Welcome, welcome to the fourth webinar for level 3-4. I do have a poll for this problem, the warm-up problem. Let me read it to you, and then I will launch the poll for everybody. How many three-digit numbers are there that have the sum of their digits equal to 5? For example, 122 is such a number because 1 plus 2 plus 2 equals 5. So hopefully that example helps you understand what the problem is asking. I'll go ahead and I'll launch the poll for you. Here's a question if you should count the example that they gave. Yes, count the example they gave. So if you're just joining, I don't know if you get the poll as soon as you join or not. But we do have a warm-up problem that we're solving. Okay, I'm going to end the poll. I don't know how many of you could see the poll. I think it depends on when you join the meeting, but we've done pretty well. Anyone else want to put a guess in there? I see some students are still putting in little answers, so that's good. Okay, so here are the results of the poll. We have about half of the students saying they think the answer is 15. The next most popular answer was 10. So let's take a look and see what we can do. So today's theme is making a list. So I'm going to solve this problem by making a list of the possible options, okay? So the first number they gave us is 122. We can rearrange these numbers and still have the same sum. So we can do 212, or we can do 221. That is all of the ones where I use a two, a two, and a one. I can also use a one plus a one plus a three to make a five. So I can do 113, 131, or 311. I can do a three, a two, and a zero. So I could do 320, 302, 203, or 230. I cannot use a number that starts with a zero and consider it to have three digits. The other options are a four plus a one plus a zero. So 410 or 401. I can do 140 or 104. I could also do 500, but I can't do any that start with a zero. So how many do I have? One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, which is the answer choice. So B is correct. If you had fewer than 15, see which ones you missed. You'll notice when I made my list, I kind of had at least some idea of organization in mind. Right? So the first row is one plus two plus two. The next row is one plus one plus three. You get the idea? So when you make your lists, I would try to keep them a little bit organized. It helps you not miss any possibilities. Okay. Let's go back to the beginning of the presentation so we can take a look at today's lesson is on making an organized list. So that is one recommended method for today. There's one or two problems where you might not need to, but you'll figure that out as we go. And we'll even have some bonus problems if we can get there today. So we always follow our four-step problem-solving strategy. You will notice that frequently as I'm reading the problem to you, I will underline the most important parts of it. That is my way of making sure I understand the problem and I know what I have to do, what I need to find. Once I know what I need to do, I can plan to solve it. Today, we'll probably try lists. I can be very careful and complete in my list, and then I can make sure my answer makes sense. So we've already done, last week was our pictures and diagrams. The week before that was patterns. And you'll notice when we did pictures, we had patterns in it. And in our lists, we might have patterns or we might have a picture. So we try to combine things all together. All right. And in organizing our lists, we have a few things to consider. The first is that we might have a large amount of data, so we want to keep it nice and organized, and we want to avoid repeating ourselves. So that's one of the biggest things is if we're just doing a mental list instead of writing our list down, that's the most important thing is we end up skipping something or we end up having things twice. So we're not, our count of how many things, how many options we have is usually incorrect when we do that. The other idea here is casework. So sometimes we'll want to split it up. And one example I use of casework is if I put on a green sweater today, what kind of pants would I put on that might match the green sweater, right? So that's a different set of pants probably than if I was gonna put on a pink sweater. What if I was putting on a shirt with stripes or with polka dots? So how many of you, when you get dressed in the morning, maybe have a kind of an example of casework? I know if I wear this top or these pants, then these are my options. Or perhaps you're thinking about the weather. Perhaps it's cold and rainy today. So you'll wanna wear certain clothes that are warm for a cold and rainy day, right? So that's examples, a very simple example of in your everyday life when you might be using casework. Or it could be on your shopping trips, right? You might buy different types of items at different types of stores. You're not gonna look for nails in the grocery store, right? So that might be, you might have a different shopping list for a different store. Just an example of real world for casework. In mathematics, it might say, like on our first warmup problem, I might say, I wanna use the numbers one, one, and two. What are the ways I can arrange the numbers one, one, and two into a three-digit number? I might wanna use a two, three, and a zero and arrange them into it. So that's casework as well. So that gives you an idea about what these concepts mean. What is the least number of matches that must be added to the picture in order to get exactly 11 squares? Okay, so these are matches. We don't play with matches, but we can use them in a math problem and a picture is no problem. So you might wanna figure out how many squares do you have to start with? How many squares do you need to add to get to 11? That would be a good starting place. We want exactly 11, not 12, not 13. I do not have a poll for this one because I wanted you to be able to see the picture for as long as possible. So put your answer in the chat. Ishan and I will try to reply to you and give you some thumbs up or some hints. All right, I will slowly start to solve this problem. I think one of the key tricks is to figure out how many squares you have at the beginning. And some students miss a few when they start to try to count these, so let's see if I can help you. So the first thing you'll notice is I'm going to count the largest squares, and those are three by three. So they have three matches across and three matches down. So you can see the first one that I'm highlighting now in blue. So we have three of those. The next one, I'm going to do a medium square, which is a two by two. And here's an example of a two by two square. And I see two of those. So there's one here, and then there's one a little bit lower to the right here, right? The next one I'm going to do is the small squares, which are just one by one. And here is an example of that. So I see one, two, three of those in a diagonal. So that gives me eight squares to start. Okay, and now I need to have exactly 11, so I need three more squares. There are a few ways that you can add matches to make three more squares. I'm going to give you a minute just to do that much, okay? See how many, then let's see if I can get some better answers in the chat now. Adam, you're speaking on mute. Thank you. So I'm going to show a couple of ways to do this. So what you can see is I've just added two matches. And by adding these two green matches, what I've done is I've increased it. I've added three more small squares. So now my small squares is 1, 2, 3, 4, 5, 6 small squares. And that gives me a total of 11. And that works great. That's one way to do it. I'm going to erase that. And I'm going to show you another way to do it. I like having video because I can change my mind. I could put a match over here and a match over here. There, here. You'll notice by doing that, what I did is I only made two more small squares. So I have five small squares. But I made one of the 2 by 2 medium squares. So now I still have 11. But I've done a different arrangement. So you could try different ways of adding either three more small squares by adding two matches. Or you could add small squares and a medium square by adding two matches. Hopefully, you like both those ways. Gregory forms two numbers with the digits one, two, three, four, five, and six. Both numbers have three digits and each digit is used only once. He adds these two numbers. What is the greatest sum Gregory can get? Again, here are the only digits we can use. He wants to make three digit numbers using each once and we want the greatest sum. And I do have a poll for this problem that I'll launch in just a second. So I have new students joining this webinar series every week. If you are new, you should know that the polls are anonymous. So there is no harm in trying to guess. No one will know if you got it right or wrong except yourself. And on math kangaroo contests, we do not penalize. We don't subtract any points if you guess incorrectly. So it's always better to make a guess than to leave it blank. Always try, try to get the correct answers if you can. But if you've run out of time, don't leave blanks, fill in something. Thank you very much. Okay, most of you have answered the poll. Thank you for doing that. And most of you are correct in the poll. So let me share that. It's a really good result here. So three fourths of you have said you think the answer is 1,173. And that is exactly correct. So if you wanted to do this problem with lists, you could try using different combinations of these numbers and see, you know, what is the greatest sum you can get adding them. But I think it's preferable in this case to use the concept of place value of your hundreds, your tens, and your ones. If we want the greatest sum possible, we will put the largest digits with the greatest place value. So that means putting the five and the six in the hundreds place. And then you would put the four and the three in the tens place. And finally, the one and the two in the ones place. Now, it doesn't matter which position on the top number or the bottom number, it absolutely doesn't. Because remember when we add, we add by place value in columns. So we're gonna get a three, a seven, five plus six is 11. So we have to carry to the thousands place. We could also ask this question as to what would be the smallest sum you could get. For the smallest sum, you would do exactly the opposite, which is you would do the one and the two in the hundreds place, the three and the four in the tens place, and the five and the six in the ones place. That would be the smallest that you could get. So then you would get an 11, you'd have to carry, this would give you an eight, and this would give you the three. So that would be the smallest value possible. Okay. A small zoo has a giraffe, an elephant, a lion and a turtle. Susan wants to plan a tour where she sees two different animals. I'm not sure that's much of a tour, but she wants to see two animals, and she does not want to start with the lion. How many different tours can she plan? So don't double count, try different ones. I do have a poll for this, but that'll launch in about one minute or so. Okay, very good. We're getting some good responses in the chat and some good responses in the poll. Remember, if you have questions as you're doing these, our TA Ishan is here and he's happy to help you. Try to answer some quick questions in the chat is a great way to ask him if you need some more help. So I'm ending the poll. I'm going to share the results. More than half of you think the answer is nine and that is correct. So that was a really good job on this problem. Let's take a look at why we get nine. So if this was just how many different possibilities are there to see two animals, we would have a slightly different answer. So if I was just asking about it, so here's my first animal and here's my second animal. So if it was just there were four animals, I would have four choices for the first one and once I visited the first animal, then I have three choices for the second animal. So that gives me four times three. This equals 12 if there was no limits, right? But we do have a limit. We have a limit that we don't want to see the lion first. So lion is not first. So we can see any of the other animals first. So that would be the giraffe. I'm not going to write it all out. That can be the elephant or the turtle. And from each one of those, I have three options. From the giraffe, I can go and see the elephant, the lion, or the turtle, right? And I have the same three choices for each of those. So now I have three first animals times three choices for the second animal equals nine. So my correct answer in this case will be that there are nine options, okay? So if there had been no limits 12, but since we do have those limits, it is going to be nine. I'm going to ask the students not to annotate on the screen while I'm annotating. It's very distracting. Other students are seeing things pop up on their screens and it's just not very polite to distract other students. Olga has three cards as shown in the picture. Using them, she can form different numbers. For example, 989, 986, and so on. How many different three-digit numbers can she form using the cards? So again, different three-digit numbers. And the interesting thing here is you'll notice that we can flip these numbers because a nine upside down is a six. I do have a poll for this one. We have polls for most of the questions today. I know the students like to have the polls. All right, I have most of you have put an answer into the poll, so we'll call that good. So let me end the poll and share the results. You did really well on this problem. Again, this is a really good group of students here. So most of you have said that the answer is 12. Remember, you might have to scroll down to see all the choices on your polls. But it is 12, so let's take a look. Remember, I like to make a very organized list, and I like the whole idea of casework. So let's do casework. Let's say that I'm going to use both cards as a 6. So both of them are going to be a 6. So I can do 6, 6, 8. I can move my 8 to the 10s place. That gives me 686. Or I can put the 8 in the 100s place. That's 866. Right? I can say that both cards are going to be in the 9 position. So I can kind of repeat this pattern. I can do 998, 989, and I can do 899. Now what's the other option? The other option is that I have one 6 and one 8. So let me make the first number be a 6 and the second one be an 8. So I get 698. I can have 689. And I can have 869. To the student who keeps unmuting, can you please stay muted? It causes a background noise. I can change it so that the first number will stay 9 and the second card will flip to the 6. 968, 986, or 896. So that is 12 different unique numbers that I can make with these cards. Okay, so that's the way to solve it by making a list. Christy has to sell 10 glass bells which vary in price. $1, $2, $3, $4, $5, $6, $7, $8, $9, and $10. Sorry I read that fast, but you know what I mean. In how many ways can Christy divide all the glass bells into 3 packages so that each package has the same price? So glass bells into 3 packages that have the same price. There is a poll. I'll give you a few minutes to read the problem and I'll launch that poll and hopefully we'll have some really good answers. want to clarify a point that I'm seeing in the chat. We don't want to make, we're not trying to divide 10 by 3. We're trying to divide the prices into three groups that are the same. So we don't have to smash a bell, but we're looking at the prices. Okay, I'm going to end the poll and share these responses. Overwhelmingly, the students say that there is not such a possible division, and that is correct. We can't do it this way. But it's not because we can't divide 10 bells. It is because when we take the prices together, we would have to add 1, plus 2, plus 3, plus 4, all the way up to the plus 10. You can do that, and you'll notice that the sum when you do that is 55. So the total price of all the bells together is $55, and 55 is not divisible by 3. So that's why it's impossible. Not because 10 isn't divisible, but because 55 is not divisible by 3. If we asked you to put them into five different groups, that would be possible. So you might want to try that as just a little exercise, how can you break them into five equal groups. Number six, and I find this one to be a tricky problem. So don't be surprised if you do too. This is a number 23. Remember in this level, there are 24 problems on the Math Kangaroo Contest. The last set of eight are each worth five points each. So this is a five point question, will come up tough on the end of a contest. In three soccer games, Daniel's team scored three goals and had one goal scored against them. For every game won, the team gets three points. For a tie, it gets one point. And for a game lost, it gets zero point. It is certain that the number of points the team earned in those three games was not equal to which of the following numbers. There's more than one way to try this problem. Okay, let me let me close the poll. I'm going to share the results, although what we found in the poll is that there isn't really a clear preference for choices. I have seven, five, and three all coming up as popular answers. So let's see which of those is actually correct. There are a few ways to solve this problem, but I think this problem, what happens is we try to interpret it, and it gets a little bit tricky. If you've ever played a soccer tournament, it would make more sense to you. I'm assuming that basketball tournaments or perhaps other tournaments might be run in a similar way. So Daniel's team has scored three goals and opponents, he's played three different teams, but the opponents have scored one. Okay, so if Daniel's team has scored three goals and the other teams against them have only scored one, we know that Daniel has won at least one game. Right, so if Daniel has won at least one game, that would give them three points for the win. Okay, and what could happen in the other games? So we know that there has to be one win, which is three points. Sorry, when I try to annotate with this type of annotations, I make more typos. So that's three points. Okay, so if let's say they won three to one, then the other games are 0-0 and 0-0. So you can have one win and you can have two ties. Right, that might be an option. So you can have one win and two ties. So let's say I won the three to one game and then the others were 0-0, 0-0 and 0-0. So two ties is another two points, so that is five points. It is totally possible. Okay, another answer students gave me was seven. Seven points. Seven points would be two wins. So we could win maybe two to one. Two to one is a win. We could win one to zero. That's a win. And then we could have a tie. A tie would be 0-0. So this is giving us two wins and a tie. Two wins and a tie is the seven points. Can we get six points? Six points would be two wins and a loss. Is it possible to do two wins and a loss? Well, just switch the zeros, right? Make this, you can do a two to one, a two to zero win, a one to zero win, and a zero to one loss. That gets you the six points. Four points is a win, a tie, and a loss. Can we figure out a win, a tie, and a loss? So we could win two to zero. We could tie zero to zero. Or we could, and then we can lose zero to one. That is a win, a tie, and a loss. So that would be three, four points. So we cannot get three points. Let's think about it. Three points is three ties, right? This is three ties. We know that we have more points than our opponents. When we have more points than our opponents, can we have tied every single game? We can't have tied every single game. It won't work. So think about it in that way. If we have more points than the opponents, we couldn't possibly have tied in all of our games. And so if you were a student who solved it very quickly, that might have been the thinking that you were doing, is you can't tie them all if you have more goals. Okay. I know that this is a table that you can use to look at this. You can think about how many games you could have won and lost. So those are possibilities. So some students are asking if you can win once. Yes, you can win once. You can definitely win once. I know that Ishan had asked if he could lead problem number seven. So go ahead, Ishan. I'm still not sure how to annotate on this. So please write along as I go. The question asks, how many three-digit numbers are there such that the sum of their digits equals four? Give them some time to solve it, Ishan, okay? Okay. This is a real short problem statement so I'll launch the poll right now. You can read the problem in the top of the poll. Okay, Sean, I'm seeing a lot of correct answers. We're gonna stop the poll and I'll let you explain it. This is similar to our warmup problem. So I think the students are getting the hang of this. So the most popular choice was 10. Almost half of them got it correct with 10. Can you help them out? So with the question, it's very similar to how you solve the first problem or I guess the warmup problem. Very similar solution. You can actually solve them the same exact way. So to restate the question, how many three digit numbers are there such that the sum of their digits equals four? So what this is really asking is how many three digit numbers, right? A number in the hundreds are there such that if you add up every single individual digit, it equals four. So one example would be 400, right? Because the first digit four plus the second digit zero plus the third digit zero equals four. And in the same train of thought, a number like 310 would work because three plus one plus zero equals four. But to make sure we don't skip over any numbers, we want to keep this in an organized list. So we can start with the 400s. The only number in the 400s that will work is 400. And we know that there can't be any numbers greater than 400 because if there was a number such as 500, for example, the sum of the digits of that three digit number would inherently have to be greater than four because the first digit is greater than four. So now we know that the three digit numbers can only exist in the 400s, 300s, 200s and 100s. There obviously can't be any numbers lower than 100 as the number must be three digits. So again, we went over the only four digit, sorry, 400, only three digit number in the 400s that works, which is 400. Next, we're going over the 300s. So the first number that works is 310, but there's one more. We can just switch around the one and the zero and we have 301. Next up, we have the 200s. Of the 200s, since the first digit is two, the next two digits have to sum up to two. All right, that means that we can first make the 10th digit two and the one digit zero. And our first number, the 200s is 220. Then for the next number, we can switch around that two and zero and then we have 202. However, that's not the only way to get the sum of two when you're adding up the second two digits. You can also add up two ones. So you also have 211. And finally, we just have the numbers in the 100s. Since we know the first digit is one, that means the next two digits have to add up to three because one plus three equals four. So we can first start off with 130. And we can do this the same way we did 310 and 301, or 220 and 202. We just switch around these zero and the three. And of course, adding three and zero isn't the only way to get three. We can also add two and one, which means instead of 103, we can do 121 or 112 in the same train of thought. And this means that we have, you just add that amount of numbers together, a triangular number, which is 10. Thank you. And now we can move on to, I think will be our last problem today. Number eight. Would you like to do this one as well, Ishan? Yeah, first I will read out the problem and then I'll give you some time. Question eight asks, Adam wrote all the numbers from one to 100 inclusive. How many times did he use the digit five? The word inclusive just means that including, the sequence is including one and 100. Okay, we're going to be running a few minutes over today, so let me share the poll. So Ishan, the most common answers are 19 and 20. I think you can help them figure out which one of those is correct. The correct answer is 20, and we'll go over why now. Just to restate the question, Adam wrote all the numbers from 1 to 100 inclusive. How many times did he use the digit 5? Again, inclusive just means that's including 1 and 100. So there's three cases we have here. The numbers are either single digit, double digit, or triple digit. But the first case and third case are done with really quickly. So the first case, if the numbers are single digit, the single digit numbers, sorry, it's the numbers 1 through 9. And of those, you only use the digit 5 once for the number 5. As for the third case, the triple digit numbers, there's only one triple digit number, which is 100, and that you never use the digit 5 in the number 100. So now we know the bulk of the numbers you have to consider are double digit, right? So from 10 to 99, how many times do we use the digit 5? And you have to keep in mind we use the digit 5 once in 1 through 9. So of these double digit numbers, there's two places where the digit 5 can occur. It can occur in the tens place or the ones place. In the ones place, that would be a number like 15, 25, 35, 45, etc. And in the tens place, that would be a number like 51, 52, 53, 54, etc. So now we know that in the ones place, the number 5 can occur from 15 through 95. And if we think of the numbers 15 through 95, where we're just taking jumps by 10, that's essentially the same as the sequence 1 through 9. And if we think of the sequence 1 through 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, obviously there's 9 numbers in that sequence. And we can think of it like this because you can just subtract the same number 5 from every single number in the sequence, right? So instead of 15, 25, 35, 45, it's 10, 20, 30, 40. Then just divide everything by 10. It's 1, 2, 3, 4, 5, 6, 7, 8, 9. And then in the tens place, you would have 50, 51, 52. You can just subtract the same number from every single number in that sequence. Subtract 49, for example. And then you get 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So you have 10 instances where 5 is used, where 5 is in the tens place, and 9 where it's used in the ones place. Now you have to keep in mind exactly what the question is asking for. It's asking how many times did he use the digit 5? If the question was asking instead how many numbers are there where there's a digit 5 involved, the answer would be different, right? Because 55 is double counted, right? We use 5 in the tens place and the ones place, but they're not asking for that. They're simply asking how many times did he use the digit 5? And in the number 55, we use the digit 5 twice. So we add up 1 plus 9 plus 10, and we get our answer, which is 20. Thank you, Ishan. Yeah, so it's pretty common that you might see on math contests where you have to consider when you have a double number and do you count it once or count it twice. And reading the problem statement very carefully will tell you which way you need to do it. Okay, so when will you use an organized list? A lot of times if there's casework. A lot of times if you don't want to miscount something. When it says how many possibilities, that's a good clue that making lists could help you solve it. I think overall, the most challenging problem today seemed from the students seem to be the soccer games, the soccer tournament with the points. That seemed to be the most challenging one. So that's the one that you might want to look back on first and see if you can get that. Then I've had a few people asking me about how they can find the past recordings. And I want to show you if you go into your Math Kangaroo account. So this is mine. Mine looks a little bit different. But this is, for example, the webinar that I teach for levels five and six. So when you go into your account, you will find in your webinars. Let's see. Yeah, in your webinars, in your My Registrations, you will find that there's a place where you can actually find the listing for the recordings. OK, so that's where you do it. Also, you will find in when you look at your registration for the contests that you have some free video resources you can look at. You have some discounts on some practice contests. These are some good places to find it. So again, this is in your registration for the contest. And this is in your registration for this webinar series. So I encourage you to look there to find the recordings if you want to review or if you missed any of this series. So thank you very much for your attention. Thank you, Ishan, for your help today. And I will see everybody next Sunday. Bye for now.

organized lists

problem solving

three-digit numbers

digit sum

systematic listing

combinations

constraints

casework

patterns

webinar