Welcome to Math Kangaroo webinar for level 5-6. This week our topic is going to be on multiples, factors, and divisibility. Probably a lot of you know what that means, but we'll go over it so you have all the vocabulary words. Don't worry about it if you don't know some of those things. You will in a few minutes, okay? There is a warm-up problem. I have polls for a lot of the sample problems today. So the warm-up is one of the ones where I have a poll, so I'll read it. And I'll go ahead and launch the poll pretty quickly because the whole question appears on the poll and it's a pretty simple question. Which of the following numbers is not a factor of 2004? So pretty important, not a factor. There's your poll because the whole question is right there at the top of the poll. So hopefully you can see that There are any students who do have trouble seeing the polls let me know in the chat That way I can make sure I give you enough time to take any notes about the problem Before I launch the chat the poll and maybe interfere with your views, okay Anybody else wants to put their answer into the poll? Now would be the time. Remember, poll answers are anonymous, so we cannot tell who's gotten it correct or incorrect. We only just get the total. As you can see, there's a couple of people saying 3 or 12, but most of you, over half, have said that 8 is not a factor of 2004. That is correct. Let's take a look at that. Some of you are new for our webinars, so welcome. I know that there were some students who registered even this week, so I hope you enjoy this experience. Takes us a little bit to get into the rhythm, and then we'll see how it works. Okay. So, a factor of a number. So, factor means that we're multiplying. A factor times a factor, any number of factors, equals a product, right? So when we're talking about factors, we're talking about things that you multiply. You also can divide a product by its factors, and it will divide evenly with no remainders. We're gonna talk about divisibility rules, but maybe some of you already know it. To be divisible by 3, the sum of your digits must be divisible by 3. So 2 plus 0 plus 0 plus 4 equals 6, and 6 is divisible by 3, so this one is okay, and we wanna know which one is not okay. The rule for divisible by 4 is to look at the last two digits, and if the last two digits are divisible by 4, then so is the rest of the number, so that is okay. The rule for divisible by 6 is that you would have to be divisible by 3, and divisible by 2, or even. And we can see that this is divisible by 3, and that it's even, so it is divisible by 6. Now, the neat thing about divisibility and about factors is, remember, 12 is just 3 times 4, right? So if you're divisible by 12, you're divisible by 3 and divisible by 4, and we've already found that in A and B. What we didn't find is divisible by 8. The rule for divisible by 8 is a little trickier. It's either look at the last three digits and see if they're divisible by 8, or it's divide by 2, and then do the rule for divisible by 4, because, of course, 2 times 4 equals 8, so we can separate it. So if I take 2004 and I divide by 2, I get 1002, and 1002 is not divisible by 4, so that would be the answer for what we cannot divide it by. Is there another way? Of course. This number's not so big. It's not so terrible. Just divide, right? It could do it, 2004. Divided by 8. 8 goes into 22 times. I get 16. I get 4 when I subtract, bring down the zero. 8 goes into 45 times with nothing left over, so then I just have a 4. 8 does not go into 4, so it's not divisible with no remainder. Okay, so let's take a look at this. Multiples of a number are the product when you multiply that number by an integer. Integer. You can raise your hand in the chat if you don't know what integer means. Integer is going to mean numbers with no fractional or decimal parts. Integers can be positive or negative. They can include zero, but they can't have any decimal or fractional parts, so 1 1⁄2 or 2.722 are not integers. If I take multiples of three, three times zero, three times one, three times two, three times three, those are multiples. Okay. Give me just a moment. Those are multiples. All right, so now we're going to have to look at, skip counting is a way of doing multiples, right? One of the easiest skip counting methods is going to be 2, 4, 6, 8, 5, 10, 15, 20. When you do skip counting, that is a way to get your multiples. And then common multiples are multiples that are shared by two or more numbers. So we wrote down the multiples of three and we wrote down the multiples of five, but which ones are the same for three and five? Which ones are in common? Well, zero, right? Because I can multiply by zero. 15 is on both lists. 30 would end up being on both lists. And you can see that these are all common multiples. And these are also the things that are multiples of 15 or three times five. So you can kind of layer them together that way. Okay, so the factor is any number that divides a product evenly, leaving no remainder. We talked about that with the warmup problems. Sometimes you'll list them as pairs. 24, my factor pairs are going to be one and 24, because one times 24 is 24. Two times 12 is a pair because two times 12 is 24. Three times eight or four times six. Once you list four times... So one of the ways to list factors is I just kept increasing this. You notice I went one, two, three, four. The next one would be five. 24 is not divisible by five. And then my next number would be six. And I already have six right here. So then I start doing the turnaround facts, the commutative facts. So I don't have to list those again. So that tells me that my list is complete and I can list them in increasing order if I like. That's another way to do it. Okay, so just alternate ways of listing factors. Okay, a number is prime. This is a new idea, maybe not for some of you, but prime number, if it's divisible only by itself and by one. The first few prime numbers are two, three, five, seven. Do you notice there's only one even number in this whole list? Two is the only even prime number right there. Because it is only divisible by two or by one, by itself and by one. One is not prime because it's divisible by one, right? So any other even number would be divisible by two. So all the primes except for two are odd numbers. You'll notice five works, but multiples of five do not. Like 15 is not in this list because it would be divisible by five. A number that is factorable is called composite. So the prime numbers, their opposites are composites. That means you can divide them. And you'll notice that we have this little trick that we can use called a factorization tree. So this is a method that is frequently used by people at your level or even a little older. So 18 can be changed to two times nine and nine can be made as three times three. Could I do this factor tree in another way? Could I do 18 is three times six? Yes, and then six breaks down to two times three. But you'll notice I got the same prime numbers from both types of trees, two, three, and three, three, two, and three, the same prime factorization no matter which way you do the original or any subsequent divisions, okay? So that's a fun thing about factor trees. If you're going out to prime numbers, you'll always get to the same result. The prime factorization for any composite number is unique. And then these are the little divisibility rules or shortcuts. Jacob, do you ever use these on your math contests or in math? I bet you've used these before. Yeah, many times. So which ones do you find the most useful? Two, three, four, five, and nine, 10, and eight. Okay, yeah, so for three and nine, we have really similar rules where you just add up the digits and you have to see if that sum is then divisible by three or nine. For divisible by four, you check the last two digits. For divisible by eight, like I said, you can check the last three digits. So if I have a really long number, I could check these last three digits and determine if those are divisible by eight. And they are because 840 is divisible by eight. I would get 105, right? So therefore, the whole number is divisible by eight. Then what I was saying earlier is double check. For some of these, you can combine the rules. So since six is two times three, you can use the rule for two and the rule for three. And if it passes both, then the number must also be divisible by six. I can do the same for two and seven to see if something was divisible by 14, or two, like three and five to see if something's divisible by 15. So don't be afraid to combine rules. You can view this slideshow. You can view this recording. And if you don't know these rules, you can jot them down at that time if you didn't get them all jotted down right now, didn't get all the notes, okay? All right, so I have polls for most of the questions because I know some students were asking for more polls last time. How many twos and fives are among the prime factors of the number 2,000? So remember, you have to completely factor it. I showed you using a factor tree. I might recommend that for this problem. I always give you a little bit of time to solve it. You can put the answers in the chat for me or for Jacob. If you have questions, you can also ask. If you need hints or some advice, Jacob is great at that. And in a few moments, I'll launch the poll. Um, excuse me, um, I think you're muted. Okay, so as you can see, I've solved it all out and yes, I was muted. Thank you for letting me know. So, you can see that we divided by 2. I was able to divide by 2. 4 times, and then I was able to divide by 5. 3 times. So, at first, when I got the 4 twos, I was like, okay, there's 2 choices that would match that, but then I only got 3 fives. So, I know the answer must be D. The other way to do that, I'm going to ask all the students to stay muted. The other way to do it is to check the answer. What if I do 4 times, 4 times, 4 times, 4 times, 5 times, 5 times, 5. Well, I like to do things out of order and make life simpler for myself, right? So, this is 20, this is 20, this is 20, and there's 4. So, 20 times 20 is 400, times another 20, it's going to be 8. This is, yeah. So, 4 times 5 is 20. 20 times 20 gives me 400. Then I'm going to multiply by another 20. How do I get 8,000, Jacob? How did that happen? Oh, 4 twos. Pardon me. These are twos. Oh, sorry. I teach a lesson before this, and so sometimes I'm still, like, doing some of those other lesson silliness going on. All right, so I'm going to get 10, 10, 10, and I have another 2. So, now I have 10 times 10 times 10 times 2 equals 2,000. There we go. Sorry about that, guys. It happens to me sometimes. You ever have one of those days where you feel like nothing you say is going to be correct? How many two-digit numbers are divisible by both 2 and 7? There are a couple of ideas to keep in mind. Read carefully. The problem says two-digit numbers. Here is your pole. The whole problem should be visible at the top of the pole too. It's like most students have completed the poll. You guys work quickly, it's nice. And we have half of you saying seven and another third of you saying five with every answer being guessed by at least one student. So let's have a look. Sorry, there's a couple of clicks I have to do in order to get from one thing to another. So two-digit numbers, those are the numbers between 10 and 99, right? Nothing smaller than 10 or larger than 99 is still a two-digit number, okay? And if a number is divisible by both two and seven, then since those are both prime numbers, we have to just take the multiple of those. So if a number is divisible by 14, it's divisible by two and by seven. Now, this would be a little bit different if these were not both prime numbers, but because they're prime, we just multiply them together. They have nothing in common basically, right? If I said, oh, they're divisible by three and 12. Well, every number divisible by 12 is divisible by three, right? So we wouldn't necessarily have to multiply them together this way. Okay, so if I take, now it turns out that the first number in the sequence is gonna be 14. So I don't have to worry about any single-digit numbers because it's already gotta be 14 or more. Then if I take my 99 and I divide it by 14, I'll find out, okay, let's see. I use a couple of approximations in my head and I'm finding out it's seven because seven times 10 is 70 and seven times four is 28, right? So I get an eight, 98, and I get just a remainder of one. So my last multiple of 14 is 98. And how many multiples are there? There are seven of them. So the answer here is B. Okay, there is no poll for this one because I wanted you to be able to see the whole problem statement for as long as you need to. Thomas, Roman, Andrew, and Michael said the following 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 Andrew or Michael is true. What number are they talking about? Kind of an interesting problem because it is combining our divisibility and our factors with logical thinking. I'll give you some time and you can ask questions or put your responses in the chat. Jacob might be able to help you as well. OK, this one's kind of tricky. Good, I'm glad. We're supposed to be challenging our students. Some of you have it, and that's wonderful, too. OK, so first thing I did is I said, there's a lot of statements here. And it says that only one of the statements by either Roman or Thomas is true. So I've got this happens when I have to, OK, so Roman or Thomas. So what does Roman say? Roman says that the number is prime. And Thomas says that the number is not. Those can't both be true at the same time, correct? Of course, because 9 is not a prime number. Then I have to look at the statements by Andrew and Michael. Andrew said that the number is even. And Michael says that the number is 15. Those obviously contradict each other as well. We're told by one statement in each group is correct. So let's see. If Thomas is correct and the number is 9, can it be an even number? No. Can the number be 15? No. So this number we know is wrong. This is wrong or it's a lie, right? So now we know that the number is prime. If the number is prime, can it be 15? No, that must be wrong. But can the number be even and be prime? I told you at the beginning there was only one even prime number, and that is 2. So by having it being even and prime, we limit it to being only one number in the whole universe that we know about, and it's 2. John comes to the computer lab every day, Carl every two days, Stan every three days, Adam every four days, Paul every five days, and Peter every six days. Today, they are all at the computer lab. In how many days will they all be there together again? I do have a poll because you can just kind of remember every day, every two, every three, every four, every five, and every six. So those are the numbers we're dealing with. 1, 2, 3, 4, 5, and 6 days. All right. Really good job on this problem. Most of you have said that the answer is 60 days. So let's take a look. Makes some sense, right? Because we have to look at what could be common multiples here. So if we're looking for the common multiples, we'd have to multiply all of these together, right? Because two and three together will give us six. We don't really have to worry about the person who's there every day, because no matter which day, that person will be there. All right. Five and six don't have any prime factors in common. So we do have to multiply five times six. And then if we look at four, it's two times two, but six was two times three. So we already have one of the twos. So we just need to multiply by one two. So five times six times two does equal 60. So every 60 days, all of the students will be in the computer lab at the same time. So it's a really kind of an important difference that we don't have to multiply them all up. We just have to look at what the common factors are and make sure we have each common factor in there together. What is the number of all the divisors of the number two times three times five times seven? It's kind of a, I don't even want to explain it because I want to see if you can figure out what that means on your own contest. So I think this question is interesting enough that I would love to see what the poll results look like without any further coaching. And then we'll get the explanation after we have responses. Okay, not everyone has participated in the poll. Anybody else want to try to guess an answer? All right, I'll share the results. There we go. We have quite a few different answers, numbers ranging from four to 210. It's a big difference, right? So let's take a look and see what is this question possibly really asking? What is the number of all of the divisors of the number two times three times five times seven? Well, the divisors are the same as the factors, right? So we need to find the factors. And they're telling, this is prime factorization, right? So they've already given us that. Now, there's two main ways to do this, well, three, I guess. You could multiply those numbers together, and then you could look for all the factors of that product. So some of you might do that. Another way to do it is you might say, well, everything is divisible by one. And then we know that this number is divisible by two, three, five, or seven. This number must also be divisible by two times three, two times five, two times seven, three times five, three times seven, and five times seven. So you notice I grouped them into where I have two of the factors together. It could also be divided where I have combinations of three factors, two times three times five, two times three times seven, three times five times seven. And I guess two times five times seven is the other one I missed. Or then you could do all of them together because every number is considered divisible by itself. And if I add all, if I just count these up, that's one, this is five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16 ways, right? So that's 16 ways. And I use the lists and the tables, which is methods that Math Kangaroo teaches at different levels, right? So maybe we've had that before, sorry. Okay, so let's try it another way. How many of you have learned about combinatoric counting? So I have four factors. I have the factor two, three, five, and seven. Whatever number I'm using as the divider can include, include, or exclude that factor. Kind of like a binary, in or out, right? It can be in or out. So if two can be in or out, that's two possibilities. Three can be in or out, that's two possibilities. Five can be in or out, that's two possibilities. And seven can be in or out, that's two possibilities. Multiplying all the different ways that those could combine combinatorics, I also get 16 different ways to combine them by leaving any factor in or out. Jacob, does that make sense if I explain that in a way that is cohesive? Yeah, that was a good explanation. Okay. All right, how many five-digit numbers in the form one blank, eight, two blank are there that are divisible by 12 and all of whose digits are different? So remember last time I made those lists? This might be another one where you kind of wanna make some sort of lists or tables or things like that and try out the different possibilities. Does anybody need a hint? Remember, you can combine divisibility rules. We need it divisible by 12, so which two rules could you combine? All right, I'm going to start working on this one just so that we can get through the problem. So I have 1, 8, 2, blank. All right, so to be divisible by 12, you have to be divisible by 3 and divisible by 4. To be divisible by 4, we want to look at the last two digits. So the last two digits must be divisible by 4, which gives me the options of having 20, 24, or 28. But it also says all of whose digits are different. So the 28 is not going to work because I already have an 8 in this number. So now I need to look at 1, 0, 8, 2, 0, or 1, 0, 8, 2, 4. And those would be the first possibilities. What number could I fill in those blanks so that now it is divisible by 3? Well, the sum of the digits must be divisible by 3 for the whole number to be divisible by 3. So so far, I have the sum of these digits. I'm just going to write it down here. The sum of the digits is 11. So from 11, I could add a 1 to get 12. I could add a 4, or I could add a 7, right? But if I add the 1, that's going to be a repeat of the first digit. So this could be a 4 or a 7. It gives me two possibilities. What about this next one? So far, the sum is 15, right? 8. So I can add a 0, a 3, a 6, or a 9 into that space and still have a sum that's divisible by 3. So if I look at all those possibilities, I have two possibilities here. None of these are repeated numbers, so I have all four possibilities here for a grand total of six possible numbers that meet all of the conditions that were stated in the problem statement. All right. Jacob, you wanted to lead this one. Yeah. So imagine 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? I'll give you guys some time to think about this. Okay, so now I guess we're going to explain it. So let's say, so I'm going to set a variable, and the variable I'm going to set is x, which is the, I'm going to set that to be the number of marbles per box. And so that means that the number of red boxes would be 1 over x, red boxes. And then there would be 180 over x, green boxes, since each box has x marbles. And notice that x has to divide both 108 and 180. So what I'm going to do is, so x has to divide 108 and 180. That means that x has to divide the greatest common divisor of 108 and 180. And so 108 can be written as 2 squared times 3 cubed. And 180 can be written as 2 squared times 3 squared times 5. So the greatest common divisor of 108 and 180 is going to be 2 squared times 3 squared. So this is the greatest common divisor of 108 and 180. So this is 36. And so x has to divide 36. But remember that we want to find the smallest number of boxes. And in order to find the smallest number of boxes, that means that we want to maximize the amount of marbles in each of the box. So if x divides 36, in order to maximize x, we want x to be 36. So the number of boxes would be the number of red boxes, which is 108 over 36 plus 180 over 36, which is the number of green boxes. And this comes out to 3 plus 5, which is 8. So the answer should be D. We had several students putting 36. And there is a 36 while you're working out this problem. So you might be on the correct path, just not all the way there. Okay, so here is the last problem and there is a poll for this one. Go ahead, Jacob. So 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? I'll give you guys some time to think about this. Okay, so I think we're running out of time, so I'm just going to explain it. Notice that the least common multiple of these numbers is 12, and that's pretty important. What that means is that for any 12 numbers, so let's say 1 to 12, the following 12 numbers, 13 to 24, will follow the same pattern of underlining as the numbers from 1 to 12. So what we really only care about is the numbers from 1 to 12 that are underlined. So John will underline the numbers 2, 4, 6, 8, 10, and 12. Peter will underline the numbers 3, 6, and 9. And then Peter will underline 4, 8, and 12. So from here, we can notice that the numbers 4, 6, 8, and 12 are all underlined. Sorry, Adam would underline 12 as well. So it's just 4, 6, 8, which are all underlined twice. So for every 12 numbers, exactly 3 numbers will be underlined twice. So what we can do is we can divide 2006 by 12, and we get 167. So there's going to be 167 groups of 12 numbers. And when we divide this, we actually have a remainder of 2. So what that means is that there's going to be 2 numbers left over, so that they kind of represent 1 and 2. So what we can do is we can multiply 167 by 3, since for every group of 12 numbers, there's going to be 3 numbers that are underlined twice. So it's 167 times 3, which comes out to being 501. And then the last 2 numbers, which will be 2005 and 2006, they won't matter since they represent 1 and 2 from the numbers 1 to 12. So our answer should just come out to being C. Thank you, Jacob. Those last 2 problems are kind of complicated ones, and you did a great job explaining them. So as you can see, the idea of multiples, divisibility factors comes up in quite a few contest problems. And it might be that it's not just divisibility. There's added-on features. So if we had the logic problem with how many factors could there be, you might have to combine. So like Jacob's problem was okay, but multiples of 12 are underlined 3 times, so you really had to find a pattern. So finding patterns could also be there. So it's really important to remember to practice doing the multiples, the divisibility rules, because if you don't have that part down, you won't be able to add on the next component to get the correct answer. How many boxes, you've got to make sure, okay, I know there's 36 is the greatest common factor, but I really need the number of boxes, so I have to divide by 36. So be real careful to read all the way through the problem, answer the correct part of the problem. Remember, 1 is always a factor. Okay, so don't leave that out. Remember prime numbers versus composite numbers. 2 is the only even prime number. And like I said, if you want to go back to those divisibility rules, those are good to have. I'll put them up here one more time. But you might want to copy those down at some point. You can watch the recording of this video or look them up on the Internet. You can find all sorts of tables of them and print one out for yourself. That might be a good thing to have when you're doing your preparation for contests. Thanks, Jacob, for doing a great job explaining, and thank everybody for coming on your Sunday. Have a good November, and I'll see you next week.

Math Kangaroo

multiples

factors

divisibility

prime numbers

composite numbers

factorization

least common multiple

divisibility rules

logical reasoning