4x4 would probably take a full hour if I want to fill one up.

Because things went longer on the logic puzzle I didn't have as much time for my main selection. I did a quick edit of my original ideas and took out the loop-de-loop exercise from: http://www.amazon.com/This-Not-Maths-Book-Activity/dp/1782402055.  As others have reported on the net this was a big hit with the kids. There were a lot of excited students who wanted to show me the different spirals and loops they had created.  I ended up shorting the time I wanted to spend on trying things out so I think I may revisit this at the beginning of next week and have a wrap discussion about what patterns everyone found.

#### Problem of the week courtesy of  Martin Gardiner:

There is a curious 5 digit number A which when you add a 1 to the end of it is three times larger than when you add a one to the front. What is A?

I told everyone to try some strategies on your own first and give this a little time if you get stuck before looking at the hint. This will definitely be harder than last week and I'm looking forward to seeing what the kids come up with.

Hint:
"∀ ɟo sɯɹǝʇ uᴉ sɹǝqɯnu ɹǝɥʇo ɥʇoq ɹoɟ suoᴉssǝɹdxǝ ǝʇɐlnɯɹoɟ"

#### Planning

Its time to try out a sample Olympiad next week!

## Tuesday, October 13, 2015

### 10/13 The Distributive Property: not just a good idea

After last week I knew that I needed to start with a review of the problem that I handed out to do at home. The great news was 8 of the kids worked on it over the week! So far naked bribery with candy is working and we will keep going with the homework experiment. At this rate, I'll be bringing in M&M's in 2 weeks. In going over the problem, this was our first group white board exercise so I emphasized listening was important before anyone did any math. To get started I also asked the entire room what the formula for the area of a circle is. As I noticed last year that tends to be a bit shaky despite being covered once a year I believe starting in 3rd grade. I then had 3 kids volunteer to  show parts of the October problem of the month (including my shyest student) which was great. I decided to narrate what the kids were doing in a louder voice to make sure the whole room could hear.

We then transitioned to a warm-up with stations. I found this great equivalent fractions puzzle on twitter from NRICH:  Puzzle  which I pre-cut a few versions of. I also brought Game of 24 cards  for the first time. I then let the kids choose which station to go to and we played around for about 15 minutes.  Generally both activities kept everyone's attention. I may alter this structure next week and do a practice Math Olympiad first but in general I still believe games are really important at this age.

Finally for the main activity I wrote up my own distributive property worksheet last night with a progressive set of problems: http://mymathclub.blogspot.com/2015/10/distributive-law-worksheet.html.   I think I did a bit better than my last attempt at worksheet creation and in general this was reasonably leveled for the kids. As I told the kids, at this stage in their math careers the distributive law is probably the most powerful tool they know.  Some general observations. There are varying degrees of emerging algebraic thinking among the room. This was great for an exercise to get kids comfortable gently manipulating variables. I should also definitely go back one step and review why a negative times a negative is a positive in one of the upcoming weeks. At any rate, this lays the foundation to talk about divisibility rules going forward.

At some point soon I need to have the kids try a practice Math Olympiad. I also want to do a pattern finding notice and wonder exercise with a group discussion and I still have a short talk about multiplying negatives to fit in.

### Distributive Law Worksheet

#### The Distributive Law

Multiplication distributes over addition: a(b +c) = ab + ac. We sometimes talk about factoring out a number using the distributive property ie. 81 + 45 = 9 * (9+5)

#### Problems:

Compute $51 \cdot 9 + 51 \cdot 31$.

What is the value of $17 \cdot 13 + 13 \cdot 51 + 32 \cdot 13$?

What is $a(b + c + d +e)$?

what is -(6 + 8)?

What is $-1 \cdot (5 - 9)$ also written $-(5 - 9)$?

What is $-(x + 1)$?

What is $−1 \cdot(a - b + c - d)$?

Find numbers a, b, and c such that $a + (b \cdot c)$  is not equal $(a +b)\cdot(a+c)$ In other words show we can NOT go the other way and distribute addition over multiplication.

Take a look at the first perfect squares from $1^2$ to $10^2$. Do you see a pattern to what number is in the units place?

Can you use the use distributive law to show why $8^2$ and $2^2$ for example end in the same digit. Hint: 8 = 10 - 2.  Try the same technique with another pair of numbers.

We can also use the distributive law to show why a negative times a negative is a positive. Pick any two negative numbers x and  y then try using the distributive law to find the sum $x \cdot y + x \cdot -y$.

Factor out x from $6x + 9x$.

Factor out x from $6x + 9x^2$

What is $x\cdot(9 +x)$?

What is $(x+1)\cdot(x + 1)$? (Hint use the distributive property multiple times)

What is $(x - 1)(x - 1)$?

*What is $(x -1)(x + 1)$?

The above answer lets you do a neat trick. Can you multiple $19 \cdot 21$ in your head?  Using the above  formula we can think of it instead as $(20 - 1)\cdot(20 + 1)$. What does that equal?

Try doing some other examples with a partner without writing anything down.

What is $(a + b)(c +d)$?

#### Multiples/Divisibility

We say a number x is a multiple of another one n if we can find another number such that
$x = y \cdot n$. Example: 20 is a multiple of 5 since $20 = 5 \cdot 4$. Can you use the distributive law to prove that if you add two multiples of the same number n that the result is also a multiple of n.
Example: 20 is a multiple of 5, 35 is a multiple of 5.  20 + 35 = 55 is a multiple of 5. How about if a number is not a multiple? For example 11 is not a multiple of 5. Is there a pattern for what numbers you can add to 11 to get a multiple of 5?

## Wednesday, October 7, 2015

### 10/6 And we're off to the races

This afternoon was finally the first meeting of the math club for the year! Things started with a rude shock when I checked out my assigned room and found it had no desks or chairs. We'll see if I can get a better site in the future weeks but for now I'm going to ask the kids to bring a clip board or some other hard surface for writing on. On the bright side having everyone sprawled out on the floor worked better than I would have expected. It does make things informal in an interesting way.

My second surprise unfortunately was that an unexpected student that was not on my roster showed up that I had to refer to the after school program coordinators to sort out. The enrollment process was not under my control and this was one of the problems I was really worried about beforehand. This later lead to what my wife called " One of the saddest facebook posts ever."  Unfortunately I'm already at capacity and can't make last minute adjustments but it is really hard seeing someone write that you've shattered their child's dreams. I'll probably have some further conversations with the coordinator about the problems in the signup process.

I decided to start the meeting with a quick introduction from me with why I'm leading the class and what we are going to do over the quarter in some general terms. I then had all the kids introduce themselves and say why they were in the club. Almost everyone said essentially they liked math which is good to have said out loud sometimes by a large group of kids.

Next I had everyone get up and form a circle for a game of Buzz.  Rules   I picked the progressive version where you add rules. First we did fives, then added sevens and ended with prime numbers. This was a great success.  Some of the kids asked about doing an elimination round which I'll have to think about for future games but I do like keeping everyone involved.

From there we did the club charter discussion I decided to add on this year. I've put some of the outlines of the areas we covered here: https://drive.google.com/open?id=18NknKDOhmR5AX09RpLEFpli1IAL9F10W2UdVRT1i9VA   But essentially I'm going to try to come back to several key points particularly around what to do when we encounter hard problems. For next time, I'm going to stress collaborating with the others in the room.

This was put to quick test because I decided to start with the 2 problems I had used on my recruitment flier (link)  although I had some trepidation about whether they were too hard. As expected they did initially stump the kids but I was able to scaffold enough to move the room through at least 1 or both of them. The key was mostly to emphasize simplifying the problems and looking for patterns. I don't think anyone reached the 3rd problem which I added for my own peace of mind to make sure no one finished early.

Finally I gave them the problem of the month from MOEMS: http://www.moems.org/zinger.htm to do at home over the week and stressed I would be giving candy out once I had received enough finished problems. I'm really hoping this proves motivating.

Going forward I'm planning to send an email out about unfinished problems since that's a bad habit of mine. I'm going to have to think some more on the best policy to take. Perhaps I should have an answer site that everyone can look at. Or perhaps I'm overthinking the issue.

## Thursday, October 1, 2015

### Reversing digits

#### Moderator Note:  What's better than one semi math-crazy after school coach? How about two of them. My friend Dan has agreed to run our feeder Middle School's Math Club and after patiently listening to me talk about the blog for a year I've convinced him to contribute to it. So without further ado:

At the middle school, the Algebra 2 teacher gave a fun "Problem of the Month" to his students. My daughter and I had such a good time with it that I wanted to share it here.

Start with any 4 digit number where all 4 digits are different.
Sort the digits in descending order as your first number.
Sort the digits in ascending order as your second number.
Subtract the second number from the first number.
Repeat.

What happens?

For instance, consider the number: 1625
Sort the digits in descending order: 6521
Sort the digits in ascending order: 1256
Subtract: 6521-1256 = 5265
Repeat:
Sort the digits in descending order: 6552
Sort the digits in ascending order: 2556
Subtract: 6552-2556 = 3996
Repeat:
Subtract: 9963-3699 = 6264
Repeat:
Subtract: 6642-2466 = 4176
Repeat:
Subtract: 7641-1467 = 6174
Repeat:
Subtract: 7641-1467 = wait a second? That's the same as the last subtraction!

We started this problem just by picking some numbers like the one above and seeing what number we'd get. We quickly found that everything we tried converged to 6174. So we started trying to organize the pattern.

The first thing we noticed is that we made some mistakes in our subtraction as we were doing it without a calculator. I pointed out that problems that involve digits frequently also involve multiples of 9. And in fact our subtractions usually resulted in a multiple of 9. This is easy to check by adding the digits to see if they are a multiple of 9. It's not hard to prove that the result is always a multiple of 9. So any subtractions that weren't a multiple of 9 were wrong.

Then my daughter noticed a pattern. She tried 4321, 5432, 6543, 7654, 8765, 9876 as starting points. Sorting the digits and subtracting turns into 3087 for all of those numbers. Then she tried 5421, 6532, etc. Sorting the digits and subtracting turns into 4176 for all of those numbers.

This suggested we find some equivalence classes. Can we describe a set of numbers that will all get the same output? It was pretty easy to suggest that the sorted digits (a+x)(b+x)(c+x)(d+x) for a particular abcd but any x would form an equivalence class. When you subtract, the xs all cancel out and you have just abcd - dcba.

Once we had that equivalence class we could start to write out the possible values for abcd. We quickly saw a more general form of the class. Any pattern of sorted digits (a+x)(b+y)(c+y)(d+x) for a particular abcd but any x and y would form an equivalence class. When you subtract, both the xs and ys cancel out.

So that means we can figure out which equivalence class any number fits in pretty easily. Start with the original 1625. Sort the digits to 6521. Use the largest possible x = 1. Use the largest possible y = 2. That leaves us with 5300. 6521-1256 = 5300-0035. Each equivalence class will contain a number with two leading digits and two zeros.

Now we can make a table of all possible pairs of digits and figure out the next number from the operation for each pair. The table can easily include the cases where the digits aren't all different, although 1111, 2222, 3333, ..., 9999 become 0000 in one step.

From the table, you can also make a graph of how each equivalence class transforms to the next class. This graph can help you find the longest chains and how many steps any particular number will take.

Here's one of the longest possible chains:
4100 -> 4086 (8200) -> 8172 (7500) -> 7443 (4000) -> 3996 (6300) -> 6264 (4200) -> 4176 (6200) -> 6175
For each repeat, I listed out the actual result with its equivalence class in parentheses. There are several other classes with equally long chains.

If you look carefully, you can find some additional equivalence classes that narrow the problem down further.

I researched this problem and it is known as "Kaprekar's operation" after the mathematician who first published it. There's a nice write up with a bit more of the algebra at https://plus.maths.org/content/mysterious-number-6174. This is a good general math blog, as long as you don't mind your math with an S at the end!