Good morning, good afternoon, depending on where you are. Welcome to the Math Kangaroo webinar, Multiple Factors and Divisibility. I'm Dr. Sarah Sagee, and we have a TA with us named Maya. We are going to be using some polls and our chat today, so make sure you're prepared for that. Your microphone will be muted during the session. And you should have paper and pencil ready to take notes and work through problems. Everything that we're doing today, you should be able to work out pretty quickly if you have some scratch paper. You should not need a calculator because you're not allowed to use a calculator for Math Kangaroo problems for the contest. You may ask as many questions as you'd like through the chat because we don't have such a large group. You should be able to get your questions answered in the chat by either myself or Maya. Maya will type your answers back. If I see a question reoccurring, I'll just answer it as we go through the webinar, okay? There will be some polls and quizzes. If your browser or your setup for Zoom allows pop-up windows, you will see those questions popping up. If they cover the question you want to read, first of all, most of them have the question written into the pop-up window, so you won't be missing anything, but you can just drag that window from the top of it and you can move it around your screen if it's bothering you, okay? So don't worry about it. You'll see the questions either way. And let's start a really great webinar this month. My name is Dr. Sarah Sagee. I'm a biomedical scientist. I live in San Diego. So when I say good morning, it is still 11 o'clock in the morning where I am. You have maybe two o'clock in the afternoon where you live. I've been teaching math and science for over 30 years. I coach an elementary school math club. I've been their head coach for six years. I do teach math kangaroo courses every week. I also like karate and swimming, and I've taught both of those. And right now my favorite creature in the entire world is my pet dog, but don't tell my four children. And I'll let Maya introduce herself. I'm Maya. I'm in 10th grade, so I'm 15 years old, and I live in Champaign, Illinois. I've been doing math kangaroos since fourth grade. And some other things that I like are sledding and weaving. And yeah, I'll be helping today. So if you have any questions, you can write to me in the chat. Okay? And I do love my children very much, don't I? Okay, let's begin. Math kangaroo emphasizes a four-step problem-solving strategy. The first step to solving any problem is determining what the problem is asking you to do. Do you understand the facts that it's giving you? Do you know what to do with it? Can you make a plan, step two, plan how you're gonna solve the problem given the information? How can you tackle the problem? Do you have more than one option? Frequently, there's more than one option for solving the problem, and that's fine. Everyone will have a preferential way that their mind likes to think about a question. If you solve your questions slightly differently than I do today, that's okay. I will try to present more than one method if I can think of one, but just because mine is different from yours doesn't mean either one of us is incorrect. Carry out your plan. It's very important as you get older in your math contest that you work very carefully. If you make a mistake in a calculation, your answer will be wrong, even though that doesn't reflect your actual ability to solve the problem, so work carefully. And then look back and check your answer. Does your answer make sense? Did you answer the correct question? Sometimes you have to be very careful about what the question is asking, and the key for that is usually either in the first sentence or the last sentence of the problem. Not always, but I would say at least 95% of the time, it's either the first sentence or the last sentence that tells you what you need to have in your answer. Do you agree, Maya? Yeah, Maya agrees. Okay, so we have a couple of things to talk about today, a little bit of vocabulary so that we're using the same terminology as we go through the webinar to avoid any confusion. And I think most of these you already know being fifth and sixth graders for the most part. An integer is the set of numbers that do not contain fractions or decimal parts. So it can be positive, it can be negative, it can be zero. Those are all integers, but we do not want 1.5 or three quarters. We're not using fractions and we're not using parts. When we look at factors and divisibility, we're looking at whole numbers, no pieces of numbers. A factor in this case is an integer that's being multiplied by another integer. And you have the example here. Sorry, I'm gonna annotate Mars now. You have the example here that a factor times a factor equals a product, right? And you've probably been doing this since about the third grade when you learned how to multiply things together, right? So when you look at any sort of multiplication, it can be five times six equals 30. Five and six are the factors and 30 is the product. So just a little bit of common vocabulary. We wanna speak the same mathematical language as we go through today. Multiples of a number are the product when you multiply that number of interest by an integer. So the positive multiples of three, zero is not really positive, but that's okay. The multiples of three are zero, three, six, nine, 12, 15. And you could have negative multiples as well. You could have negative three, negative six, negative nine as well. An easy way to find multiples, you probably learned when you were really little, it's skip counting. I know one of the kindergarten teachers at the school where I help, she loves to have the kids practice skip counting. So she'll have kindergarteners skip counting by fives, fours, tens, twos. And those give you all the multiples of a number, the positive multiples anyway. So five, 10, 15, 20, 25, 30. Those are multiples of five, it's skip counting. Common multiples are numbers that are shared from two or more numbers. So if you notice, I did the multiples of three here and the multiples of five. And if I wanna find the common multiples, they have to be divisible by both three and five. So they actually wind up being the multiples of 15. So this is gonna be an important idea that the common multiples of three and five are gonna be the multiples of 15 in this case. Okay, hopefully that all makes some sense. You've seen these types of things before. The factors of a number are the opposite. Before we were multiplying the number, now we're gonna be breaking it down into its factors. So the factors of a number are what divides it evenly, leaving no remainder. So if you remember, I said, we're not using fractions today. We're not using parts of numbers. We want to have no remainders. So the factors of 24, for example, sometimes we'll list them in pairs. It makes it a little bit easier. Our one in 24, because one times 24, two in 12, three in eight, and four in six. In this case, when finding the factors of 24, I know that I have finished because the next factor, you notice anything about these numbers? I like to make my factors increase. Right? Five is not a factor of 24. And the next factor of 24 is actually gonna be six. And that gets me down to six and four again, right? Not 64, but six and four for the comma. So once I get a turnaround fact in my pairs, I know I'm done. There are no more factors. So I call these turnaround facts. Four and six, and six and four are turnaround facts. And I know that I've completely factored my number 24. So if I listed them in numerical order, the factors of 24 are one, two, three, four, six, eight, 12, and 14. What we're gonna focus on today is using factors and multiples to solve our Math Kangaroo word problems. Okay? I see in the chat that this was already asked a little bit about prime factors. A prime number can only be divisible by itself and by one. And here are the first few examples of prime factors. Two, three, five, seven. You can't divide seven by anything except seven and one. Nothing else will go evenly without leaving a fraction. Same for 11, 13, 17, and 23. So a number that is not factorable, oh, pardon me. A number that is factorable and not prime is called composite. So composite and prime are kind of like opposites of each other. A number that cannot be factored and a number that can be factored. Composite, or some people have a different pronunciation of it, that's okay. So sometimes you'll be asked to find the prime factorization of a number. And one useful tool is a factor tree. So if I wanna factor the number 18, I say, oh, okay, 18 is divisible by nine and two, it has these factors nine and two. Two is a prime number, so I can't divide that by anything else. Nine is divisible by three times three and three is prime numbers. So the prime factorization of 18 is two times three times three, or two times three squared. If you're familiar with exponents, anytime you multiply a number by itself, you increase the power. So the two here means that there are actually two threes multiplied. So this would be a prime factorization. And when you're doing a prime factorization, it does not matter which way you factor it. So if I had instead started with 18 and decided, oh, you know what? 18 is divisible by three and six. I can further divide the six into two and three. Do you notice I get the same exact prime factorization? So when you have a prime factorization, it is unique for a single composite number. Does that make sense? It does not matter which order you use, you'll get the same prime factorization and it will represent the same composite number. All right, that's a lot of our background information. Now, some of these are little tricks. I've been using the word divisible a little bit already. So hopefully you have a hint about what divisible means. If a number is a factor, so we started with, for example, 24, right? And we had some factors of 24 that we listed out. Let's see if any of these rules will apply to 24. If a number is divisible by two, I think you all learned pretty young that that means it's even and that the even numbers are one, two, four, six, and eight. So 24, the right-hand digit, the one's place is a four. It is an even number and it is divisible by two. And if you remember, two was a factor. If it's divisible by three, the sum of the digits will be divisible by three. Well, what is two plus four? Two plus four does equal six and six is divisible by three. So therefore 24 is divisible by three. The last two digits are divisible by four. Is 24 divisible by four? 24 divisible by four? Of course it is, it equals six. So it's divisible by four. Is the last digit a zero or a five? In this case, no. So we are not divisible by five. Is it even? And the sum of the digits is divisible by three. Now that's kind of interesting. Why is that rule for sixes so long and complicated? It's basically a combination of the rules for two and three. When we talk about some other parts, common factors, least common multiples, different things, we'll discover that if something is divisible by, because six is two times three, if anything is divisible by six, it's also divisible by two and divisible by three. So we can combine our divisibility rules. And some Math Kangaroo questions will ask us to just do that, to combine our divisibility rules. So we'll get a good practice question with that. Divisible by seven is cross off the last digit, double it and subtract it from what remains. Now you and I know that 24 is not divisible by seven, but let's practice this because this is a strange rule. So 24, the last digit is a four. So we're gonna cross it off. We're gonna try to subtract eight, which is double it. We get negative six. Is negative six divisible by seven? No. And we already knew 24 was not divisible by seven. Eight, the last three digits are divisible by eight. Now, if you don't know three digit multiples of eight, that is okay. What you could do is you could divide by two and then check to see if it's divisible by four because two times four is eight. So you could actually separate those rules if you want. Divisible by nine is a handy rule. The sum of the digits just has to be divisible by nine, just like the sum of the digits is divisible by three for the threes rule. And I think all of you learned a long time ago that if the last digit is a zero, the number is divisible by 10. So those are the main rules that you need to know. There are others for other numbers. If you'd wanna look them up, you can look these up and you can find even more rules. Can I go over the rule for seven again? Well, let's look at a number that is divisible by seven. Let's try something like 777, right? We all know that's divisible by seven, but we'll just use that as an example. We're gonna cross off this last digit and double it. So seven doubled is 14. We're gonna subtract 14 from what remains and we're gonna see what we get. Seven minus four is three, seven minus is 63. Is 63 divisible by seven? It's seven times nine, right? But even if you didn't know that, you could cross off the six, double it, it's a six and you get zero. And zero is considered divisible by everything. So yes, this number would be divisible by seven. So there's an example. You should have this table in your handout. So Maya, if you can link the handout one more time, not everyone has the handout. This table is in that handout and it will be helpful for you as we solve the questions. If you don't have the handout, check the chat right now. The link is there. You should be able to open it. If you can't print it right now, print it after class. It's handy to have around. Okay, which of the following numbers is not a factor of 2004? You can go ahead and use those divisibility rules to figure out which of these is a factor of 2004. If you don't have divisibility rules, this is one of those unusual, Math Kangaroo gives you answer choices. You can test them. It's not the preferred method, but you'll notice that Maya has put up the poll and what you can see is the question is still in the poll. So even if it's popped up in front of your window, you can still see the question. So you're not missing anything. If you do not have the pop-up window for the poll, feel free to answer the questions in the chat. Because I know not every browser, not every download of Zoom, not every device works with the pop-up windows. So feel free to put the answers into the chat. That way Maya and I will know that you're following along and we'll be able to help you if you're not, if you have questions in the chat anytime you'd like. It's interesting that this time I am seeing the poll and usually I don't. How about we try those divisibility rules? If we sum up the digits of 2004, we get 6, right? And 6 is divisible by 3, so this one works. Is 2004 divisible by 4? We checked the last two digits. 04 is definitely divisible by 4. For 6, it had to be divisible by 3 and be even. 8 is divisible by 3 and it's certainly an even number, so it's divisible by 6. To be divisible by 8, the last three numbers need to be divisible by 8. Is 4 divisible by 8? No. To be divisible by 12, it has to be divisible by 3 and divisible by 4. We're combining the rules and we've already determined it was divisible by 3 and by 4. So we are divisible by 12. We'll move on to the next question. All right, how many two-digit numbers are divisible by both two and seven? Do you remember we were talking about if something is divisible by 12, it's divisible by three and by four. So how can we understand this question? How can we find numbers that are divisible by both two and seven? We can use the idea of a least common multiple. Two and seven are both prime numbers. They're not divisible by anything else. So the least common multiple is actually just going to be two times seven, which is 14. So we just need to look at multiples of 14. We want two-digit numbers. So 14 is a two-digit number. We want to stop when we get to, look at the question, we don't want any three-digit numbers. So if you get to three-digit numbers, you've gone too far. How many multiples are there? It's not asking, remember, it's not asking you what the multiples are, it's asking how many are there. This is part of make sure you answer the correct question. So they're not asking us to just make a list, which we could do, 14, 28, 42, right? This is what you should be doing on your scratch paper. Is there another way? So I can make this complete list. So this, that's seven numbers, right? That's seven. Is there another way to do this? Because I said there's frequently more than one way to solve a question. What might be another way? I want two digit numbers, right? So what if I take the first three digit number and I divide it by 14? That will tell me how many there are that are less than 100. We know the first one is a two digit number, so we're okay. So we can do 100 or we could even do 99 since that's the largest two digit number. So 99 divided by 14 is seven with a remainder of one. So this would be another way to find out if there were seven multiples of 14, which is two and seven together. Hopefully it makes sense and it was good to have an option of another way. Okay. We're gonna go on to the next question, Maya. Have you shared the results of this one? Okay, how many twos and fives are there among the prime factors of the number 2,000? Remember prime factors means you can't divide it any farther. You've gone as far as you can. And remember that every composite number has its unique prime factorization. And it does not matter which way you factor it as long as you go to completion till you get to prime numbers. These, I have a question. Do you see these little numbers here, MK 2,000 number seven? That means that this was an actual math kangaroo contest question in the year 2000. It was a number seven question. There are 30 questions on the math kangaroo test. So if it's in the first 10, they're usually an easier level question. The teen numbers are usually the medium questions. And if it's in the 20s, it will be one of the more difficult math kangaroo questions. So hopefully that answers the questions. Could this be on a kangaroo test? It was on a math kangaroo test. So remember we can make our little factor trees. You can do 2,000 and you could maybe divide it by 10. 10 can be divisible, right? Two and five, those are prime numbers. What can I get here? How about 20 and 10? You do not have to do it in this order. It does not matter. As long as you get to twos and fives, you're answering the question. I'm gonna move to the next slide. I factored it in a slightly different way on the next slide. So this gives you an alternative. I just started factoring out twos in this slide. I'm gonna come up with the same answer. So the question is asking how many twos and how many fives? And you'll notice the poll question is a fill in the blank question. So you're gonna say so many twos and so many fives. In this case, I see one, two, three, four twos. So the answer here would be four twos and three fives. But if I was writing out the prime factorization of the number 2,000, I could write it as two times two times two times two Or if you're familiar with exponents, I have used the number, I've multiplied the number two four times, so it's two to the fourth. And I have the number five three times, so it's five to the third. If you're familiar with exponential notation, that's fantastic. If not, you'll probably learn it next year. All right, we'll move on, hopefully. This is a combination problem. So Math Kangaroo sometimes likes to combine things. So far, we've done questions that were quite simple in terms of being multiples or divisible factor questions. This one combines a little bit of logical thinking. And I think this is a fun type of question to understand those ideas of factors, divisibilities, and primes, but to have to use your little bit of logical thinking to come up with the correct answer. Let's read it together. Thomas, Roman, Andrew, and Michael said the following about a certain number. So we're thinking about a certain number. Thomas said, this number is equal to nine. Roman said, this number is prime. Andrew said, this number is even. Michael said, this number is equal to 15. Only one of the statements given by either Roman or Thomas is true, and only one of the statements given by either Michael or Andrew is true. What number are they talking about? So remember at the beginning of the class, I said that usually the first sentence or the last sentence tells you what we need. Here, we need the certain number they're talking about. Now we have to figure out what is all of this stuff in the middle? What is it actually asking us? So I'm gonna do a little bit of highlighting for you. This sentence here says only one of the statements by Roman or Thomas is true. So if we combine, this is Thomas's statement, and this is Roman's statement. One of those is true. Then we have the same sort of idea where it says that only one of the statements by Andrew or Michael is true. So these two, only one of those is true at a time. So this is a case where we do have the multiple choice answers. You can test each answer and see if it fits the condition, or you can read through the clues that they're giving you. If the number is nine, like Thomas said, is the number prime? No, it's not, but that's okay because only one statement needs to be true. Andrew said this number is even, so that would be false if the number is nine. And the number is 15 is also false. And remember, we can't have both of them false. So we know that the answer is not the nine. What about if the number is 15? Then this one is, is the number equal to nine? No, that's false. Is the number prime? No, that's false. We can't have both yellow statements be false, so it is not 15. Remember, you can always send your responses in the chat if you don't get into the poll in time or if you can't see the poll. I'll let you think about the other three choices for just a moment more. One way to think about this question rather than testing answers, can it be equal to 9 and prime at the same time? No, it cannot be equal to 9 and prime. Can it be 15 and equal to 9 and prime? Definitely not. Could it be even and prime? Is there a number that's even and prime? Because that way, one of each statement set is true. What number is even and prime? Even and prime should give you choice B. John comes to the computer lab every day, Carl every two days, Stan every three days, and Adam every four days, Paul every five days, Peter every six days. Today, they're all at the computer lab. In how many days will they all be there together again? This is the type of question that's called a least common multiple. We wanna know what the least common multiple of every day, two days, three days, four days, five days, and six days might be so that they can be there again on the same day. So you could list out the multiples. That would be one way to do it. Could list out two, four, six, eight, 10. That's the multiples of two, multiples of three, three, six, nine, multiples of four, multiples of five, multiples of six. Might take you a little while, but you could certainly find the least common multiple by making a list of these numbers. There is another way. We can look and see which things are common to each of these. So six, I wanna switch colors. Six is divisible by two and three, right? So if something is divisible by six, we don't have to worry about two and three. We know it's already divisible by two and three. Four days is two times two. So one of the twos is in the six, but the other is not. So we need to keep that two. And then we still have this five days. The one we don't have to worry about for every day. So if we take six and we multiply it by five for Paul, and we still have a two left over here, that is gonna be the least common multiple. So how many days will they be together again? Given that would calculate your least common multiple. Now we're starting to get in into the chat, very good. So every 60 days, everyone will be in the computer lab together. Now there are multiples that are common. So example, on the 12th day, you would have two of the students there. On the eighth day, you would have another two students there, but we wanted all five students in the lab at the same time. Well, actually it's six because there's another one here. We want all six students in the lab at the same time. So every 60 days, they will all be together. Okay, this is an interesting question. The number of all the divisors of a number, two times three times five times seven is equal to? Now this is a prime factorization, isn't it? Right, these are all prime numbers. So we could multiply these out and get a nice big number. And then we could find all of the divisors of it, all of its factors. That's a perfectly acceptable method to do this. There are a few other methods as well. So if I took all of this and multiplied it together, I'm gonna get 210. So you could start factoring 210. You could do two and 105. Is it divisible by three? It is divisible by three. You could do three and 70. Is it divisible by four? No, it's not. Is it divisible by five? It is, so you could divide it by five and you would get 42. And you could keep going until you have all of the divisors or factors. Divisors is another way of saying factors. Are there other methods? Is this method great? It is, but I'd like to show you at least two other methods today. One method is you have these factors, two, three, five, and seven, and they can either be included in the divisor or not included in the divisor. For example, six is a divisor, right? So this is included, this is included, this is not, and this is not. 10 is a divisor, so this is included, this is not, this is, and this is not. 14 is a divisor. Are you understanding that I could be? So I can use different combinations of factors and they can be included or not. And if you've done a little bit of combinatorics, which is another field, so this is another way that Math Kangaroo is incorporating multiple types of fields, multiple types of mathematics into a question. This has two possibilities, included or excluded. So that's two possibilities. The three is included or excluded. The same with the five and the seven. So if I wanna know how many possibilities there are, I just multiply these all together and I get that there are 16 possibilities based on if it's included or excluded. There is another way as well, which I've done on this table. You can make a table. If only one of the factors is in the divisor, you have one, two, three, five, and seven as possibilities. You can include any two of the factors to the divisor. So this would give you like six. This gives you 10. This gives you 14. So these are all the combinations of two of these. So starting with the two, you can have two and three, two and five, two and seven. Starting with the three, you can have three and five, three and seven, and finally five and seven. How many ways can you incorporate three factors? Two, three, five. Two, three, seven. Two, five, seven. Three, five, seven. Those are the way you can incorporate three factors. And there's only one way to incorporate all four factors. And if you add these up, you'll find out that there are also 16 different combinations that you can use using a table like this. So I've shown you at least three different methods that all come out to being that you can come up with 16 possible factors of this number. Remember, this is a number. It's just broken into its factors. The question does not say which method you have to use to solve it. So that's been an interesting part. We want to know how many divisors. Remember, divisors is another word for factors. Right, so how many factors are there of this number? And this is unique. It's a prime factorization, so it is unique. We calculated it a moment ago. It was 210. So this question is basically could be reworded to say the number, how many? So the number of could mean how many? How many factors of the number 210 are there? And there are multiple ways to solve it. The interesting thing is because it gave us the prime factorization, we don't necessarily have to multiply it all together. We could make a table or we could have done the combinatorics where this could be in or out, included, excluded, and you'd get two times two times two times two is 16. I hope that explaining it in multiple ways didn't confuse you, but instead gave you options. My goal was to give you options, not to make it more confusing. I hope that worked. Here's another question. Together, Ella and Ola have 70 mushrooms. 5 9ths of Ella's mushrooms are brown and 2 17ths of Ola's mushrooms are white. How many mushrooms does Ella have? So as is typical, the question, the answer is, excuse me, the question it's asking you is usually at the beginning or the end. We already know they have 70 mushrooms altogether. We want to know how many of them does Ella have? So we can say that Ella's mushrooms plus Ola's mushrooms is going to equal 70. And then they give you a little bit of information. 5 9ths of Ella's mushrooms are brown and 2 17ths of Ola's mushrooms are white. What information does that tell you about the mushrooms that Ella and Ola have? Does it matter if we're using brown or white mushrooms? Does the question ask you about brown mushrooms or white mushrooms? So this question is giving you a little bit of extraneous information, confusing you with colors. You'll notice it's a number 22 question. So that's one of the more difficult questions. We don't have to worry about the color, but what is interesting is it's telling you that Ella's mushrooms, the number she has is divisible by nine. And the number of mushrooms that Ola has is divisible by 17. So we need to look at the multiples of nine and 17 and somehow sum those up to be 70. Does that help? Give you a few moments to think about it. Love to see some answers in the chat as well if you can't use the poll. This little star just means it's a multiple of. We don't actually need to find that number, but just to be able to write it out a little bit more. Ella is a multiple of nine, and Ola is a multiple of 17. So when you get your final answer, and it's Ella's, it needs to be a multiple of nine. So that's something you can use to double check yourself. If you're not getting a multiple of nine, then you have made a mistake someplace. Remember, the fourth step is always to look back and check your work. Let me go to the next slide, because I've organized my work in a table. It's a good way for you to do these kinds of questions. Ella's mushrooms are multiples of nine, so I've written some multiples of nine. And Ola's mushrooms are multiples of seven, so I've written some multiples of seven. And then this table just shows the sum. So with different combinations, I've been able to make this sum. And you'll notice that if Ola has 34 mushrooms, and Ella has 36, then the sum is 70, which satisfies all the conditions of the question. But what does the question ask? Always very careful to answer. It's asking how many does Ella have? So our answer is going to be the 36. Again, you do not have to make a table. You could list it out in some other way. But a table is a great way of organizing your options so that you don't lose your place. Because if you remember, once you have a plan, which my plan was to come up with the multiples of nine and 17, and then see if they add to 70, then my next step is to carry out my plan very carefully. And the table helps me be careful. Somebody's asking if they could have a fractional number of mushrooms. There's no indication in this question that there's a fractional number of mushrooms, so I would assume a whole number of mushrooms. Okay, how many five-digit numbers in the form 1, blank, 8, 2, blank are there that are divisible by 12 and all of whose digits are different? So we need to know, again, remember they're asking us some tricky questions. They're asking how many. They're not asking you for an example. They're asking you how many examples are there. So you're going to have to find all of the examples and figure out how many that is. How do we find a number that is divisible by 12? We talked about that a little earlier. If you don't know the rules for divisible by 12 on five-digit numbers, how about the rules for divisible by 3 and divisible by 4? If it's divisible by both 3 and 4, it will be divisible by 12. That's one of the magics about factors and being able to combine them. If you're still struggling, I might suggest that you try divisible by 4 first, because divisible by 3 is the sum of the digits, but divisible by 4 was that the last two digits are divisible by 4. So it might be easier to determine the last digit first and then determine the second digit by what's divisible by 3. What numbers can you put in that last space? 20 is divisible by 4, so that would be an option. What would be another option? 24 is divisible by 4. 28 is divisible by 8, but there is another little piece of this question we always have to read carefully. It says all of the digits are different, so because the digits are, we have a repeated 8 here and here, we're not going to use this choice. So we're just going to start with these two, and now we have to make it divisible by 3. So what numbers can you put into the blanks so that it's, they're divisible by 3? And then how many choices are there altogether? So you can look at the sum of the digits that you have so far. 1 plus 8 is 9, 2 plus 4 is 6. So this is already divisible by 3, so you just want numbers here that are also divisible by 3. So we can get 1, 0, 8, 2, 4, because 9 plus 6 is 15, we're okay. 1, 3, 8, 2, 4, that's going to be 4, 12, 14, and 18 is divisible by 3. Okay, can you fill in the second? Can you figure out which could go in the blanks here? We have 1 plus 8 is 9 plus 2 is 11. So we need to be able to make, put in some digit here so that the sum is divisible by 3. 11 is not, 12 would be, but if we put in a 1, we'll be repeating a number. How about if we choose the 4? And then we can go up another 3. So in the end, how many did I get? Hopefully you've gotten there too. I have six numbers that are working. This question is a little bit tricky. You'll notice it's a very high number on Math Kangaroo, but now the good news is you know how to solve this one now. So if there was a similar question, you'd be all set. It's not about being able to answer it before you started the class, it's about being able to answer it once you've completed the class. So if you feel like now you understand what I did and you could do it, then you're right on track. That's why you came to class today. Okay, here's our next question. Imagine that you have 108 red marbles and 180 green marbles. The marbles have to be packed in boxes in such a way that every box contains the same number of marbles and there are marbles of only one color in every box. What is the smallest number of boxes that you need? So let's do a little bit of understanding the question. Okay, so they give you two numbers and then you want to pack these marbles in boxes with the same number and one color and then you have the smallest number of boxes. How do you get the smallest number of boxes? You have to put the greatest number of marbles in each box. Does that make sense? If you want to use fewer boxes, put more marbles per box. So this gives us a greatest common factor problem. Greatest meaning the largest, right? Because we want to put in as many marbles as possible. Okay, so we have to look at the numbers 108 and 180 and try to see what is the greatest common factor. So give me a moment for that. Use those divisibility rules or maybe start dividing these numbers. Try to see if there are common factors between the two of them. I noticed right away that they have the same numbers in a different order. And if you look at them, they're both divisible by nine, but I don't think that's the greatest common factor because they're also both even. So nine times two is greater than just nine, and there's even a little bit more that you could do. Let me give you a little time to experiment with the numbers. But did you see, I was just using the divisibility rules, I found nine and two by using those divisibility rules pretty quickly. So 18 is a good starting place. Seeing a little bit in the chat, somebody is saying, what about divisible by four? Remember the rule for divisible by four was the last two digits had to be divisible by four. And 08 is divisible by four and 80 is divisible by four, so we actually have a number that's divisible by nine and by four. Are there any others that are larger? What's nine times four? Don't forget to answer the question that's being asked. I'm getting you partway there, but not all the way there. The question is being asked. If I take 108 divided by the 36, because I determined that 9 and 4 are factors of both of them, they're common factors, 108 divided by 36 is 3, that's prime. So I know there is no larger factor. 108 divided by 36 is 5. So again, 5 is prime, and there is no larger factor. But be very careful to read the question. It's not asking you how many times is it divisible, is it? It's asking you, how many boxes do you need? So hopefully, we'll be getting to the boxes, you're going to have 3 of red marbles, you're going to have 5 of green marbles, and that's going to be 8 boxes total. So there were a couple of steps on this question. You had to follow all the way through in order to find out that you're going to have 8 boxes total. However, it was important that you had the same color in every box. So 3 boxes are red, 5 boxes are green marbles, and you have 8 all together. How do we do? We have only one more question, which is just about the right amount of time for the webinar today. But this is the extra challenge question. All the whole numbers from 1 to 2006 were written on a blackboard. John underlined all the numbers divisible by 2, Adam underlined all the numbers divisible by 3, and Peter underlined all the numbers divisible by 4. How many numbers were underlined exactly twice? So we have a lot of information here. We have 1 to 2006, we have divisible by 2, divisible by 3, divisible by 4, and we want underlined exactly twice. Remember, this is an advanced question. The questions that we did before, remember our computer lab problem when we were trying to figure out how many days would all 6 students be in the computer lab at the same time, and we needed to find the least common multiple? This question is different than that. Because in the least common multiple, you would have 3 underlines. But we only want 2 underlines. But we are not going to start writing out numbers all the way up to 2006, are we? I don't think so. So we should look for a pattern. The least common multiple of 2, 3, and 4 is 12. So if there's a pattern, it's going to happen every 12. So can you figure out what happens between 1 and 12, 13 and 24, 25 and 36, and then repeat that pattern without having to draw out the entire thing? So math kangaroo problems like patterns as well. So again, you can see they're combining the idea of multiples and finding a pattern. It's a good problem for keeping you thinking. I'm not getting too many answers right away so I am going to give you a little more help. If I write out the numbers one through ten and I underline in different colors every two and then I do every three and then I do every four. How many of those are underlined twice? So now we're starting to get a pattern, right? There are three underlined twice for every 12. But how many times is this pattern going to be repeated? I'm going to move to my next slide where I've done this a little bit more. We know that there are three for every 12 and we have to figure out how many 12s there are in 2006. Turns out that there's 167 with the remainder of two. So if I take that 167 and I multiply it by the three that I know are in every 12, I'll get 501. Then we have to ask ourselves about the remainder two. If you remember the pattern from what I did before, are we going to underline the first two numbers twice? So within the first two numbers, nothing is underlined twice, right? So we don't have to worry about this remainder two. We don't have to worry about it. The answer is simply going to be the pattern is repeated 167 times. It's repeated every 12 and there are three that are underlined twice. So the answer is C, 501. Hopefully now that you've seen it, you can go ahead and put some more of the questions that you have in the chat. I know this was a complicated question and Maya has been doing an amazing job of answering your questions in the chat. So big applause to Maya. She's been on top of all of your questions. I hope she's been very helpful. The combination of getting some of that individual attention from Maya and the instruction from me helps. Question about why it's 501, not 167. In each 12, there were three times that I underlined twice, right? So the four is underlined twice, the six is underlined twice, and the eight is underlined twice. So the pattern repeats 167 times, but the pattern is three of the selected numbers meet our condition. So using the least common multiple, we knew that the pattern was repeated every 12. We knew that dividing 2006 by 12 was 167. And we knew that within those 12, there would be three numbers double underlined. So we had to take 167 times three. It was a very tricky question. That's why I chose it. I didn't want to choose all easy questions. I wanted you to give a real taste for what math kangaroo problems would be like. I hope you appreciate that I gave you a combination of different levels. I don't want you to be surprised when you take the contest. The contest is designed to give you a really good test. Okay, so I just want to wrap up our four step problem solving method. Understand what the problem asks. Look for those first sentence, the last sentence, because they'll really tell you what answer do you need to provide. Look at the information that's in the question. Do you have all the information you need? Make your plan. Is there only one plan to solve a question? No, you saw on some questions I solved it three different ways. The good news was I got the same answer three times. So that answer is correct, right? So don't worry if your plan is a little different than mine. Today our theme was multiples, divisibility, and factors. So I tried to use that type of a plan for all of the questions, but that doesn't mean it was the only one. Carry out the plan very carefully. If you make a mistake in multiplying or dividing, you'll have the wrong answer. So check your answer. The math kangaroo contests are multiple choice. So if your answer is not one of the choices, that's a really good clue that you need to go back. Okay, again, as you could tell, I had a big variety of questions to choose from. I would say that on each math kangaroo contest, there were at least two or three questions that were factors and multiples. So it's a really good topic to get a good control over so that you'll get at least a few easy questions on your math kangaroo contest. I'd like to thank Maya. She's been doing a fantastic job. She's been helping me all day. And I want to ask you, there is a survey that will pop up after this webinar closes. Please, yourself, your parents, please complete the survey. It helps give me feedback so I know how well I did and what I can do better and how I can make these webinars the most helpful for the students. So thank you for your attention. I hope you enjoyed the class. I hope you had some new challenges. And remember, it is supposed to be challenging. So thank you, Maya. And thank you, everybody.

Math Kangaroo

factors

multiples

divisibility

problem-solving

least common multiple

prime factorization

divisors

interactive quizzes

greatest common divisor