Welcome, everybody. Good afternoon. Welcome to the Level 5, 6 webinar number 5. So sometimes we have students joining us for the first time. If you're here for the first time, welcome. We will be working some really cool problems today. The topic is making a table or making a list. So in all of our problems today, you'll probably see something like, how many ways can we do this? And the only way to count up how many ways there are to do something is to figure out how to do that, right? You'll see. I think it'll make sense as we go through the problems. For today's lesson, there is no handout. So you don't need to worry about any sort of handout, but you do definitely want to have some scratch paper where you can work through these problems. And we have polls for most of the problems today. OK, so welcome, Jacob. He's our teaching assistant. You can send messages to Jacob or to me. Anytime you need to during the class, we will ask you to put your answers in the chat or in those polls. Here's our warm-up problem. We add together two different numbers chosen from the numbers 1, 2, 3, 4, and 5. How many different sums can we get? Remember, small words or big specific words are very meaningful in Math Kangaroo. So if I add 5 and 1, I can get 6. If I add 4 and 2, I can get 6. But those are the same sum, not different sums. And I will go ahead and launch the poll right away because I think you should be able to read the problem in the top of that poll. And the correct answer is not there. That's strange because I just saved that. I'm going to end that. I'm going to have to check all my polls and make sure they got. Oh, you know why? Because that's not the warm-up. I'm going to go ahead and end that. Warm-up problem. There we go. All right, does anybody else want to put an answer into the poll? I'm going to end the poll. Most of about 70% of you have participated. So if you want to put that answer in right now, let's do it. OK. So most of you who answered the poll have said you think that there are seven different sums that we can get. That is the solution that I have as well. Can show you how I did it. I made a table. So I went ahead and 1, 2, 3, 4, 5 across the top. And 1, 2, 3, 4, 5 coming down. I have to do two different numbers. So I can't add 1 plus 1. But I can add 1 plus 2. And I get a sum of 3. 1 plus 3 is 4, 5, and 6. I don't want to repeat coming down. So I'm basically going to be able to cut this into a diagonal. I'm not going to do 2 plus 2 because those are not two different numbers. 2 plus 3 is 5, 6, and 7. 3 plus 3, don't want to do that. So 7 and 8. 4 plus 4 is 8. 4 plus 5 is 9. And 5 plus 5 is, again, not two different numbers. So the first problem here is I have to add two different numbers. I don't want to waste my time repeating combinations because 1 plus 2 is the same as 2 plus 1. And then I want different sums. So you'll see some of these sums repeat. So I only want to find the unique sums, which would be 3, 4, 5, 6, 7, 8, and 9. So that gives me seven sums. The other way to think about it is that I can get all of the sums from 2 plus 1, which is 3, to 5 plus 4, which is 9. And how many numbers is that inclusive? 3 to 9. That would be seven numbers if I include the endpoints. So that's another way to do it if you don't want to make a table. So you can think about what's the smallest possibility, the largest possibility. And I can get every number in between as well, especially because I'm adding ones. So there you go. You'll find on math kangaroo problems there's frequently more than one way to solve them. Sometimes I have time to present the options. Sometimes we just stick with the theme of the day, which today is making tables and lists. So we're going to be very careful in reading our problems to determine what is it asking. There it was different numbers for addition and different sums. Our plan today, our primary plan, will be in some sort of list or table. That's not the only way you could solve these problems. I'm going to be very careful when I make my lists and tables. And I'm going to check my answers to make sure they make sense. And that can be by trying another method or to see if there's another way to go back and see if I've made any mistakes. So hopefully you'll like today's lessons. We often make a list when we're doing things like casework. Casework is if I have a complicated problem, I might want to say, if I do a certain part, what are the endpoints that I can get there? And a very simple way to think about this is to go to the real world. If I put on a purple sweater today, what color pants would I wear with the purple sweater that won't clash? I might have a whole different set of pants I want to wear if I had put on a green sweater. So there's just a real world example of the types of things you might do in your own daily life where you're making a first decision. And that first decision then determines what the second choices will be. But so I don't get confused, I'll list them all out in these math problems so I don't lose track of how many options I had. So there's an example of casework. The next thing we have is making data tables. So sometimes we might have more than one characteristic. So a table will help us separate out different characteristics. So they can help us find patterns. They might help us, let's say we were separating things by the days of the week. That might be a reason to make a table. Let's say that there's a famous problem that you'll see, like you have so many animals, so many birds, and you have so many cows, and you have so many, and they have different numbers of legs. And so you might be adding up the different numbers of legs, and so a table can help you with something like that. So we'll keep in mind these two strategies as we go together. We're using them interchangeably where some student might make a list, another one of you might make a table. So don't be surprised if sometimes they're interchangeable. How many two-digit numbers are there which can be expressed using only different odd digits? In this case, 53 is a different digit, is a different number than 35. And when they say different, the 3 and the 5 are different from each other. So they would both qualify here. Does that help? So 53 and 35 both count because the digits in the number itself are different, right? So as you can see, I've been drawing. If you didn't know what to do, how to start this problem, maybe that's a hint. Looks like a lot of you know what to do. I have most of you have answered the poll, and the answers are pretty good. I'll end the poll here and share that with you. So most students answered the poll, and almost 70% of the students who answered the poll said 20. They even have a student who said, I clicked the wrong answer, I really meant 20. So you did really well with that one. Good job, everybody. Let's just make sure we can see what we needed to do here. So the numbers that would fit in this diagonal are going to have the same. So we need to be using different odd digits, but this 33 would be the same. So that wouldn't count. But every other one of these would count. So that gives us 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 options on this half of my chart, and also 10 options on this half of the chart. So just to give you a few examples, this is 51, 53, this would be 73. That's how I'm doing my chart, right? So we do have 10 on the top and 10 below the diagonal, so that is a total of 20. So sometimes we don't even have to make the list of all the possibilities if we know how many there will be, and it's just asking how many. For some of you, making the whole list of possibilities might be your favorite way to do it. So you might start listing 13, 15, 17, 19, 31. You might go ahead and make the list. Is that incorrect? No, that's perfectly a fine way to do it. If that's the way your brain wants to process through it, go for it. But just be organized so you do all the ones that start with a 1, all the ones that start with a 3, and you'll get the same answer at the end. Adam wants to read a 290-page book. He decides to read 4 pages each day except Sunday and to read 25 pages each Sunday. He started to read the book on a Sunday. How many days will it take him to read it? I do have a poll, I'll go ahead and launch that. Anybody else? Oh, some people say more time, sure. Okay. So, we have most of you answering the poll. We still have quite a few who probably didn't get through all the work, but that's okay. What I'll do is I'll share the results. So, 70% of you, and again not everyone answered the poll, but say 41 days. So, let's take a look at that. In order to speed up the explanation, I went ahead and started a table. So, I started this. So, we start to read on Sunday, and on Sundays we read 25 pages. And then on Monday through Friday, we read 4 pages every day. So, 4 times 5 days is another 20. I don't know why Saturday is separate. On Saturday, we read another 4 pages. So, the total pages in any one week on a Sunday through Saturday week is 49 pages. So, we can go through week by week by week, and we can add 49s. That is a great way to do it. So, 49 plus 49 is going to give us 98 for the second week, etc. So, that would be a great way to do it. And what we're going to find out is at the end of week 5, if we take 5 times 49, so 49 times 5, we have 245 pages read. That's after 5 weeks. Now, we need a 290 page book. So, in week 6, on Sunday, we would read 25 pages. That means we have only 20 pages to go in our book. And we can read the 20 pages in the Monday through Friday. So, we're going to read those 20 pages here. This takes us to Friday. Friday is the day we get to finish our book. So, let's see how many days that will be. We have 5 times 7 for the 5 weeks we read entirely from Sunday through Saturday. Then, we have another Sunday and Monday through Friday. The only day we did not read in that 6th week was Saturday. So, we're going to have to add 6 more days. And that gives us 41 days to finish the book. So, hopefully you did some variation of that using those types of calculations to solve this problem. And we can move on to our next one. The leaders of the math camp in Zakopane decided to divide 96 participants into equal groups where the number of participants in each group is between 5 and 20. In how many ways can this be done? So, 96 participants. You're putting them in groups that have between 5 and 20 people. Everybody still working? I'm not seeing very many answers in the chat. About half of you have answered the poll. We do have quite a few questions today, so we'll have to pick up the pace a little bit. Remember, if we're going too fast and you want more time for any problems, all you have to do is come back to the recordings of these, which will be available later today or tomorrow. And you can look at the problems all over again and take as much time as you need to just by pausing the video. So, I'll end the poll. So far, I realize not everyone has answered, but you're doing really well. The answer is four. Let's take a look at how we did this question, what the question kind of means. So, if you're dividing 96 into groups, you're basically looking for the factors of 96, right? So, obviously, 1 and 96 are factors of 96. We can divide 96 by 2, but that is going to be less than 5, and the larger factor is also going to be more than 20. So, it's not going to be very helpful. This is divisible by 3 as well, but that gives us a 32. And again, that's not in between the targets. We can divide it by 4 as well, but that gives us a 24. Again, the 4 is too small. The 24 is too large. So, all of these are factors, but they're not group sizes that we want. So, great that we did it, but they're not part of our answer. Let's look at what comes after that. We can divide it. It's an even number divisible by 3, so we can make a group of 6. We can make groups of 6. If we do that, 6 goes into 9 one time, 36, so we get 16. Okay, well, 6 and 16 are both factors between 5 and 20, so that's great. We can also divide it by 8. 8 times 12 is 96, and that is the factor pairs that we can get out of 96. So, 4 of these would be size groups between 5 and 20. So, the correct answer here is 4. So, a list. Factors are lists, right? So, we used our strategy, and we solved the problem. We have 6 segments with lengths 1, 2, 3, 2001, 2002, and 2003. In how many ways can we select 3 of these segments to build a triangle? Hint, there is a rule about building triangles and about the segment lengths in triangles. So, there's your hint for that. Can I build a triangle that's a 1, 2, 3? So there's the question is can you repeat some of these segments and the answer is no. You have six pieces, six segments, and you have to choose three of those six. So you can't use two of the twos or anything like that. Okay, we're gonna have to end the poll. I know not all of you have gotten through this. Remember, if you are having some questions, you can try me. If I'm too busy to answer, also try Jacob. That's why Jacob is here. He's here to help you out. He's been doing Math Kangaroo and other math contests for a long time, and he's got a lot of good, helpful suggestions for you. So this is an interesting one. We have every single answer choice was selected this time. A slight lead for six possibles, but it's not too convincing, right? It's not anything like over 50%. It's just one third of you saying that. So let's take a look at the problem in a little more detail. I actually have a second slide where I've already started the table, so that might make it a little easier. Okay, so in a triangle, there is a rule about triangles that the sum of two sides has to be greater than the third side. I can draw a little picture here to demonstrate that. Where's my little cursor? If I have a length of three, and I have a length of one, and then I have a length of two, I'm gonna label this in just a second. There we go. So if this one is one, if this one is two, and this one is three, you'll see when I try to make a triangle, the top of my triangle collapses down, and I basically just get a double segment. There's no way to form the triangle because one plus two only equals three. It's not greater. So the shorter sides have to be greater than the longest side. So one, two, three does not work. So can I do any of them with a side segment length of one? 2,001 plus one equals 2,002. I have exactly the same problem. So one is not going to occur in any of my triangles. I can use the two, however, right? We can't use example two, three, and 2,001 because I would have a big long line and then tiny little ends that would never touch each other, right? So we can't do that. But we could, for example, use the 2,001 plus two, because that's 2,003. I couldn't use a 2,003, but I could use the 2,002. I can use the 2,002 plus the two, and I can use the 2,003 because 2,002 plus two is a four. I can use the 2,001, and I can add three, and then I can use the 2,002 or the 2,003, right? Because 2,001 plus three is 2,004. So I can still get a triangle even if I use the 2,003 side. I can get a triangle if I use 2,002 plus three. That's 2,005. So I can use the 2,003 here. I can also use the 2,001 plus the 2,002 and make the long side 2,003. So those are the six options that are possible for making an intact triangle with these segments. I hope that makes sense, but just remember the two shorter sides have to be at least as long when added together as the long side. Otherwise, they'll either never meet, like this one never met, or they can make just a flat segment if it's exactly equal. All right. Here's another good problem. A two-by-two-by-two cube is to be constructed using four white and four black unit cubes. How many different cubes can be constructed in this way? Two cubes, meaning the two-by-two-by-two cubes, are not different if one can be obtained by rotating the other. So this is gonna take some three-dimensional spatial visualization on your parts. Let me try to help a little bit. I'll do my best. I'm not bad at drawing cubes. I can't draw many fancy pictures, but cubes I've practiced. The idea here is that this whole cube would be black. Maybe this whole cube is black. So all the faces on it are black. We can do the same with this one. So if I rotate this, I have half black and half white, no matter which way I rotate it, that's always one two-by-two cube, two-by-two-by-two cube. I hope that helps. I have more precisely drawn cubes on the next slide, but you have to start somewhere, right? I have some students already asking for the poll, and I do have one here. Let's see if it works. So, to give students a place to start with this, you might try casework. So, you might think about the cube as being constructed with a top layer and a bottom layer. If I have two black cubes in the top layer and two black cubes in the bottom layer, what are the options? If I have three black cubes in the top layer, three black cubes in the bottom layer, what are the options? That's an example of casework, how you might divide this up and make sure you're not missing any possibilities. Okay, I know some of you are going to be furious with me because I'm going to stop you now. You can keep working and you can ignore me. That's always an option, but I'm going to end the poll and I'm going to go over the solutions so that we can get to some more of these problems today. So I'm ending it here and we have every answer choice. The most popular one is eight. Let's take a look at this and let's use the casework that I was talking about, okay? So I'm going to clear this because I have a slide that shows all the drawings and we'll go over and explain it in terms of casework. So if you look at this first one, that's basically the one that I drew where you have all four black blocks on one side and all four white blocks on the other side. Where I had it with them in the front, the black in the front and the white in the back, but that is just this block turned on its edge. And they said, don't count it if all you've done was turn it on its edge. So this is a half and half cube of half top and half bottom. Now the other way to think about it. So this is this cube where this is the top layer and this is the bottom layer. So if you make three, that's a B. If you make three blacks on the bottom, that gives you one white on the bottom. So here's your one white in each of these cases. So that means like back there in the back where you can't see it, there is a black. And then you would have one more black cube that you would put in the top layer. You could put it over the white. You could put it over one of the blacks on the wing here. You could put it over the black that's in the middle of the L shape. You can put it over the black that's on this side. So that basically kind of gives you a right-handed version and a left-handed version. The difference between these two versions is where the top one is. So basically we have taken this and rotated it 90 degrees to four positions. So these are 90 degree of just the top. The bottom did not move. So we're not rotating the whole cube. You can almost think of a Rubik's cube if you rotated one layer of a Rubik's cube. That's what you've done here. So you've changed the position of the top black with respect to the bottom white. The other way to do this is to do two and two. So you can have two blacks in the front and two blacks in the back, which corresponds to this one. The difference between this one and the first one, even though there's two and two, is that we have whites over blacks and blacks over whites. Then the only other cube that you can construct is if you do opposite corners. Opposite corners here. So if you count up these possible cubes, we only have one, two, three, four, five, six, seven possible cubes. Now remember I said casework, list? So the options were on the top I have no blacks. On the top I have one black, but I can move it around 90 degrees into four positions. On the top I have two blacks. And there's two different ways to do that so that it's still unique from a rotation of the first one. I know it's pretty tricky. This was a really tough problem. Again, this is a five point problem toward the end of a math kangaroo contest. So this is why it's really important that you start getting those visualizations. Make sure you're practicing. Number six. Using only one color, in how many ways can you color three different cells in the strip shown below so that no two neighboring cells are colored? I do not have a poll for this one because, you know, the answers require kind of looking at this. So let's think about it. I can color the first one, the third one, and the fifth one. Remember I'm coloring three, so don't color the seventh, right? What other options are there? How many options are there? And I think one way to do this is to just go ahead and go through and doing it together. So on my next slide, I think we can go ahead and work through it together and come up with the answer all together. And then hopefully we'll be able to move on to another problem. Okay, so I think I can just use a nice fat marker so I can color the first, the third and the fifth. If I color another first, I can do the third and the sixth. I'm trying to use a pattern. I'm gonna keep the beginning the same, but there are three options there because I can do the fifth, the sixth or the seventh. Now I can also color the first and the fourth. That means I can not color the fifth, but I can color the sixth or I can color the seventh. Is there any other options? I could color the first and the fifth and the seventh. I think it covers all of the options where I have colored the first square. So you see the pattern, the case work that I've done, the first and the third, the first and the fourth, the first and the fifth. So what happens if I color the second square then I can color all the evens, right? The second, the fourth and the sixth. I can do the second, the fourth and the seventh. I can do the second, the fifth and the seventh. And that takes me all of the options where I colored the second one and skipped the first. I think there's one more option where I can do this, the third, the fifth and the seventh. So how many options was that? That's one, two, three, four, five, six, seven, eight, nine, 10. So the correct option would be E. You can do this with coloring or you can make lists, right? You can do one, one, five, seven, one, five, you know, one, three, five, one, three, six. You can make a list with the numbers as well. We want Jacob to have a chance to lead a problem today, so we're going to skip ahead a little bit because I know Jacob prepared this bonus question. Was that one of them, Jacob? Yeah, so the bonus question is how many five-digit numbers, which uses digits 1, 2, 3, 4, and 5 with all different digits, have the following properties when looking at each number from left to right? So the first two digits form a number divisible by 2. The first three digits form a number divisible by 3. The first four form a number divisible by 4. And the five-digit number is divisible by 5. So I'll give you guys some time to think about this, and then we'll go over it. Hey, Jacob, I think we better explain it because some students will need to leave. We're about out of time. All right. Could I get permission to? Yes, yes. All right, so. So we know. So I'm saying the window, the five digit number with five different place values. Right. And so then. So since five digit numbers of five. And that means that this number has to be five. Right. Because, well, you can't have the numbers of five and the last digit has to be either zero or five. But we only have five as an option. So the last digit has to be five. Now, when you look at the two properties, say that the first two digits have to be divide two and the first four digit digits have to be divisible by four. OK, so that means that this digit and this digit has to be even. So this is either two, four, and this is two, four. And they can't be the same since the digits are all different. So now, well, since those two values occupy two and four, then that means the other two values have to occupy one and three. OK, so now we can try to figure out all possible five digit numbers, and luckily there's not too many. We can only have one, two, three, four, five. One. Sorry. Let's say three, two, one, four, five. One, four, three, two, five and one. Sorry, three. Four, one, two, five. So now what we want to do is we want to check the conditions for divisibility by three and also divisible by four because we only said that the first four digits is divisible by two yet. That's the only thing that we guarantee so far, but we haven't yet guaranteed that the first four digits form a number divisible by four. OK, so let's see if the first three digits is divisible by three. And remember that in order for a number to be divisible by three, then the sum of the digits has to be divisible by three. So in one, two, three, four, five, the first three digits sum to six because they're one, two, and three, which is divisible by three. So this one satisfies the divisibility by three rule. Now the next number, three, two, one, four, five, is also satisfying the divisibility by three rule since the first three digits are three, two, one. So that also works. But for the next two numbers, their first three digits have some permutation of one, three, and four. But one, three, and four sums to eight, but eight is not divisible by three, so these two numbers do not work. So now let's check the divisibility by four, and notice we don't have to check these two numbers since they both don't work. But remember what the divisibility by four rule is, and it's that the last two numbers have to be divisible by four. But 34 is not divisible by four, it's just not divisible by four because 34 is two more than eight times four, so 34 is two more than a multiple of four, so this does not work. And same thing for 14, 14 is also not divisible by four, so neither of these numbers work. So the answer should be A, there's no such numbers. Thank you, Jacob, that was explained very nicely. So today's lesson was some of these problems where you actually have to make lists and tables can take you a pretty good amount of time. So be very careful, that is the one drawback to these types of questions. But having those tables or those lists is going to help you stay very organized and not miss any of the possibilities. So I think the biggest challenges is the amount of time that these problems can take. So try to find any sorts of shortcuts, use this casework. Sometimes you notice on my first problem I said, oh, the top is symmetric to the bottom, so I only needed to count up one. So hopefully you have found these challenging but helpful, and you've been able to learn how to do some of these problems. Don't forget you should be practicing some past Math2Kangaroo contests to get used to solving all of the different types of problems for our contest coming up in March. And can you reply to me in the chat, there was a student who was having trouble finding, yeah, so you're still here. Okay, so we had a student asking how they can find the recordings. So I'm going to go over that. If you already know how to find the recordings of past webinars, feel free to take off and come back next week. I'm glad I'm seeing so many students week after week. Jacob and I enjoy being able to explain these challenging and fun problems to you. So if you want to stay for how to access it, I'm going to go ahead and do that now. Okay, so the how to access it is you're going to have to go into your Math2Kangaroo account. So mine might look slightly different to yours because I obviously have a teacher's account. But if you go into the courses, you should find your registration for this webinar. So right here, you can see this is webinar set B. That's the webinar that you are enrolled in right now. If you go to the contents tab here, you will see that it has the recordings that are already prepared and ready for you to view. Obviously, today's will take a little bit of time to have the beginning and the ending edited if it needs to be, but you'll find it here. Okay, the other thing you might find is the handouts for the classes are available here where it says event info. Let me show you one more time. The webinar registration, event info, and here you can see the handouts. All right, so you'd be able to download any handouts with figures if you haven't done that already. One other really neat thing that you might want, this is, for example, I'm leading a session. When you are enrolled in a Math2Kangaroo contest, this is a contest at a school that I supervise, but if you look at the event info, you can find some discount codes to be able to practice some past contests and to look at some video solutions. That's a really great way for you to do some solutions.

mathematical problem-solving

combinatorics

number theory

systematic approaches

lists and tables

casework problems

constraints

permutations

divisibility rules

2x2x2 cubes