Hey, good afternoon. I imagine we're going to have some more students joining us, but we'll start right on time with the students we have. Thank you for joining on time. This is webinar number 10 in the set A series for level 5, 6. So this is the last of our webinars for this session. I will come back to this at the end, but there is another session of webinars that will begin in January. So if you enjoy this experience, you can come back and do it with completely different problems, not repeating problems, again in January. OK. So today, we've gone through, in our past nine lessons, we've come up with a lot of tools you can use to solve math kangaroo problems. All of today's problems will either ask you to use more than one of those in order to get to the answer, or you might have options that you could use more than one type of tool to get to the answer. So what one of you might use to solve this problem might be different than what another students will use. Sometimes, I will explain it with more than one method as well so that you can see the options. I do have quite a few polls today, and let's get started. Jacob, let me know, our TA Jacob, let me know that he is not feeling well and he is at the doctor. So I don't know if he will be joining us. If he does, that would be great. If not, we will still have a wonderful lesson, and I'm sure he would be kind of sad to miss the last one because he's been such a good help, such a good TA for this. OK. Here's our warm-up problem. There are stools and chairs in a room. Each stool has three legs, and each chair has four legs. Altogether, there are 17 legs. How many chairs are there in the room? Okay, I will launch the poll. I was waiting a couple of seconds because we still have some students joining and I didn't want them to miss the problem in the poll. All right, so here's the poll for our warm up. Remember the question is asking how many chairs are there in the room. Answer for the number of chairs. All right. Anybody else want to add their answer to the poll? Not everyone has answered. Okay, that's most of you. I know some of you have just joined in the last few seconds, so here we go. We have a few people who have said five chairs, and most have said two chairs. So let's see which of those is correct. It says there are stools and chairs. The stools have three legs, and the chairs have four legs. This is s-t-o-o-l. Okay, stools three, chairs four. And we know that all together there are 17. So we could look at the multiples of three and four. So we did multiples and divisibility, so we could look at 3, 6, 9, 12, 15, 18 is too many. And then we can look at multiples of 4, 8, 12, 16. Now we need to find a sum of some number of stools and some number of chairs that is going to give us a total of 17. So I don't know, let's see, 6 plus 12 is too many, 8 plus 6 is 14, 8 plus 9, that's going to give us the total of 17. So this is three stools and two chairs, d. So multiples is a good way to do this. You could have done a guess and check. You could have made a table with different numbers of stools and chairs, could have organized your guesses and your data in a table. So just a few methods for solving this problem. All right, so just because I want to review all the different types of problems and methods that we've used, because today remember we will be combining them. So look for patterns, use your algebraic thinking, ratios and proportions. We just did one with multiples and factors that could be helpful. We're going to definitely see some times or calendars today and a little bit of geometry, 3D or 2D. 2D also included the spatial reasoning and how things turn and fold. Hands-on, that was last week's fun lesson where we were cutting out papers and making shapes and different things. So also there were some nets that we could fold up in different ways. Keep those techniques in mind. You can try those on the problems we solve today. All right, so sometimes you need more than one tool to finish a job or there could be more than one tool that will serve your needs. Okay, so don't forget we could combine them or we could apply multiple techniques. And one of the nice things about being able to apply multiple techniques is you can use the second technique to check your answer that you got using the first techniques. You could have multi-step problems where you apply a different technique to each portion of the problem, right? So an example here is if you are fixing your faucet, you might need a wrench to hold a part and then a screwdriver to tighten a fitting. Or if you can't find a wrench, maybe you could use a clamp or a plier. So be creative with your tools. There's not just one size fits all for these. So keep in mind how you do it and how another student might do it could differ, but that doesn't mean that either of you is incorrect. Just means you're using your own methods. How long will it take to print one million forms if it takes one minute to print 100 of these forms? And we'll launch a poll after a few minutes to allow you to calculate. Here is your poll. Again, the whole problem appears at the top of the poll, so I think this one is one we can launch right away. I still have some people who haven't put an answer in the poll. If you're still working, that's fine. If you do have an answer and you can put that in the poll, that would be great. All right, I'll end the poll and share the results. This matches what I'm seeing in the chat, so it at least gives us a good idea. So I see this in the chat as well. I see some students selecting A and some students selecting B. It seems like B is the more popular answer. 2 3rds of you have said B. So let's take a look. What techniques is this one going to use? Well, the first part is we know that it takes one minute to print 100 forms, and we want to print 1 million forms. So we want to know, so this is one minute, and this is forms. So if we finish this proportion, this will tell us how many minutes it takes. You notice our answers have hours in them, so we're going to have to use a second technique, which is our time conversions, knowing that there are 60 minutes in an hour. So first, let's see. How do I get from 1,000 to 1 million? I like to take shortcuts, so I might say, OK, well, the 100 is going to knock off two zeros from here, so that would give me 10,000. So I'm going to take 10,000 minutes. Then I just have some long division. 10,000 divided by 60, and I can cheat again. I can get rid of one of those zeros, and then I can put my decimal point right here. So 6 goes into 10 one time, gives me a 4. 6 goes into 40 six times, gives me 36, gives me again a remainder of 4, goes in again six times, 36, gives me a remainder of 4. And you can see that I could continue to carry this out, adding in as many zeros as I want, or I could take my remainder at this point would be 4 sixths, and 4 sixths is 2 thirds. So I have 166 hours and 2 thirds of an hour. So 2 thirds of 60 minutes is 40 minutes, because if I take the 6, that would be 2 times 20. So the 166 hours and 40 minutes B is the correct answer. So yeah, you can do long division. You can find shortcuts, make it a little bit easier for yourselves. OK, here's a little bit of geometry. Mark has sticks with the lengths of 1 decimeter, 2 decimeter, 3 decimeters, 4, 5, 6, 7, 8, and 9 decimeters. If you don't remember what a decimeter is, a decimeter is 1 tenth of a meter, or 10 centimeters. Using these sticks, he builds triangles with each side made of one stick. How many triangles with a side of 1 decimeter can be built using these sticks? I do have a poll that I'll launch in just a moment. And the notes that I put there are just some background information. You don't really need it for this problem. Everything is in the same units. You don't have to convert anything. But it is good to remember what a decimeter is. And here's the poll for this problem. Don't be shy about answering in the polls. They're anonymous, so we can't see who's answered which answer. There is a good question being asked, which is, can we repeat sticks? You could from one triangle to a next repeat sticks, but you can only use those sticks. So I can't use like two eight decimeter long sticks in the same triangle, because I only have one eight decimeter long stick. Does that make sense? I guess you would have to repeat sticks because you have to use the one decimeter stick as many times as you're making the triangles. OK, so here is the end of the poll. I'll share the results. OK, we have 45% of you saying sticks. We have some saying 3, 1, and 0 comes in at 27%. OK, so a mix of answers. No one said 2, which is interesting. I wonder why 2 is lonely. All right, so I'm going to clear off what I put here about the units. That was just to give a little bit of education on decimeters. So I'm going to draw, here's a 1 unit long stick. OK, I am also going to draw, I will do a 3 unit long stick. That's too long for 3, isn't it? I don't know if I make this one a little bit longer. 1 and a 3, and then here is a 2. So you can see, if I tried to make a triangle, if I connected these sticks at the ends, I would end up with just a segment. I'd end up with a 3, and then this would be the 1 stick, and this would be the 2 stick. And it would just lie down right on top. I wouldn't be able to get any height on my triangle, would I? You can try this with any of them. You're going to end up with the same problem. There is a triangle property. I'm going to type it, because I think that will be easier to read. The sum of any two sides of the triangle must be greater than the third side length. Why is that? Because if I, for example, take a side of 3, and I take a side of 2, and I take another side of 2, I'm going to be able to get some height. If I lie these down on top of each other, you'll see that the 2's are going to overlap. This overlap allows me to get the height. So the answer here is that the sum of any two sides So the answer here is 0. If I use any of the ones with a 1, I could try to use 1, 8, and 9. But 1 plus 8 exactly equals 9. It's not greater than the third side length. So in this case, it won't work. So you can see if I use 3, 2, and 2, 3 plus 2 is 5. So 3 plus 2 equals 5. That's greater than 2. Or I can do 2 plus 2 equals 4, which is greater than 3. So the blue triangle works, but nothing that I've done in the green will work. So the correct answer here is 0. And it's based on this important properties of triangles. You can experiment with that yourself. OK. I have to click there. OK. For snails, fin, pin, rin, and tin are moving on identical rectangular tiles. The shape and length of each snail's path is shown below. How many decimeters has snail tin gone? So you can see we want to know how far has snail tin gone. That's the question here. And they've given us quite a bit of information to use. There is no poll because I didn't want to remove this nice big version of the figure. So you can send your answers to me in the chat. Okay, having a good discussion with some of you over the chat. I do appreciate when I see you working really hard. I'm going to start explaining it. If you're still working you could tune me out I guess. So what we see is if we look at the first snail, Finn, we see that Finn has gone one, two, three, four, five diagonals. And the five diagonals have a total of 25. So each diagonal is five. Right? Let's take a look at what Finn has done. Finn also has one, two, three, four, five diagonals. So there's five diagonals. But then there's one, two, three, four verticals. And then we see that this is a total of 37. We can figure out how long the verticals are by subtracting the five diagonals, which was 25. Right? So four lines, vertical lines, must be 37 minus 25, which equals 12. So then one vertical line, we'll do it like this, one vertical line must equal three. Okay? So, so far we have these tiles. I'm going to draw part of the tile. So far we have a tile. We have diagonals on the tiles. We know that the diagonal is five and we know that the vertical is three. Now, it's a rectangle with a right angle. So some of you might recognize right off the bat that this is a three, four, five right triangle. Or you can use right triangle. Or you can use your Pythagorean theorem where five squared equals equals three squared plus four squared. You can double check that. The other option is to look here at what Rin did. Remember I said there could be more than one method for solving these. Right? So we don't even need to use Pythagorean theorem if we look at what Rin does. Rin has one, two, three, four, five, six verticals, six up and downs, plus one, two, three, four, five horizontals side to side. And we know that that, I'm going to come over here, that equals 38. So we know that the verticals are three. So we have six times three plus five of the horizontals equals 38. So we can say that five horizontals is going to be 38 minus 18 is 20. So then one horizontal equals four, which we already knew from looking at this Pythagorean theorem. But we could find it even if you didn't recognize the right triangle. So again, two methods. So now if we look at what Tin has done, we can see that Tin has a diagonal which is five, a vertical which is three, the horizontal four, another vertical three, another five, three, five, three, four. And if you add all of that up together, so five, three, four, three, five, three, five, yeah, five, three, four. If you add all of that up together, you are going to get 35. So you can see a little bit of combinations of methods, which is really fun. I hope you like this problem. I liked it. That's why I selected it for the lesson. Okay. A wooden cube with the, sorry, little mouth mistake there. A wooden cube with the length of its sides equal to three feet was painted with 0.2525 hundredths gallon of paint. The cube was then cut up into unit cubes with a side length of one foot. How much paint is needed to paint the unpainted sides of the unit cubes? I do have a poll for this one. I'm going to start doing some drawing as you're working so that we can make it out of here on time today. But remember, we're looking for how much paint is needed for the unpainted sides. So like basically how much more paint you don't have to repaint any sides that have already been painted when the cube was the larger size. Hey, almost everyone answered the poll, so thank you for that. If you're just guessing, that's okay. Let's learn how to solve it together. If you knew how to do it, that's wonderful. So here we go. We have about half of you have said half a gallon, 0.5, and there's some other answers here as well. So let's take a look. So basically, this problem is asking, we have painted the entire external surface, the surface area of the large 3x3x3 cube. So let's figure out how many sides we've painted in terms of faces of small cubes. So you can see that we have painted 9 here, and 9 here, and 9 here, right? So we've basically painted 9 faces times 6, because this whole cube would have 6. So if we were looking at the small cube, we have already painted, I can't multiply today, we have already painted 54 faces are painted. If we're talking about 1x1x1, right? Cubes. Okay, how many of those small 1x1 faces are there when we take the entire thing apart? Well, we have 3x3x3 cubes, right? That equals 27 little cubes when we take it all apart. And each of those 27 cubes has 6 faces. So that's a 42, 12. See if I can get all my numbers right. 162 faces in total. But we've already painted 54. These are like pre-painted, right? So if we do a little subtraction, 12 minus 4 is 8. We have 108 need to paint. Okay, so if we painted 54 faces with 0.25 gallons of paint, how much would we need for 108? Well, this is times 2, right? So we need to do times 2. So we need 0.5 gallons to finish up the painting. So what methods did we use? We used some three-dimensional geometry. We used cutting it apart. We used quite a bit of rational calculations here. We used a proportion. There's quite a few things we used, right? So that's what I mean when I say a combination of methods. All right, the numbers, oh, this gives us some logical thinking. I like this one. The numbers 1 to 8 are written like this. The numbers 1 to 8 are written on eight cards, a different number on each card. So there are eight cards and we've numbered them 1 through 8. The cards are divided into groups A and B into such a way that the sum on the cards in group A is equal to the sum of the numbers on the cards in group B. If we know that group A contains only three cards, then we know for sure that A, exactly three cards in group B have odd numbers on them. B, four of the cards in group B have even numbers on them. C, the card with 1 is not in group B. D, the card with 2 is in group B. Or E, the card with 5 is in group B. Only one of those, A through E, is a true statement. Or an always true statement because it says we know for sure. I'll give you a few minutes. There is a poll that I'll let you see the problem big before the polls are a little bit smaller print. I'm trying to launch the poll. Okay, we're in the poll. I know not everyone's answered, but we need to move along a little bit. So, actually more than half of you do have the correct answer. The correct answer is the card with two is in group B. But since there are some other answers, let's take a look. So, you can see what I've done. The hint was, what is the sum of the numbers one through eight? Well, the sum is 36. You can add them sequentially, or you can do the little shortcut, which is you add the first and the last, which is nine, and there would be pairs, four pairs of those, so it's 36. And then we know that half of them, half of that value is in group A and half is in group B. So, here's group A and here's group B. We know that A contains only three cards. So, let's see, we could do eight and seven is 15. We need three more, so eight, seven, and three could be in group A. And that would mean that one, two, four, five, and six are in group B. Okay, what if we do eight and six and four? Then we would have one, two, three, five, and seven in group B. Okay, what if we try to do it without the eight? Because we could try eight and five. Eight, five, we're not going to be able to get there with eight and five, right? So, eight plus five is 13. We would need another five, so that's not going to work, right? So, seven, six, and five will also work and then we get one, two, three, four, and eight in group B. I believe that's the only combination of three cards that can give us a sum of 18, right? You can try to work out any others, but I think you're not going to be large enough or you're going to have to repeat a card and we only have one of each card. So, now what you'll notice is the card with number two is always in group B, right? The card with five can be, if we look at E, here we have a five, so that fails E. The card with one is not in group B, but the card with one is always in group B. The four of the cards in group B have even numbers on them and you can see that that's not always true. Okay, and then same with A. There is one other thing to think of. Remember with our logical thinking, our reasoning cards, we can sometimes look at the answers. What if we do not put number two in group A, right? So, if this, if this number two is, if D is false, if we put two is not in group B, then two is in group A, right? For 18, we need to have a 16. So, two plus 16 would give us our sum of 18 and there's no way with the cards we have to make a 16. We would have two eights and we don't have two of them. So, therefore, the two has to be in group B. So, there was an alternate way of solving this. You can either make the groups that are possible or you can start to analyze if any of these won't work. All right. Eve wants to make a square using only pieces like the one in the picture. What is the smallest number of pieces she needs to make a square? Remember a square. She has to start with this and the square should not have any gaps in it. It should be completely covered. I'm already seeing guesses. I do have a poll and the picture is pretty simple, so I think you can probably use it through the poll. You may not flip this piece in a mirror image. You can rotate it and turn it around. Remember we kind of had that with our hands-on lesson where we couldn't flip things over, only rotate them. Remember, leave no gaps. And you can't separate those five little squares into other pieces. It has to stay looking like that backwards L shape, or whatever you would call it. Okay, I'm gonna show you two ways to solve this puzzle. Okay, let's take a look at the poll results. So we have that almost half of you think that the answer is 12 pieces. The next most common answer I guess is 20 pieces. Let's take a look. We'll try two methods. One is we can see that there are five squares here, right? I think there's no one who could argue that there are five small squares. So if we have to use units of five squares and we wanna make a bigger square, then we could try five times five, which is 25. Or we can go up a multiple of five, 10 times 10, which would be 100 squares. But because we have to be in multiples of five, we really don't have any choice other than that. So if we take that L shape, one, two, three, four, five, and we try to come up with a way to put a fifth square over here. I'll change the color actually. We try to come up with a way to stick a fifth square over here. We can do that if we put another piece on top like this, one, two, three, four, five, that would work. But this is five along here, can we stack it up five high? And the answer is we cannot stack that up five high. We can only stack it multiples of two, right? Cause this is times two. So we are not able to do it with the five pieces. So let's see if we can do this so that we have a 10 by 10. We can go ahead and we can do this same block of 10 squares over here, right? Make it the exact same way with two pieces, right? So this is two pieces and this is two pieces. So this gives us a total of 10 going across the bottom and we could stack that up. We would need, this is two pieces. So we would need two times five in order to get the 10 by 10 square. So let's see how many of these five pieces we have. Well, we have 100 small squares. And if we divide that by five, that equals 20 of those puzzle pieces. Or you could count them up that this is, there's four in this layer, this is four pieces. And we need to have four more layers of that, right? So it would be four times five. So the correct answer is E, 20. And majority of you missed that problem. So this is a good one to have had some practice so that the next time you see something like that, you'll be able to build it correctly. Here's some more, another geometry problem for you. The perimeter of the rectangle ABCD is 30 centimeters. So that's to these points here, ABCD. See those points? Three other rectangles are placed so that their centers, that's important part, are at the points A, B, and D. The sum of their perimeters is 20 centimeters. So the sum of what? The sum of the three other rectangles. What is the total length of the thick black line? Again, this is a five point, well, this is a four point question at this level. Oh, I'm seeing answers in the chat, and we don't wanna leave you without an answer for this. So let's take a look. This is kind of an interesting problem. You can see that what we've done when we overlap the rectangles is we lose what's in blue there, right? That gets lost. What is that? Well, we know that A is the center of this small rectangle. So we can extrapolate, somebody asked me if we can use the push-out method, and you kind of can. So the part on the large rectangle that's missing is the same as what we would draw to finish rectangle A, right? Because this is the same as this, and this is the same as this. Those are, means they're equivalent length, those little dashed lines, right? And because we cut out 1 4th of that rectangle, we've lost 1 4th of the perimeter. So we lose 1 4th of the perimeter of A, B, and C. So if we know that A plus B plus C equals 20, that's not C, it's D, pardon me. But we've lost 1 4th, then we're gonna actually have 15 because 1 4th of 20 is a five, okay? So now we know that, let's see what bright color I can use. We know that the orange, the orange part is 15. And we know that the part that is bright green that's missing, the green missing equals five. So now if we take this information together, we can see that we had the 30 minus five plus 15. 30 is the original perimeter minus the five we lost in green. So this is original, the original big rectangle. And then this is the three pieces of rectangle. When we add that together, 30 minus five is 25, 25 plus 15 is 40 centimeters. Okay, little interesting problem there, isn't it? I hope you like it. All right, that is all we are going to have time for today. So we are going to wrap up. This is the last of our webinar series. So I hope you've enjoyed the Level 5-6 webinar. Remember our four-step problem solving, we've gone over that before. Remember that you have the ability, you have the resources and some codes to practice. So use video solutions, use some practice contests. I do also want to let you know that we will be continuing. We don't have, we don't need to worry about summer course, but we do have courses in January. At the same time, we will have webinars for Set B, which is all different sample problems. And I will still be the one instructing those webinars at the same time, starting after the New Year's break, we will start in January. So if you'd like to continue to do these webinar classes, I'd love to see you again, ask your parents to sign you up for the next one. Okay, I'm sorry that Jacob wasn't able to make it. I know he told me that he's really enjoyed working with you. He likes it when you guys chat with him. And he's saying that you don't have to practice an entire contest all at once. You can also just practice a few problems here and there. So it's good to practice a whole contest to know about the pacing, but even practice one or two problems is that's all you have time for on a particular day. When I leave the Zoom, a survey will open up. The survey will let us know what you liked about the class and how we can make it even better in the future. I hope you have enjoyed it. I certainly have enjoyed teaching all of you. And I hope you have a wonderful winter holiday season. And I hope I see a lot of you again in January. So let me leave so that that survey will pop up for all of you. Thank you so much. Bye bye.

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