This is the official start time for the Math Kangaroo Level 5-6 webinar series. Today's lesson is about finding numerical patterns. I have a warm-up question that's open right now. It says, find the sum of the missing digits. So why is this about patterns? Well, because there are certain patterns when we multiply. When we multiply, we actually repeat certain numbers. Multiplying by 5 means we repeat a series of 5's and 0's, right? 5, 10, 15, 20, 25. So using that pattern is going to help us solve this problem. I'm going to go ahead and solve this one. So 3 times 5 is 15. So we know that B... Sorry about that. The wrong tool. We're going to ask all students to turn off their videos. So we know that B has to be a 5. So if I have 3 times 5 is 15, I've carried a 1. I can see that the 6 times 5 would normally be 30. So somehow I've managed to carry a 4 into that hundreds place. So whatever is A times 5 has to begin with 4 and has to add a 1 to be 46. Well, 9 times 5... Sorry, if I do the 9 here, 9 times 5 is 45. Plus the 1 gets me 46, right? So if I do that, and then I want the sum of the missing digits, 9 plus 5 equals 14E. Okay, so I saw some of you get that in the chat. Thank you. So today you will notice that sometimes I have chat. I have a poll for the problems where you can answer A through E in a poll. And other times you can put your answers in the chat for me or for our teaching assistant, Jacob. Jacob, would you like to say hello to everybody? Luca, I don't hear you. Want to say hi? Oh, sorry. Can you hear me now? Yes. Great. Hi, everyone. Don't worry. I was having tons of troubles with some of my stuff earlier today. OK. So go back one slide. You can see our title slide for today just shows us that we're going to be doing number patterns. So let's take a look at what this means. Every time we do a math kangaroo problem, we want to start by trying to get through our four basic steps. To understand the problem, this may mean reading and rereading the problem a few times to make sure we know what it is asking us to do. We're going to have a plan to solve the problem. Today, I want you to think about patterns, patterns you know or maybe you could learn. Or can you make this a simpler problem? Sometimes we want to just go ahead and make it a simpler problem in order to find out how to solve it. So see if you can simplify it. Work your plans out very carefully. I encourage you to use your scratch paper. Use plenty of pieces if you need to. You may find that it helps to write number one and then work your number one and then make another space for number two. Because you do want to check back and reflect and see, did I get it right? Did I get it wrong? And if your scrapbook is very messy, you're going to have a hard time checking out your work. So welcome to the webinar. You're going to be keeping your cameras off because this is being recorded. And we don't want to accidentally have students' faces show up in those videos that other people can view. So remember, cameras off. If you have questions, try Jacob while I'm speaking. If Jacob is speaking, you can try me. Go ahead, put your answers into the chat and into the polls. Hopefully, that gets us to where we need to be to start doing more fun and challenging problems today. OK? So what is a pattern? Sometimes a pattern is going to be the arrangement or the repetition of numbers, shapes, and colors. Always repetition, right? Because for example, the pattern here, 2, 4, 6, 8, is not repeating anything. But there's definitely a pattern. This is called a sequence. And what we're doing is we're increasing by 2. This is an arithmetic sequence. There are other types of patterns. You might, can you figure out the pattern 1, 2, 3, 4, 1, 2, 3, 5, 8? How about that one? There's definitely a pattern. The pattern is you add the two previous numbers. So the next number would be 13. And then 21. This is called the Fibonacci sequence. The pattern can be related to any type of event or object. So you can even have a pattern for your class schedule at school. Maybe you always have PE on Tuesdays. If Jacob is in high school, maybe he has math class first, and then science class second, and then English class third period. And it's the same every day. Or you can be in a school with a block schedule where on every other day, you repeat. You might have certain classes on one day, and then other classes the next day. You might have to wake up at a certain time every day to get ready for school. And when you wake up, you might have a pattern of what you do to get ready for school. There's a lot of ways we can think about patterns. So just keep that in mind in your life. You might have a pattern in the flooring in your kitchen. If you have tiles somewhere in your kitchen or in the bathroom, there could be a pattern there. Lots of places we have patterns. So the reason we're going to emphasize patterns today is because sometimes when a problem is complex, if you identify the pattern, you may be able to simplify it and maybe even reduce the number of calculations you have to do. And this can save the time and can make you more likely to be correct. Instead of adding 25 numbers, if you find the pattern and you only have to multiply two or three numbers, you'll probably do that more accurately. So you might try, this is the hint, try a simpler question first. We're trying to sometimes find those simple questions. Today, I will give you a warning. These are mathematical problems. And you might find patterns in multiples. Take a look for those sometimes. OK. Out of how many small squares will a figure be made if it is made the same way as the figure in the picture but has 10 steps? So that's, again, math kangaroo problems. We sometimes have to read them twice. So what is the figure? Oh, the figure is going up one step each time. So this figure has 1, 2, 3, 4, 5 steps. We want 10 steps. And I do have a poll that I can launch for this problem. So you can put your answers into the poll. Polls are anonymous. I cannot tell who has given which answers, if you're correct or incorrect. And neither can anybody else. So don't worry. Also, math kangaroo problems that contest themselves, we do not subtract for an incorrect answer. So do not worry if you get some wrong. If you leave them blank, you'll definitely get 0 points. So you would rather guess than leave it blank in a math kangaroo contest. Okay, anybody else want to put their answer in the poll? Remember, the surest way to be incorrect is to leave it blank. All right, so overall, the students did really well. 55 is the correct answer, and that was the response given by 84% of you who did the poll. Now, not everyone participated. Sometimes people don't see the poll, and that's okay. Remember, if there is no poll, go ahead and put your answers in the chat for me or Jacob so we can help you out. So I want to show a couple of things here. So if I look at this, if I look at it vertically, there is 1, 2, 3, 4, and 5 squares going up and down vertically, okay? So if I was going to make this, but I had 10 steps, then I would need to add up all the squares, 1, 2, 3, plus, plus, plus, plus, plus, plus, plus, until I get to like 9 plus 10, all right? I'm going to teach you a shortcut for adding up all these numbers, all right? Some of you may know this. If not, this is a great shortcut. You can use this a lot. So you're going to take the first number and the last number, and the sum is 11. If you take the second number and the second last number, the sum is 11. Guess what? You'll keep getting that. You'll keep getting that the sum of the first number and the second number is the same. Now if we do this, how many sums will we make? Well, we'll have half as many sums as we did numbers to start with. So we had 10 numbers in our sequence. So we would take 11 times 10 divided by 2, or half of 10, which is 5. So that gives us 55. So this is a little trick, a formula that you can use for adding the numbers in an arithmetic sequence. Arithmetic means that you go up by the same amount or down by the same amount in between each number. All right? I bet Jacob has seen that before, right? Yeah, he's got his head nodding, yes. Okay. 9 is the ones digit in the product 49. 3 is the ones digit in the product of 7 times 7 times 7. I should have said 7 times 7. We know it's 49, right? I was kind of giving you a little hint there, right? So we know this is 49, and so that does agree that 9 is the ones digit. And 3 would be in the ones digit of 7 times 7 times 7. What is the ones digit in the number, which is the product of 100 7s? Now you can bet we do not expect you to do 7 times 7 times 7 times 7 times 7 100 times. You cannot use a calculator in Math Kangaroo. So see if there's a pattern. OK, I'm getting quite a few answers. I don't want you to feel like I'm rushing you, but I do want to get through a lot of problems today. So remember, if you want to slow down and see these again after class, you'll be able to get to the recording. It might take a few hours for them to post it, but you'll be able to see this all over again and try the problems if you needed more time. OK, so if I do 7 x 7, I'm going to get 49. So the 1's digit is a 9. And if I take this 1's digit and I do the 9 x 7, 9 x 7 is 63. So the next 1's digit is a 3. Now, if I take that 3 x 7, I get 21. So the next 1's digit is a 1. You might be thinking, what are you doing, Coach Siggy? Well, remember, anytime, even if we take a really big number and we multiply it by something else, the 1's digit is just dependent on the number that's in the 1's value, right? So the 1's digit of the product is going to be 4 x 7 is 28. So it's going to be an 8, no matter what else is to the left of that, right? So you only need to see the 1's digits to know the 1's digits of the products. Now, if I take that 21, 1 x 7, I get back to 7. And then again, if I take the 7 x 7, I get back to the 49, right? So I get 9. So you can see that if I have, I'm going to erase this just to give me a little space. If I have only one 7, then I'm going to get a 7 in the 1's place. If I have 7 x 7, then I'm going to get the 9. If I have 7 x 7 x 7, I'm going to get the, the next one is the 3. 7 to the 4th is going to give me the 1. 7 to the 5th is going to give me a 7 again. So after 4, this is a single repeat. You'll notice that after that, it repeats. So remember, we're talking about patterns and repeats. So now I have to look at that 100 7's, and I'm going to divide it by a group of 4. Right? So 100 divided by 4 goes evenly 25 with no extra, no extra factors, no extra multipliers or remainders. So it's going to be an even group of 4. The answer will be that it ends with the 1's digit of the number 1. So by identifying the pattern, we're able to multiply by the number of repeats and come out with however many it takes. Alright, number 3. In a two-digit number, A is the 10's digit, and B is the 1's digit. Looks like this, AB. Which of the conditions below ensures that the number will be divisible by 6? Some of you will know a pattern for divisibility and will know how to make sure something is divisible by 6. because this problem says we need to ensure that the number will be divisible by 6. Let's think about what ensure means. It means that if I use this and it is ensuring it, there could be no exceptions, right? So how about a plus b equals 6? What if I do 1 plus 5? 1 plus 5 equals 6. Is 15 divisible by 6? No, it's not, right? So therefore, this one is not correct. Let's also look at what divisible by 6 means. Divisible by 6 means that has to be even, right? Because it has to be divisible by 2. And also that the sum of the digits, the sum of the digits would be divisible by 3, right? Because even means it's divisible by 2. The sum of the digits divisible by 3 means it's divisible by 3. If you're divisible by both 2 and 3, you're divisible by 6. So let's look at this. If a, let's say a is 1, then b could be 6. Is 16, 16 this way a multiple? Let's take a look at b. What if we put a 1 here, then we have a 5. Is 15 divisible by 6? No. How about this? If we put a 1 here, then our b is a 2. Is 12 divisible by 6? It is. So this one, we're not crossing out right away. Let's take a look at e. If we put in a 1 for a, no, if we put in a 1 for b, we would have a 2 for a. So that gives us a number like 21. 21 could be odd. So this one doesn't work. So let's take a look at b a little more in depth. If the sum of the digits is divisible by 3, then the fact that we have a ratio of 1 to 2, right, because b is always twice as large as a, so that gives us a ratio of 1 to 2, which means a total of 3. So this b ensures that our numbers are always divisible by 3 because, again, the 1 to 2 is a total of 3 parts. And b must be an even number. So d is going to give us our correct answer because it's always even and it's always, the digits would always have a sum of something divisible by 3 because you have 3 parts together. All right. Number 4. Barbara is creating different squares using sticks of equal length in the way shown in the picture. She labeled the squares with the numbers 1, 2, 3, and so on. How many more sticks will she use to create the 31st square compared to the 30th square? So we want to know the difference in the number of sticks between the 30th and the 31st. I'll give you a minute. I do have a poll. We haven't had a poll in a few questions, but for this one, I do. Hopefully, you've taken a look at these figures. Here comes the pole. That's not the right pole. because I don't have the right pole. Yeah, I'm seeing a lot of answers not so many of them correct because this is a pretty tricky problem I want to point out that this is a five-point problem There are 30 questions on your test and we do think that they get a little trickier as you go on So number 25 will be pretty tricky Okay, so let's take a look I'm gonna take a few notes here. I think this will help if we try to find the patterns So what I notice is in that one I have one going across but I have two of them and I have One going up and down, but I have two of them If I look at the two, I have two going across and you'll notice that I have three groups of that, right? so it's kind of like I have three times two and Then also vertically I have that same three times two sticks If I look at three I have three of them and I always have one more So I have one two, three four horizontal rows So it would be three sticks times four the same with the vertical There's three sticks in the vertical column, but there's not three columns. There's four columns So I have three times four So for each of these you notice it would be three times four and then times two because I have to add them together so on my 30th square I Would have 30 sticks either going across or up and down but I would have 31 rows of those or 31 columns of those So I have two one for the rows one for the columns times 30 times 31 for the 31st I would have 31 Columns or rows of 32 sticks Okay, so those are the two combinations. This is for the 30th square 30th one and this is for the 31st So now what can I do here perhaps I can do a little bit of arithmetic I want to do 2 times 31 times 32 That's one group and I want to subtract 2 times I'm gonna put it in the same order 31 times 30 Okay, so some of you have learned the distribute I bet you have all learned the distributive property, right So I'm looking at the distributive property. I have these are the same so I can factor out 2 times 31 is 62 Right, so I could do that. I have 62 Times 32 Minus 30 Great because this is my I'll even try another color for you. I Have the 32 Minus 30 and that's right here So now I have 62 times Which is 124 So using a little bit of arithmetic a little bit of understanding the patterns of how many sticks and how many rows Hopefully that helps Okay Remember if you need to see it again, you will have the recording Okay Jacob you had wanted to lead to this one, correct? Yeah, so How many three-digit numbers possess the following property after subtracting 297 from such a number? We get a three-digit number consisting of the same digits, but in reverse order. So I'll let you guys think about this for a bit Okay, so it seems like there's a little bit of confusion on what the word consisting means, so I'll just suggest that. So basically what this poem is asking is that we take a three-digit number, so... I'm not sure if I have typing or, like, writing permissions. Yeah, sorry, I had to take them away from everybody because not everyone was sharing nicely. Only Jacob may annotate. Okay, good. So what we have is that we have like a three-digit number, so we're going to say this number is ABC, right, and when we subtract 297 from this number, we get another number that consists of the same digits. So they're going to consist of ABC, but in reverse order, so it's going to switch. So this becomes CBA when we subtract 297. So hopefully you guys can try to build off of this and hopefully solve it, but I'll give you guys a bit more time to think about this. I think we will launch the poll just to see how the students are doing. I think Jacob and I are finding that you're finding this problem challenging, and that's okay. It is good to be challenged by our problems. That means you're learning, and once we explain them, hopefully the next time you see something like this, you'll have a better idea of what to do and where to start. Okay, so we're going to go over this now. So I ended the poll, so hopefully you can see that now. So yeah. First, what I want to do is I want to try to find a different way to write ABC because ABC, just like that alone is a little bit hard to deal with. So what we can do is we can say ABC. You can write this as 100A plus 10B plus C. Then we can do the same thing for CBA. So CBA can be 100C plus 10B plus A. Now, when you subtract 297 from ABC, we get 100A plus 10B plus C minus 297 is equal to 100C plus 10B plus A. Now you can notice that the 10B's cancel, so that's pretty nice. And then what we have is we can move the C to the right-hand side and then move the A to the left-hand side. We get 99A minus 297 is equal to 99C. Now, maybe this is a little bit hard to notice, but 297 is 99 times 3. So what we can do is that we can divide the whole equation by 99. We can divide the whole equation by 99, and so then we get A minus C is equal to 3. So now we want to figure out what possible digits satisfy this equation. And remember, A and C are digits. So if A is 9, then C has to be 6. If A is 8, then C is 5. And then we can continue doing that over and over until we get 1 and 4. And notice that C cannot be 0 because, well, it's a three-digit number. So this has six different possibilities for A and C. So you might think that the answer is 6, and I got a few of those answers in chat. I also got 7 because I think some of you guys thought that C could be 0. But you also have to notice that B can be any number, basically any digit, because we canceled it out. It basically doesn't matter. So B has 10 possibilities because it could be anywhere from 0 to 9. So we have to multiply 6 by 10. So 6 times 10 is 60, so D is the right answer. Yeah, I think that this is a challenging problem for fifth and sixth graders. So I think one of the things to see is I'm sure everybody knows that, for example, the number 25 is two 10s plus five 1s. So we're kind of doing that same thing. Any hundreds number, any single digit in the hundreds place would be 100 times that number. And any single digit in a tens place is 10 times that number. And then the rearrangements, I think, is some algebra that not all students might be familiar with yet. But it might help them if they subtracted one from the other and put the 297 on one side. So see if that helps. I think that Jacob's explanation that you can put any number in the middle for B is absolutely right. So that gets you up to 10 times as many choices. Okay. I told you we would have challenging problems. If all the problems were easy, it wouldn't be a really interesting contest, would it? What is the 2007th letter in the sequence kangaroo, kangaroo, kangaroo, kangaroo, kangaroo, kangaroo? This one is pretty approachable. And I do have a poll for this one. All right, I have a lot of answers in the poll and it's going really well. I'll share that. Look at that. Compared to the last one, I think this one was much more approachable by all of you, so that's good. So even if you come across a hard problem in a math kangaroo contest, that doesn't mean the next one is going to be tricky for you. So it's worth looking ahead and trying the next problem and then going back to problems that were tricky for you. Okay, so it's always worth that. That's a good kind of a good test-taking strategy for a lot of different tests. I remember when I was in college, if I was taking a test and I really couldn't do a problem, I would turn the page and try a different problem because points are points, right? You want to try to get as many points as you can in the time you have. So we want the 2007th letter and what we have is a sequence of eight letters. Kangaroo is eight letters long, right? Eight letters. So if I take 2007 and divide it by 8, I'm going to get 250 and then a remainder of 7. I think you can all do that division or you can divide 2000 by 8. 2000, you know, is a multiple of 8, so there's a remainder of 7. So then all we have to do is count 1, 2, 3, 4, 5, 6, 7. The 7th letter will be 0. Oh, it's not a 0. It's not math. This is letters. Okay, so that one was good. All right, in a 4 by 2 table, two numbers are written in the first row. Each of the rows below contains the sum and the difference of the numbers written in the previous row. So, for example, here you can see that 10 plus 3. Oh, my mistake. Wrong little tool. Okay, 10 plus 3. 10 plus 3 is 13 and 10 minus 3 is 7. Each of the rows. So in a 7 by 2 table filled in the same way, the numbers in the last row are 96 and 64. What is the sum of the numbers in the first row of the table? So it starts, it's not the table you see on the right. It is a different table, but it's filled in with the same sum and difference pattern, right? 13 plus 7 is 20. 13 minus 7 is 6. Give you a few minutes for this one. Right, this seems to be a tricky one because Jacob and I are not getting a lot of answers So we think that you're taking your time to work through it. Let's take a quick look at it If I was to write this another way, I Have ten I'm gonna I'm working on the table here on the right hand side So if I have a ten and a three I'm gonna add them to get thirteen and subtract them to get seven I then do the same thing. So Let's say I Substituted in the variables X and Y right? So if I call this X and this Y My 13 is X plus Y and my 7 is X minus Y Then 20 is going to be this one is going to be the sum so X plus Y plus X minus Y See I don't need the parentheses right because of the Associative property so I have X plus X and Y minus Y So this gives me 2 X And I'm gonna end up getting 2 Y for the 6th Now if you don't know what I'm doing with X and Y, that's okay Let's take a look at the pattern. Let's just take a look at the numbers. I see some question marks. That's okay Let's erase So you have it on the video if you want to see what I was doing there, that's great But let's just look at it with patterns without algebra because that's alright, too Do you notice? That 10 and 20 is a doubling and 3 and 6 is a doubling 13 times 2 is 26 and 7 times 14 is also times 2 Let's see if that would work for filling in the next pair 26 minus 14 26 minus 14. Oh 26 plus 14 my mistakes 26 plus 14 is going to be 40 26 minus 14 Is 12 so you notice that pattern of doubling is holding So every other Row is double the one 2 above So if we do that here and we go have Z's right so now we'd have 32 And we're gonna go have Z's here so we have 48 Let's go have Z's. But remember I'm skipping right? So I have 16 and 24 and I'm going have Z's again. I have 8 and I have 12 And Now I want to know what is the sum of the numbers in the first row? Well, well plus 8 is 20 So I can prove it using my X's and Y's Or I can observe it By knowing what doubles and halves are which I'm sure all of you know how to double or how to take half of something Right. So there's a really nice question for you Okay Jacob I'm gonna let you go ahead and lead this question that you wanted to lead Because in terms of time, I think we should probably do this bonus question before the other one if you want to do it okay, so How many natural numbers are there for it's a quotient and plus 41 over n plus 5 is a natural number? I'll give you guys some time to think about this question before we go over Can you define a natural number for them and make sure they understand quotient Yeah, so a natural number is just basically The numbers like 1, 2, 3, 4 so they're integers, but they're all positive And then the quotient is like the result that you get after dividing n plus 41 and n plus 5 Hopefully that clears up some confusion Jacob, I think since we have to finish up soon, we better explain this one. All right, so what we want to do for this question is, the n in the numerator is quite annoying. So what I'm going to do is, I'm going to write it like this. So I'm going to take out one of the n plus fives from the numerator. And what this does is, you can, instead of having n plus 5 plus 36, you can have 1 plus 36 over n plus 5. So that means that n plus 5 divides 36. So in other words, n plus 5 is a factor of 36. So in order for this to be a natural number, that means that n plus 5 has to be like 36, 18, 12, 9, or 6, unless I missed anything. And since n is a natural number, that means that n has to be greater than 1. So n plus 5 is greater than 6, sorry, greater than equal to 6. So if I didn't miss anything, then there should only be five possibilities. So that means that n is 1 of 31, 13, 7, 4, and 1. So those are the only values of n that give n plus 41 over n plus 5 as a natural number. So there's only five values, so it should be d. I think this is another challenging problem. I encourage students to look at this step right here, right? I think we can follow that 41 is 5 plus 36. And then just think about how you add fractions, right? If you have a common denominator and you're adding fractions, then you just add everything in the numerator. So that's what we've done. We've added everything in the numerator. So in essence, Jacob undid this. So he said it was n plus 5 over n plus 5. And then he added 36 over n plus 5, right? So since we just keep the denominator the same, you would add the numerators. And anything over itself is a 1. And then we get to this part where we, okay, now we need to be able to divide 36. And I think the students know how to find factors of a number. So you can find the factors of 36. But this combines a lot of pretty advanced skills at your grade level. And you're going to find that math kangaroo contests are designed to be challenging. They are designed so that without preparation, you probably wouldn't do better than 50% on the math kangaroo contest. So that is why you are here is to get this preparation. Please don't be discouraged. Just know that that's why you're here is to get this type of experience. All right. So you can view the recording. You can see all of these again on your math kangaroo account. If you go to where you have registered for the class, you will be able to see that link. It does take a few hours for the administration to get these edited and put up there. But you'll find it by tomorrow. So make sure that you might be able to find a pattern in a problem and use that pattern. Sometimes you'll find that that helps you figure it out. It was really, really difficult to do the problem with the sticks, right? The squares and the sticks until you figured out that there was a pattern that you could follow. All right. Work very carefully. Use the four-step process. Read and reread your problems until you can figure out what it is they're asking and how you can make sense of those words into actual problems. Take notes, like the one that Jacob was doing with the ABC minus 297 is CBA. Well, until you wrote it down with ABC and CBA, it didn't really make a lot of sense. But once you start writing what you're reading, that usually does a lot to help you clarify what it means. Thank you for attending with us today. Jacob and I look forward to doing another lesson with you next week. Have a great week. Bye, everybody.

numerical patterns

arithmetic problems

multiplication patterns

problem-solving strategies

arithmetic sequences

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reverse patterns

math webinar

pattern recognition

simplifying calculations