afternoon. Welcome to the Math Kangaroo Level 3-4 webinar. This is webinar number 6. So I hope you are all enjoying this webinar series. If you're new, I hope that you enjoy today's. Sometimes students ask me how they can get recordings. If you stay till the end, like a few minutes later, I can help you take a look at how to get the recordings. I'll open my Math Kangaroo account and let you see where those are. Today, webinar number 6 is making a table. I don't know why. Ishan, my waiting room thing is not working properly. So if you can continue to admit students, that would be fantastic. I don't know why it's not showing properly. Okay, so during the webinar, if you have questions, please feel free to send a chat to me or to Ishan. He is our teaching assistant. We're both co-hosts on the meeting, so you can reach out to us. And we will have polls for a lot of our questions today, so you'll be able to put your answers to questions in the poll. And that is anonymous, so feel free to do that. Okay, we have a warm-up problem. Our class has 30 students. The number of boys is four times greater than the number of girls. How many girls are there in our class? Doing really well in the poll, doing really well in the chat. I'm going to give you three seconds to click on an answer if you haven't done that already, and we'll take a look. So I'll share the results. So a few students say 24. 24 is the number of boys in the class, but the most of you say six, and that is the number of girls in the class. So I'm going to show you one way to solve this problem. Remember, this is just one way. We want to know the number of girls in this class, and we could make a table. You see the next slide if I have a table on it. I do. Okay. So if we say that there's one girl, there's going to be four boys, and that's a total of five. If we say that there are, let's try, three girls, three times four is 12. That would give us a total of 15 students, but we know that we need 30 students, and that is, we have to multiply this, the whole thing, times two, because we need to get to 30 students, right? So what if I did six girls, and then six times four, I'll write that here, six times four equals 24, and six plus 24 is 30. So that gives me that the number of girls is six. Now, some of you might have done this a slightly different way, a version that doesn't maybe need a table, but it's the same idea, is you might have said that the ratio, if you had looked at a ratio, perhaps, the ratio of girls to total is that there's one girl out of every five students, and then you might have said, okay, I need 30 students, and since this was times five, I have to divide by five, and I will get six. That is another very good way to solve this problem. So any version of that, you could fill in like a lot more of this table. I noticed that I needed to double my row, and tables are really good for helping us find patterns. So that's one of the things you might notice when we use tables, is that it helps you see a pattern. It's a really good visual of patterns. So remember, we're going to be very careful today. We have several questions where Math Kangaroo thinks they're being very tricky, and Math Kangaroo is going to ask you, like, which of these is not possible, or things like that. So be careful that you know what it's asking. Read really, really carefully there. Our plan today is going to be to use tables. Now remember, this is just one of the tools you can use to solve Math Kangaroo problems. We like to give you one tool, nice and clear each week, and give you some practice, and then when you do a whole contest, you can decide which tools you like to use, which ones fit the problem, and which tools fit your personal style. But we have to practice each one of them in order to get pretty good at them. When I do my tables, that's a really good way of helping me be careful, right? I definitely have to be organized when I do a table, so that's one of the reasons that tables are so great. And then if you can find another way to solve your problem, that's a really good way to make sure that you have the correct answer, right? Look back. I did it another way. I did it with proportions and ratios. I got the same answer. I must be all set. Okay. So, we are on lesson number six. We're over halfway through. I hope you're enjoying it so far. All right. A table is information. It's a set of facts, and we arrange them into rows and columns. Usually, we like to have labels on them. If you're doing your table just for yourself and your notes, your labels can be pretty sketchy. It's okay if you just wanted to do an S for Smith and a J for Jones. It's no problem. It's your table. You'll notice that sometimes we can just use tally marks or tick marks or things like that. Sometimes we're only going to have names and numbers in the columns. Everything like that will work, okay? The little boxes are called cells. So, if you ever hear somebody talk about the cells of a table, that's one individual box. We like to use tables when we have data that we call has more than one characteristic. And a really famous problem that we don't have an example today is like the legs on animals problem. So, you might have to figure out how many of different animals you have and how many legs you have. For example, a fly has six legs and a cow has four legs, right? So, you might be filling in your table to know characteristics of the different animals. Sometimes they'll give you data and you have having a table will help you organize it. Sometimes you're going to have to figure out the numbers and the data for yourself, okay? So, we'll have some of those as well. You'll see we have a lot of blank tables we're going to work with today. Tables are great so that you don't repeat yourself. And again, like I said, you can find patterns in your tables. That's a really helpful tool. So, for most of them today, I might have a blank slide with no table. And then we'll switch when I'm solving it to a slide where I've drawn an outline of a table. It'll just make it a little faster for us. But we have polls as well. So, I didn't want to give you all the tables because on a math contest, you won't have them. But if you're struggling and then you see mine, that might help you. A certain dance group started out with 25 boys and 19 girls. Every week, two more boys and three more girls joined the dance group. After how many weeks will there be the same number of boys and girls in the dance group? Be careful. What do we need? We want how many weeks. Not how many there are, but how many weeks. Not how many dancers. And we want the same boys and girls. Same boys and girls. Not how many boys or girls. Not the total number of students, but the number of weeks. Just a few more seconds for those who want to answer the poll. Remember, it's completely anonymous, so even if you make a mistake, it's no big deal. It's all about the learning process. It's okay to make errors if that's part of how you are going to learn to solve. If you didn't know where to start your table, I've put the table up on the screen, and all I've entered so far is the information that's given in the problem. Okay, I'm going to end the poll. And you've done very well on this poll. Almost three-fourths of you, 73%, say the answer is six. That is correct. And this is one of these tables that's going to help us detect a pattern. So when we start, we can call that week zero, because it says after how many weeks. After zero weeks, we have the starting amount. After one week, we're going to add two boys, so that's 27, and we're going to add three girls, so that's 22. You might also notice that the difference between 25 and 19 is 6, and the difference between 27 and 22 is 5. Let's try one more week. After two weeks, I'm going to get two more boys, 29. I'm going to get three more girls, 25. And again, you're going to notice the difference is now only 4. So if the initial difference was 6, then it would take me six weeks before the difference is zero, right? So when I get to the six weeks, it will be 6 times 2 is 12. It'll be 25 plus 12. That gives me 37. It'll be 19, and 6 times 3 is 18. That's also 37. So you can fill in the whole table all until you get to 6. You can notice a pattern that it's decreasing by one each week, and the initial difference was 6, so you need six weeks. So there's kind of like a little bit of a hybrid, right? Part table, part reasoning, kind of like if some people have hybrid cars, right? Part electric, part gas. Okay, problem number two. A kangaroo noticed that each winter he gains 5 kilograms of weight, and each summer he loses 4 kilograms of weight. During the spring and fall, his weight does not change. In the spring of 2008, he weighed 100 kilograms. How much did he weigh in the fall of 2004? You might also notice that this is a working backwards problem, because we have 2008 is at the end, and we want to go back in time to find 2004. And be careful, there's a difference between spring and fall. And you can think about the seasons and which order they go in a year, because that does count in this problem. So I'll leave the problem this way for a little bit. Then I will launch the poll, and I'll advance to the next slide, which shows the beginning of a table. So you can figure out how fast, how much help you need, if you want to wait. Do your best, though. Always try as hard as you can, because this is all you'll get when you see the Math Kangaroo Contest. We won't give you a table. You have to make it yourself. All right, I'm going to share the results of the poll in just 10 seconds, so if you want to put in your answer real fast, remember blanks are worth zero points, correct guesses are worth, in this case, four points. Okay, so this is interesting. We have half of you think the answer is 96 kilograms and about a third of you say 92 kilograms. So let's take a look at the problem and see why we're getting those two responses and which of them is correct. So if I work backwards, I think everyone can see the table, all I did so far is if you think about the seasons of the year, right now we're in winter, the next season will be spring and then the summertime and then fall you'll start a new school year and then you get to winter where we are now. That's the way the school, that's the way a year goes, right? So if it's the beginning of 2008 right now, how much does he weigh here? You can fill that in. He weighs 100 kilograms, right? 100 here. So if we go backwards and since he gained 5 pounds over the summer, he must have weighed 95 pounds before that gain and since he had lost weight in the summer, he must have weighed 9 more pounds or 99. So when we work backwards, we do the opposite operation. What you might look is if you see across, you can see every year kangaroo has gained one pound because lost four but gained five. So every year he's gained a pound. So you can do it that way, you can fill that in as a pattern, right? You can also, from the 99, he's gained five so he was five less in the fall. These are also going to follow that pattern. Oh, sorry, it's a little typo on my part. We've got to subtract and then we subtract. So a lot of students put their answer is 96. This is the weight for the spring but the question specifically says the fall. So the answer would be 92 because the fall comes after the spring in the year 2004. So over summer, lost weight. The correct answer is 92. 96 is here. It makes some sense that a lot of you put that as your answer but be very, very careful with our math kangaroo problems. We like to trick you. Okay. Number three. Ishan, you wanted to lead this one? Okay. So the question states in the expression 2002 and blank square 2003, blank square 2004, blank square 2005, blank square 2006, either plus or minus can be written in the place of that blank square. So which result is impossible? We'll give you some time to think about that and I'll open the... I don't think we have a poll for question three. So, in one more minute, I'll just explain the answer. Yeah, there was no poll because the symbols were difficult to get into the zoom poll. I will explain it now. Don't worry if you didn't get it. This one is a little hard to solve if you don't have some prior experience. So the first thing I did to solve this problem was I decided I would try trial and error. So first we have to figure out how do we get a result that looks like that. So we have five numbers and we can either add the numbers or subtract the numbers. So what should be sort of intuitive is that you have to add two numbers and you have to subtract two numbers. So we could add 2002, 2003, and 2004 and if we subtract 2005 and 2006. So maybe saying add two numbers is incorrect, but you put two of those blank squares should have addition signs and two of those blank squares should have minus signs. So now let's see what's the largest possible number we can have. The largest possible number we could have is if we added the 2005 and 2006. So we put the plus signs in between those and then you put the minus signs in between the 2002 and 2003 and 2003 and 2004. And that would give us a number that looks approximately like 2010. So how do we get a number smaller than this? Well, to get a number smaller than 2010, but still in that range, in the 2000s range, we would have to switch around one of the pluses and minus. So if we switched around the plus and minus between the space in between the 2003 and 2004 and in between the 2004 and 2005, we would notice that the number drops by two. That would give us 2008, which is nice, but not exactly where we want it to be as 2008 is still bigger than any number in the answer choices. So starting from the 2002 plus 2003 minus, sorry, plus 2004 minus 2005 minus 2006. Let's switch the number in between or let's switch the plus sign in between the 2002 and 2003 with the minus sign in between the 2005 and 2006. Sorry, Sean, I can't take notes as fast as you're speaking. Sorry, I'll try to slow down. So let's try it out once more. 2002 plus 2003 plus 2004 minus 2005 minus 2006. Right, we're doing the opposite. So this way we get the smallest possible bound. And this will give us a number that looks approximately like. Five, nine. Nineteen, nine. Sorry. Nine. 1998, which is one of our answers. If I did the addition correctly. So that means that A is obviously impossible. So now we want to find a number that's bigger. So we can switch the plus in between the 2003 and 2004 with the minus in between the 2004 and 2005. And that would give us a number that's approximately one bigger. So 1999. So knowing this, we want to increase it some more because 1999 is not any one of those answers. So now we can switch the plus in between the three and the four with the minus in between the five and the six. And from 1999, we would have 2001, which is sorry. I'm not getting the 1999 because if you switch, you're going to change it by at least two each time. Yeah. Three, five. Sorry, 1998. No, I don't know where 1999 came from. Sorry. What we notice is that because we're switching numbers, like 2002 with 2004 or 2003 with 2005 and 2004 and 2006. Because we have the two plus signs on the left and the minus signs on the right. We're switching numbers that shift by two. So every time we switch the plus sign in between the 2002 and 2003 with the minus sign in between the 2004 and 2005, we would increase the number by two. And that follows for every single time we switch something. So if we switch the plus sign in between the 2003 and 2004 and the minus sign in between the 2005 and 2006, that would increase that number by two as well. Which means that no matter what result we have, it would have to be a number that ends in a multiple of two. Since we found that if we have 2002 plus 2003 plus 2004 minus 2005 minus 2006, we have 1998, which is an even number. Since we're just bumping the numbers up by two each time, the only possible numbers must end in two or four or six or eight and be even. Which means the impossible result is 2001. I'm not sure if I did a great job explaining that. So if you want to explain that to me. Yeah, let me try again because it's hard for me to take notes since you're talking, Ishan. So there's a couple of things to consider here. The first is 2001 is the only odd number. OK, so I have an even, an odd, an even, an odd, and an even. So if I have two odd numbers and the rest are even, if I add two odd numbers, I get even. If I subtract two, I get odd. And if I add an odd and then subtract an odd, it goes back to even. So 2001 is the only odd number and that's the one I'm not going to be able to make. Now I have a couple of tricks that I like to get for the others. So for example, if I want to get a 2004, I'm going to use this one in red. I'm going to keep the 2004 here. Now can I make the others kind of cancel each other away? So I can, for example, say that 3 and 5 make an 8 and 2 and 6 make an 8. So if I subtracted the 2003 and 2005 but added the 2002 and 2006, I would have, I'd have a positive 2, a positive 6, a minus 3, and a minus 5. And you can see that that leaves me with the 2004. So 2004 is possible. So maybe this approach will help some students figure this out. I could do the same thing. Let's say I want to make 2006. I'm going to leave the 2006. Can I make these go away with each other? Well, I can do a 3 and a 4 could be a minus 3 and minus 4 is minus 7. And then plus 2 and plus 5 is a plus 7. Sorry, this should be a 4. There you go. So that's one way to be able to see that you can make these other numbers. And then I really do like the hint of the evens and the odds. There is no way to get an odd result. What Ishan is saying about moving the pluses and minuses around and seeing what differences you can make, that's a great strategy. Just keep in mind, if I switch a plus and a minus, if I add one and subtract, I'm going to go by twos. So if I keep a 2003 but subtract 2004, my number will change by two every time. So hopefully that has made it a little more clear and given you a few strategies to try this problem on your own. It's a really tough one. All right, Ishan, will you read number 4? So question 4 states, one number was chosen from the numbers 51, 52, 53, 54, and 55, and the digit 0 was placed between the digits of that number. What is the difference between the new number and the number which was chosen? I'll open the poll now, give you some time. I encourage the students to try this with a few of them. This one is not nearly as difficult as it likes to think it is. I think we'll, or I'll explain now and I'll close the poll. Well did very well, around 60% of you got it right. So just to restate, the question asks, what number was chosen from the numbers 51, 52, 53, 54, 55? And the digit zero was placed between the digits of that number. What is the difference between the new number and the number which was chosen? So a question like this is very easy to brute force. You can just list the numbers, like Dr. Sikki did. 51, and the number with the zero in between the digits is 501. And the digit is, or the difference is 450. And with the next set of numbers, you have 52 and 502, and the difference is 450. Now, if you want, you can continue that with all the numbers or you could just realize that every single time you increase your number by one, the new number, which is that number with zero in between, also increases by one. So no matter what, provided that those numbers have the same tens digit, if you do this pattern with, take the number and then you put the zero in between them and then find the difference, every single number within the same tens digit will always have that same difference. So knowing that, or you could just brute force, you'll know that the difference between the new number and the number which was chosen must be the 450. Thank you. Thank you, Ishan. I think this question is much clearer than the other question. It's a simpler one, but it's also a five point question. So just because a question is five points on Math Kangaroo, don't assume that it's gonna be too complicated or too difficult. A little bit of brute force or a little bit of reasoning about place value, and you'll have this one really easy and really quick and be able to move on to other problems. In one year in March, there were five Mondays. Which day of the week below could not appear in the month five times as well? I shouldn't give more explanation, but think about a month. There's normally seven days in a week, always seven days in a week, but the number of days in a month is not a multiple of seven, right? So some days will appear four times and some days will appear five times. So if we want the ones that are not five times, they must only be four times. Big hints I'm giving today. Some of you might be asking, why is this in a table lesson? Well, a monthly grid looks like a table, doesn't it? And if you didn't know how many days there were in the month of March, you know that February is the only month with 28 or 29. The others have 30 or 31. So you could always use 30 or 31 and try to figure this problem out. That will get you there. But if you know the knuckle trick, put your hands in front of you. Right in front of you. You can see that March has 31 days. Okay, we're gonna end this here. A lot of you answered the poll, so thank you. Over 40% of you say Thursday, but we have answers for every day of the week. So that's pretty interesting. You'll notice that I've started making some notes already. We know that there are 31 days. We know that there are, if I do four times seven, cause seven days in a week, so four weeks is 28 days. 31 minus 28 gives me three days that must be repeated a fifth time. Okay, so I know Monday is one of them. So if I started my month on a Saturday, one, two, three, that makes Monday the third day. I can add seven, so I get Monday would also be a 10th. Monday would be a 17th, Monday would be a 24th, and then Monday would be the 31st. So Saturday works. Clearly I could make Sunday the first day and that would work as well. Let's try, can Tuesday be repeated? Well, if Sunday is the first day of the month, let's change that around. If Sunday is the first day of the month, each one of these is one less, right? So this would be the 30th and then the 31st would be Tuesday. So Tuesday would also work. Wednesday is an interesting one. I'm gonna actually clear off what I have on the calendar so I can make it a little easier, not run off the bottom of the calendar here. Let's say I wanna try to make it, ah, come back here. Perfect. Let's say I wanna see Wednesday. Okay, if the first day of the month is a Monday, this is Tuesday, this is the second, this is the third. Remember, three days get repeated five times. So that would make this the eighth, ninth, 10th. This would be plus seven is 15, 16, 17. You'll notice that this comes out being the 31st. So Wednesday is possible, but Thursday, and also it's not on here, but Friday are too far away from Monday. Another way to think about this is to count three days before including and before and after. So let me try that way because I like to present a few different methods to help students out. If we know that Monday has to be there, we have three days, it could be Saturday, Sunday, and Monday could be your three days that are repeated five times. It could be that Sunday, Monday, and Tuesday are the days repeated three times. It could be that Monday, Tuesday, and Wednesday are the days repeated three times. So any day, Saturday through Wednesday could be repeated three times depending on what the first day of March happens to be that year. But you can see that Thursday and Friday are never within those three days. Hopefully that helps. A couple of ways to look at the problem. The ladybugs, which live on an enchanted meadow are either red and have thick spots or yellow and have 10 spots. Various animals, including both red and yellow ladybugs came to a birthday party for the dragonfly. Sorry, I don't mean to cross it, I mean to underline it. The dragonfly noticed that the total number of spots on the ladybugs at the party was 42. How many ladybugs came to the dragonfly's birthday party? Now, remember, just keep it in mind, read the problems exactly. Both red and yellow ladybugs come to the party, not just one type. And we wanna know how many total ladybugs, not red or yellow, but total ladybugs came to the party. Another one strange thing that Math Kangaroo did with this problem is you notice the answer choices are not in descending or ascending order. They're kind of mixed up, so choose carefully. Okay, most of you have answered the poll, you've gotten through this problem, and this question is a much more successful one. The correct answer is five, and you must have been very careful about selecting the correct answer, even though they're out of order, so good work. Let's take a look at it. I had a student tell me that, ah, I know this is easy. If we have seven red ladybugs, seven times six is 42. We're done. Sounds good, right? That would be seven. Unfortunately, it says that there are both red and yellow ladybugs, and this would be zero ladybugs, so this breaks the rule. Then I had another student say, okay, I know that the answer is 42, I need to have a two in the ones position. What multiples, the 10 will never get me a two in the ones position, it's always going to be a zero, so that two has to come from the six, and that's a really good observation and a kind of a shortcut on this problem. If I do two times six, that's going to get me 12, right? And then to get 42, I'm going to need another 30, so this would be three times 10 spots, so that gets me the 42 spots, and it gets me five ladybugs. You can fill out the table a lot more, but when there is a shortcut, it always is a good idea to try to use it in Math Kangaroos so you don't run out of time. Okay, let's try to get on to number seven. There are 13 coins in John's pocket, and each of them is either a five-cent coin or a ten-cent coin. Which of the numbers below cannot be the total of John's coins? So 13 fives or tens? Here's a hint for some of you who haven't answered yet. Think of this as a maximum and a minimum problem. Nishan was trying to introduce the other problem with maxes and mins. What's the greatest and the least value you could have with these coins? OK, so if I end the poll here, you did very well. Of those who answered, 3 4ths of you are saying the answer is $0.60, which is exactly right. Let's take a look at it. If I have 10 of the $0.05 coins that's $0.50, 11 of them is $0.55, then I would have to have either $0.03 or $0.10. So you can see, I can get every value that differs by five. The smallest amount I can get, the minimum, is to have 13 of the $0.05 coins. That would be a $0.65 value. So I can get nothing that's lower than $0.65. And the greatest amount that I can get is all of them are $0.10. So I would get $0.130. So any value that is a multiple of five, that's between $0.65 and $0.130, I can get. So this one is too small. Can't get that. That would be only 12 coins. All right. I think we can move to a bonus problem. Oh, we have number eight. We're not up to the bonuses yet. Cool. Let's do number eight. In his wallet, Stan has one $5 bill, one $2 bill, and one $1 bill. $2 bills are pretty rare. Some of you might not have seen a $2 bill. Which of the following amounts can Stan not make using the bills that he has? There's been a lot of these nots in today's lesson. So if I take the $2 bill and the $1 bill, I can get $3. If I take the $5 bill and the $1 bill, I can get $6. If I take the $5 bill and the $2 bill, I can get $7. And if I take all of them, the $5 bill, the $2 bill, and the $1 bill, I can get $8. But there's no way to get $4. All right. I know some of you love to see bonus problems. So I wanted to get you to see a bonus problem today. It's a tricky one. The sum of two digits, one from inside the square and one from outside the square, is greater than 10. How many such pairs can we make? So, important parts of this, one from inside, one from outside, and a sum greater than 10. So for example, 8 plus 7 is 15, that would count. Here's a table that I've made to help you. I took the numbers that were outside because there were fewer of them, and I put them in the first column. If I use the 1 from outside, the greatest sum I can get would be to add the 9 from inside, and that's still not greater than 10. So that's going to be a big 0 there. There are none of them that will make me greater than 10. If I use the 2 on the outside, 2 plus 8 is 10. Only 2 plus 9 is 11. So that's only one combination. What if I use the 5 on the outside? Then I need all numbers that are greater than 5, right? The 6, sorry about that, the 7, the 8, and the 9. So there would be four possible combinations. If I use the 8 on the outside, then any number greater than 2 is a possibility, right? Because 8 plus 3 is 11. So 4, 5, 6, 7, 8, 9. That's 3, 6, 7 possible numbers. If I use the 7, then any number greater than 3, so 4, 5, 6, 7, 8, and 9, that is 6. Now if I add that together, there's 18 possible combinations. There's quite a few pairs that will give me a sum greater than 10. Why did I make a table for this problem? To keep it very organized. Because if I tried just making lists, I could repeat myself or get confused because I have numbers inside and numbers outside. So this is what we're talking about with two different characteristics. Some of these numbers are repeating, and I could get very confused. So the table is a really good way to organize and not miss any possibilities for this particular problem. So I'm going to come back up. When will you use the table? Use the table when you have different categories or characteristics. Use the table when there's a lot of data to sort through and you don't want to skip anything. What are the challenges of tables? A lot of students find it difficult to know, what do I put in the different columns and the different rows? And we might have slightly different opinions. The way I might draw the table might be different than the way you drew it, might be different than the way Ishan draws it. And you can switch the columns and the rows. So what I might do up and down, you might do going across horizontally. And that's perfectly fine. Don't worry about that. So the biggest difficulties, like I said, there's a little bit of like, how do I initially set it up so that it's going to work really well and not take too much time? But these are typically problems that if you try by another method, you're going to get a little bit of confusion. You'll lose your place. And so being really organized and having that table will end up paying even though you took a little more time to make it. So I hope this is a good strategy for you in the future. Do I have anybody who needs to stay and figure out how to get past recordings? You let me know in the chat if you need that. All right, it looks like we're good then. All right, I will see everybody next week. I hope you have enjoyed the lesson on making tables today. Thank you, Ishan, and thank you everyone for coming. Enjoy the Super Bowl this afternoon. Bye.

Math Kangaroo

Level 3-4

mathematical problems

making tables

data organization

interactive components

logical reasoning

arithmetic sequences

word problems

problem-solving skills