Drinking party problem

To start with, I had no idea how to solve this problem. I asked Vincent a bunch of questions and still had really no idea of how to solve it. After talking to Ainsley, she gave me the hint of looking at it from a computer science context. From there, I was able to come up with an answer! Although, this question is really interesting because it seems to have many ways to solve it, which would be really interesting in a classroom context to develop logical thinking.

Once I started thinking of looking at the combination of rats that get sick to figure out which bottle is poisoned, it all made sense!

With 2 rats, you can check 4 bottles. If no rats get sick, it’s bottle 1, if rat A gets sick, it’s bottle 2, if rat B gets sick, it’s bottle 3 and if both get sick, it’s bottle 4.

Extending this to 3 rats, I found
         Rat A.    Rat B.      Rat C
1.       0.            0.             0
2.       1.            0.             0
3.       0.            1.             0
4.       0.             0.            1
5.       1.            1.             0
6.       1.            0.             1
7.       0.             1.            1
8.       1.             1.            1

These are all the combinations with 3 rats, therefore there are 8 bottles that we can check with 3 rats. It seems to be going up by powers of 2, since 2^3=8

2^10=1024, so we would be able to tell 1000 bottles by using the combination of rats that got sick to identify the poisoned bottle.

Unit plan with lesson plans

My unit plan is for Workplace Math 11 on probability and statistics. I did it in excel, so I'm not sure how well it transferred to google sheets. I can email the original if need be. The 3 lessons that are complete have been highlighted in yellow.
https://docs.google.com/spreadsheets/d/18we1w4WGhGPGMGs0cKE9dIlVYk49-06-qmtqTRRE8g4/edit?usp=sharing

Math textbooks response

As a former student, I don't know how much I agree with the article. In my experience, for the most part, reading a textbook was enough for me to be able to answer most questions correctly. When I was younger, this wasn't true and I needed more guidance, but from about 5th grade onward, I was able to start doing my math work as soon as I knew what section we were doing. However, as a teacher, I view things differently. I know that my experience was unusual. I got my passion for math from my teachers, but my understanding came easily. As a teacher, I think giving students real context for their lives and getting them to see themselves doing math is really important, which seems to be something the textbook examples lacked.

I think that the explanation of the concepts should come from the teacher. As the teacher, you should be aware of your students interests and to some extent their goals, so that you can bring their lives into math and get them to see a reason to do it. I like the idea of textbooks for work problems though. I think having some consistency with problems that students are doing throughout the province is a really good thing. And having a little review in the textbook of how to do the problems in case the students forget their notebooks at home would also be great. But I think it should be clear that it is review only and that the actual lesson should be taught by the teacher.

Scales problem

My first thought when I read this problem is that 1 should be one of the numbers. My second thought was that in order to get all the numbers, I would have to use the difference between 2 weights to get all the numbers between 1 and 40.

I started with
a+b+c+1=40      which means that
a+b+c=39

I also thought about the lower numbers. I realized that with 1, I would either need 2 or 3. I realized that 3 gave me more options, and so I figured that one of my numbers would have to be 3, which gave me
a+b+3+1=40     which means that
a+b=36

I also can say that:
a+b+3=39
a+b+3-1=38
a+b+1=37
3+1=4
3=3
3-1=2
1=1

From here, I thought about how to make 5. Since I could only use the weights once each, I would need a larger number and subtract 3 or 1 or both from it. My first thought was that since 1 and 3 are small, it would be better to use the largest number possible, which is 9. If one of the last 2 numbers is 9, the other is 27 from a+b=36. Now I have:
27+9+3+1=40
27+9+3=39
27+9+3-1=38
27+9+1=37
27+9=36
27+9-1=35
27+9+1-3=34
27+9-3=33
27+9-3-1=32
27+3+1=31
27+3=30
27+3-1=29
27+1=28
27=27
27-1=26
27+1-3=25
27-3=24
27-3-1=23
27+3+1-9=22
27+3-9=21
27+3-9-1=20
27-9+1=19
27-9=18
27-9-1=17
27-9-3+1=16
27-9-3=15
27-9-3-1=14
9+3+1=13
9+3=12
9+3-1=11
9+1=10
9=9
9-1=8
9-3+1=7
9-3=6
9-3-1=5
3+1=4
3=3
3-1=2
1=1

I'm sure that there are more ways to do this problem, although I'm not sure of what they are. Hopefully someone in class found a different solution! I think that I don't understand this problem deeply enough to come up with an extension for it. This problem was the most challenging to me of the ones we've done so far, it took me by far the most amount of time. I would have to spend more time understanding why powers of 3 work in this case to understand how to extend it.