for the test. Maybe whoever comes first can start this question for find the warm-up machine, the wetlings or others. That's really nice. There's 105 numbers written in a row. 1 and 2, twice, 3, 3 times, 4, 4 times, 5, 5 times. Each number n is written exactly n times. How many of these numbers are divisible by 3? So if you've just showed up we're just doing the warm-up problem because no one else is here. Put it cynically. Yeah, that's the correct answer, that's the correct answer, so let's just quickly do it. We have n numbers, right, so we can use arithmetic sequence, we can see how many numbers are there. We've got f from 1 plus 2 plus 3 plus all the way to n. Then we would have n times n plus 1 divided by 2, that gives us 105, so n times n plus 1 is 210, and that means n is equal to 14, so we write all these numbers from 1 to 14, and the ones that are divisible by 3 are 3, 6, 9, and 12, and each of these numbers appears, you know, exactly 3 times, 6 times, 9 times, 12 times, so we have to add all of them together, and we get the answer, 30. Oh, I just realized something. Yes? Right now the Super Bowl is happening, I think. Oh, I see. We're not going to have that many people today. I see. Thank you for being here, you two, we appreciate this. Yes, at least we have some small statistics. Brain over brawn. Oh, I see, that's the thing. Okay, so thank you for the hardworking students. All right, I think we're just going to start then. Yeah, we get started now. So we're going to talk about sequences and patterns today, and what it means by that. First, we review the two common sequences that you probably, you know, most likely seen before, the arithmetic sequence in which the difference between two consecutive terms is a fixed constant, and the geometric sequence in which the quotient between two constant, between two neighboring, two consecutive terms is a constant, right, and when we have this sequence, we can calculate the, you know, biggest number, smallest number, the d, the distance, r, the ratio. We can calculate the sum as well in a fast and easy way because we have a closed form formula. For example, the sum of all the terms in the geometric sequence here is just equal to 1 minus r to the power n plus 1 over 1 minus r, and you have arithmetic sequence, you have a formula. Most of the time, you can just get away with adding the biggest is the smallest and second biggest is the smallest. So, I mean, formulas are important. Many times you can just do it from scratch, but it's good to know this. But many problems that involve arranging a sequence of numbers or objects, you know, sometimes they're just like segments or object figures in a certain configuration and according to a certain rule. So, what I mean by this is like this. You have a number that, you know, may not be arranged in a row, but in a circle, a closed circle, or an array, like a two-dimensional, a table, a pyramid, all kind of different configurations, and we are asked to, and the rules are specified of how to arrange these numbers, and many times we are asked to find the missing numbers. And then we have another type of problem in which we are given an initial list of numbers, and then this list goes through an operation, and then again, the output of this operation becomes the input of the next one. So, this operation could be the same, but many times they are just the same operations, repeated operations, and then we have an initial list as an input and a final list at the output. So, what are the questions associated with this type of questions? So, some common questions involve like finding out the missing numbers, finding out the smallest and biggest values, and value of certain numbers after some repeated operations, and then sort of sometimes more complex questions ask for whether or not a particular value can be obtained, whether or not a certain arrangement is possible, and sometimes like what is the minimum number of operations we need to do to get to a certain result. So, you know, it could be anything. For example, like if we specify a rule to paint a chess board, and the question may ask is it possible to finally obtain a configuration with like the black and the white squares arranged in a certain way. We come to more concrete examples, but that's the idea. And usually for these questions, you know, we need to spend a little bit of time in the beginning to explore the questions, make sure you fully understand the rules and the constraints, and these are the problem-solving strategies we talked earlier that could be very helpful in this question, like solve a smaller version to find a pattern. Sometimes we can actually find an explicit formula to construct the sequence, and then that allows us to find all the missing values or to determine whether the bound of the values, the smallest and the biggest possible value, and try and error helps a lot. And a lot of the time when we are asked to find out whether a construction arrangement is possible or not, we need to use parity, and I would say a lot of the time you have intuitions that you find the right answer, but the path to prove that this is like the smallest or the biggest value could be difficult. So let's go to problem solving. So numbers 1, 2, and 3 are written on the circumference of a circle. Afterwards, the sum of each pair of neighboring numbers is written between them. In this way, six numbers are obtained. 1, 3, if we start with 1, 2, 3, and we would end up having 1, 3, 2, 5, 3, 4. This operation of writing the sums of two neighboring numbers is repeated three more times, resulting in 48 numbers on the circle. What is the sum of these numbers? So for a question like this, I think we can just start it right away. I'm drawing a circle, and then we put three numbers on it. It could be 1, 2, and 3. Just make sure we understand the rule, right? And they say we add two neighboring numbers, and right in between here, 1 and 2, then I have a new number 3. 2 and 3, I have 5. 1 and 3, I have 4. So I finish one operation, and I have six numbers. I repeat the same thing, just add two numbers neighboring, and write the sum, the sum in between, and do it three more times. Okay, so we have 48 numbers in the end. And so let's just keep maybe 30 seconds, one more minute. So one thing you can do is that you think that we continue it and we see the pattern, but the pattern here is not clear. It might not have any pattern like 1, 2, 3 is okay, 3, 4, 5 is 3, 4, 5, also three consecutive numbers, but on the circle itself, you know, 1, 3, 2, 5, 3, 4, doesn't look like any patterns. Oh, great, everyone got the correct answer so we can get started, right? And for this question, we can find the general formula actually. So we can start it with, if I start with ABC, the new numbers would be A plus B, B plus C, and C plus A, right? And so the sum of all the numbers is 3A plus B plus C, right? And now, of course, I can forget about this notation ABC, CVA, I can just start with ABC, I mean, it doesn't have to be in this order, DEF, and I do the same thing, but most importantly that the sum, every time I perform one operation, the sum increases by a factor of 3, right? So that is the most important observation, not the number itself, but the sum of the numbers. So if I do 3 more operations and have a 1 in the beginning, so that's 4, so the sum would be 3 to the 4th power, and then of course I have to multiply by the sum of the 3 numbers in the beginning, so 3, that is 6 times 48, and you all got it correctly, 486, right? Okay, so question number 2, Dave, Dave, you're going to have us do this question. All right, so, there's a numerical sequence that consists of 200 zeros, yikes, just saying. The sequence is transformed to a different sequence of 200 terms in the following way. At the beginning, we add 1 to each term of the sequence. In the second step, we add 1 to all terms numbered with an even number. In the third step, we add 1 to all terms numbered with a number that's divisible by 3, and so on. After 200 steps, we obtain a sequence with the 120th term equal to what? So that is the question. Now before we get into this, I want to make something clear, which is this is a kind of weirdly written question. Let me just clarify this real quick, because I really think that the competition to be held accountable, when it says the numbers are numbered, it means that you've got these numbers from 1 to 200 that represent which zero it is, okay? It doesn't mean that we add 1 to every number that has 1 in it, because if we did that, I'm just going to say, if we added 1 to every number, and then we added 1 to every number numbered with an even number, and then we added 1 to every number that's divisible by 3, then that would never happen, because 1 isn't divisible by anything. So we would never change any number ever again. But we're not doing that. When it says numbered with a number, we mean like which of the zeros it is, just to clarify. So with that in mind, let's go. Otherwise, it's pretty much impossible to solve. Yeah, so the first step, I mean, we just do add 1 to everything, right? And then the second, we add 2, so this one becomes 3, and this one stays as 1. We still add 1, though. So it becomes 2. Oh, add 2, but we only add 2 to the even number, right? But we add 1, so it will become 2. So like the fourth one will become 2. Yes. And then the third one would become 3. No, the third one would become 2 the next round. Yes, to add 2, yeah. Just go ahead and do it. As you can see, this is a weird pattern, and that's partly the question's fault, if it's just confusing. But yeah. So kind of the trick for this, the first thing to realize is that if we add 1 to every term, that's divisible by 1, 2, 3, and so on. That means that we add 1 for every divisor, right? For every divisor of the number, if it's divisible, then we can add. So then we add 1 to it. And that tells you how many times we're going to add 1 and how big the 128th term is eventually going to be. Okay, so let's begin reasoning our way through this. Okay. So, because okay, everyone's answered and it's all perfect. Yeah, excellent. So, if you have what happens here is that you start with zero, right? And then you add one for every number that's divisible by one, right? Each term of the sequence means each term divisible by one. That's the same thing because everything's divisible by one. Then each term divisible by two, each term divisible by three, and so on. Which means that for every single place, every single zero in here, every single place, we add one for every divisor, right? I hope that makes sense. I know it makes sense because everyone got the question right. Both of you got the question right. You add one because if it's a divisor of the number, then that number divides by definition. So you add one, right? So the question just becomes, it reduces to, well, if we're looking at 120 and we add one for each divisor, how many devices does it have? And that's where it's a great time to have been to our number theory section because we can prime factorize this. It's two to the third times three times five. Again, hopeful. I don't know if you guys have actually been doing what I said and doing your highway exit prime factorizations, but either way you got it. So, you know, it doesn't matter. Anyway, this means that the total amount of divisors of 120 is four times two times two is 16 and that would be A, which everyone got. So yeah, problem 26 crushed. I think this is going to be a good exam this month. Yeah, great. A lot of time we have to use the tools of number theory and all we got with algebra to solve these type of questions. Okay. So let's do the next one. Okay. Now we have numbers in a circle. At each vertex of the 18-gon in the picture, a number should be written, which is equal to the sum of the numbers as the two adjacent vertices. Two of the numbers are given. What number should be written at vertex A? So we don't connect yet. I mean, we connect them. It looks like a circle, but because you have so many vertices. So each one is equal to the sum of the number as two adjacent. That means someone left the Super Bowl. Okay. So A would be equal to the sum of this number and this number. That's what the question means. And we are given, so we are given a very clear rule of how to put the numbers here. We are given two of them. We are asked to find A. question number one we were able to find an explicit formula for the sum after every single operation for this one we might be able to maybe not formula but maybe you might be able to find a pattern to start with um you know to start with two numbers for example you can start with a and whatever you want like a and b and then the one in the middle is a plus b and sometimes doing a specific number helps and sometimes it's using letters actually when I have a see the pattern more clearly so suppose we have a and a plus b as the three consecutive vertices if we can continue this way let's see if we're going to continue putting using new numbers or this number we repeat because in a sense that we have sort of two degrees of freedom so if you even know two of this number two vertices we can uniquely find this vertex right and these two if they are uniquely determined we are known than this one we also depend on that would be uniquely determined so in a sense we only have two degrees of freedoms right and you would sort of think that oh I don't need more variables than this so that's one way to think about it Let me continue to get some correct answer, but let me continue. Suppose that I have a and b, one vertex apart, and so the one in the middle, a plus b, and now b is the one in the middle, so one is left, a plus b, so the next one has to be minus a, right? Let's just work my way like that, so make sure it's b, and now minus a here, that means next one has to be minus a, minus b, right? And that means next one is going to have to be, if I make a mistake, David, please correct me, this is b, so minus a, so this one has to be minus b, right? So the sum is going to be, and this one has to be a, and now I need to have a plus b, and now it's good because once I have two numbers that's repeated, that means the third one is going to be the same, and the sequence repeats, so I know that, of course it goes in a circle, but I know that after six numbers, the sequence repeats, and yes, we only need to have two variables to create all the numbers in the sequence, so we know that we can start from here, the known number, 18, we can put 18 equals to a, right? I start from here, so 18 is a, and I don't need to calculate b all the way, but I know that I can skip one, two, three, one, two, three, so this one's going to be minus 18, that's minus a, right? I know that this is a and minus a, and then minus a, so that one's minus 18, and then it's going to be 38, and I know that the sequence repeats after six numbers, so I can just, you know, go backward from a, one, two, three, four, five, six, I mean, good luck, this is 38, so I can fast forward six numbers, and I got 38 as a, right? And don't calculate everything, no need to try to find for b, or calculate every numbers in between. So this is a question about, I mean, I think knowing algebra, right, would help in this case, like knowing how many variables we need, sort of like how big picture thinking would help in this case. Okay, good. Next question, I think, David, would you go first with this question? All right, problem 27. For each vertex of a square, one positive integer is placed. For any two numbers joined by an edge of the square, one is the multiple of the other. However, for two diagonally opposite numbers, neither is the multiple of the other. What is the smallest possible sum of the four numbers? So now this is one where you can't really calculate everything in the sense that you're not finding the general rule, you just want the smallest one, so that should help out a little bit. But try to optimize, right, try to see what is the smallest possible thing can do, and why that's the smallest. Got it. So one thing we can do here when we're approaching this problem is we can start by just looking at each corner and the reason I'm saying corner is because really what this is telling us is you have a relationship between three numbers right if I look at just these three guys I know that these have some sort of multiple correlation and these has a sort of multiple correlation but these guys don't right there's nothing here which means that I've got four options for a B and C either a is multiple B or B is multiple of a and either B is multiple C and C is multiple for B that's a lot right so what you can do is you can either just do casework on all of them or hopefully you can find a way to restrict which of those four options is possible so let's look at that a little bit right we one option is a is a multiple of B and B is multiple of C right then there's a is a multiple of B and C is a multiple of B I'm just gonna draw the arrows where the arrow means multiple of then there is B is a multiple of everything and then there is finally B is a multiple of A and C is a multiple of B right these are like our four options here now we also know that C is not a multiple of A and A is not multiple of C right so when we look at this problem the logical thing to ask ourselves is okay we know that it has to be one of these things if we don't know the generally the idea that I'm using here is if we don't really understand which is the best way or we don't understand the constraints of the problem just try everything and see what starts happening see what starts not working because for example if A is a multiple of B and B is multiple of C then A is a multiple of C right that cannot happen so this one's out this cannot be a possibility if C is a multiple B and B is a multiple of A then C is a multiple of A again that can't happen that's out so either B is a multiple of everything or A and C are multiples of B right so let's just assume kind of simply without this is called without loss of generality it basically means no without um we're just going to choose one of them because they're basically the same thing if B is a multiple of everything here's why because say that we have our square let me just figure out to raise all right excellent so either B is a multiple of everything or B is a divisor everything so now we know this say that B is a multiple of everything right so B is a multiple of A and C right that means that D will similarly also be a multiple of A and C right so it's gonna look like this basically B and D are big A and C are small if we made B a divisor we just switch it around so A and C would be big and B and D would be small it's the same thing so because of that we're just going to make A and C as small as we can and then make B and D as small as we can so it's pretty obvious that the smallest numbers here for A and C are 2 and 3 right we want something small so let's just do that the reason we're doing this is because you know we can't have 1 as a number because everything's a multiple of 1 the diagonals are not going to not be multiple of each other but 2 and 3 work they're relatively prime then we need B to be a multiple of 2 and 3 and D to be multiple of 2 and 3 right what does that mean it means they're multiples of 6 any multiple of 2 and 3 is also multiple of 6 so let's say that we made B 6 right let's try that we want things to be as small as possible what happens now we have a problem because no matter what we make D, D has to be multiple of 6 right it's multiple of 6 by definition which screws over when we look at the other angle obviously this fails so instead we're gonna try the next thing what if we make B 12 then we can make D 18 and we find everything works so 12 and 18 are not multiples of one another and actually if you notice 2 3 12 18 is the same right it's just 2 and 3 multiplied by 6 so the two diagonals are the same smallest numbers but we multiply by 6 so that this relationship works and that leaves you with the sum of all those numbers which is just 35. Yeah thank you Davis as I said in the beginning it's not like heavily on number theory anything sometimes this could be very elementary right but it requires a lot of thinking and reasoning okay great next one a certain automatic mathematical machine works as follows it adds 1 to the number that is entered or it multiplies a number by 2 so you can choose one of the two operations after performing any operations it follows the same rule again the number 0 was entered into the machine after a certain number of operations the result become 100 what is the smallest number of operations that must have taken place to get this result so we can we have an input a and the output coming out it could be either a plus 1 or 2a and then we feed one of these into the next step and again it's going to input one of these two we are allowed free to choose any of these operation and we start from 0 and the question asks like how many steps do we need minimum smaller number of operations And I want 100 as the output. This is similar to the previous problem. It could be very elementary in terms of, you know, algebra. Good thing that they already given us, like, you know, some suggestions here, so we know it cannot be smaller than 8. And you may try, you know, from left to smallest, the biggest to see if you can make 8 or you can do 9, or you need 10. Oh, this is the, I'm sorry, this is the old poll. We haven't launched the new one. Because we got answers, the 28 I would think would be too big. I was wondering too. Sorry. Technically I believed it because if you just go from zero to 25 and then multiply it by two twice, you do get 28. Yeah. Okay, so we- Okay, they just added one all the time, but no. Okay, nevermind. Instantly it's in. That's amazing. All right. Nice answer, yeah. So we know that it could be, you know, because we got hundreds or 50 or something. We don't need too many. So you start from zero, right? We're not going to multiply by two. We're going to stay at zero. So the first operations that I have to add is one. There's no way around it. So then I have to add one, right? So I get one here. And for one, either you add one or you multiply by two. It's the same thing. Doesn't matter what goes here. You have two, right? So the first two steps are fixed. There's no way. And you don't want to waste anything. The first two steps are done. And now we need to get from two to a hundred. Let's first do a little bit of bounding. Like how many steps do we need? And remember, we talk about, you know, power of two. So we now have a two to the seventh is 128, right? Two to the sixth is 64. So sort of, you know, we need maybe not too many. Of course, we want to hit exactly a hundred and not like, you know, 128 or 64. So around that, but we sort of maybe think the number should be around here and not around here. And we can do a lot of that with trial and error. So maybe I can just do a hundred. I go backward. I would get, what, 50. And then what's next? 50, I think maybe I do this faster in here. A hundred. And then now, you know, by two, we have 50. 50, and then what can we do? 25, 25 is an odd number. So I have to subtract one. Maybe I get 24, because I want to use as many multiplication as possible. Right? It grow quicker than addition, only when I cannot use multiplication, then I do addition. So 24 divided by 12, 12, and then, well, I can do. And for this question, you actually need to construct the sequence because, I mean, with only reasoning, we do not know whether we can get a hundred or not. So we have to make sure that we can actually construct a sequence. So six, gonna divide by two, so that would be three. I think I'm good here because I know I have a two. So that's gonna be two and one and zero. So that's one, two, three, four, five, six, seven, eight, nine, nine steps, right? So I have an answer, it's nine. And for this question, you sort of have an idea. You can feel that nine is correct answer and we need nine steps. But we might, I mean, in the exam, I think we can get away with that nine is correct answer, but since there's a B, so in reality, maybe at home, we want to make sure that you cannot, eight is not enough, right? And the way to, I probably would not spend a lot of time in here, but you can use a bounding here, two to the seven. Here you already spend two steps, right? So can we go from two to 100 using only six steps? So that's the question, so I need eight. And for this one, you can use bounding. If you use six step with all multiplication, we're gonna go up to 128, that's too much. And then if we have to replace one multiplication by additions, then the biggest we can get is two plus one, right, because when you add first and then you multiply later, you go faster than you multiply first and add later. So two plus one, and then we have only five more steps left to do the fifth, that's three times 32 is 96. So this one's smaller than 100, so either you under hit or you over hit with only six steps. So that means it's sort of like the proof that we cannot do it with this step, and we are sure that it's nine steps. Okay, good, I hope we can go to two more questions. Let me see how, okay, this one. How many 208-digit numbers exist whose consecutive digits form two-digit numbers which are divisible either by 17 or 28? Short questions, but see what it means, 200-digit numbers and any two consecutive digits form a multiple of 17 or 28. So again, for this question, probably it's no magic trick, or, no magic trick, or, you know, finding a general formula. We probably, just like question number five, we need to construct these numbers to a certain extent, right? We do a few steps to construct these numbers. I think I get started, everyone's doing. I just list down, I'm not doing anything. I just list down multiples of 17, which I'm sure is what everyone is doing. 51, 68, 55, 32, and multiple 23. It's going to be 23, 46. So somehow I have to create a 2008 digit numbers, and then any of the two consecutive digits has to be, you know, in this set. So there are quite a number of steps in this question, so I think I just start with the steps, the obvious steps that everyone probably is doing. If we start with 17, then we are in bad luck, right? Because we just stopped there, so there's no way we can extend it. So for example, I start with 34, and then I can go to, good thing they have a six here, 46, right? 46, and then I go to 69, and then 69, and then, okay, so maybe I just write 34, 46, 69, and then if I go to 92, I can go to three, and here I realize that actually I can go back to 34, that is what I want, right? It has to be cyclic, and it is continue so that I can have a 208 digits. So, I mean, once you realize is that probably 80% of the problem, but there's a trick over here, so I hope at this point you can, ooh, got one answer, okay. When we also read 34, 46, the thing is that you go to 60, instead of 69, we have a 68, right? So go to 68, if we go to 68, and then we would go to 85, and 51, and then 17, and then it terminates here, it terminates. So we can use 68, but don't use it in the beginning, we can use it at the end, right? Because the sequence terminates. So this one is cyclic, and this one terminates. So knowing these two facts, we have you solve the problem, and I think that's what most people have been doing, but going from here to the answers, it takes a few steps, so maybe one more minute. So let's go to the next one. Oh, that's interesting. I got sick at the answer and I don't understand why we got sick. Oh, I can see I can see why I got sick. Okay, good. Let's go to. This is a long question. So I type up the answer ahead of time, but still, it's a good thing to discuss. So let's discuss the thing. So we realize that we have this cyclic sequence with these numbers, right? 22, 23, 34, 46, 69, 92. I can do from smallest to the biggest, right? So you think that each of these we can do, we can start with one of these numbers and we have we have five choices. And then maybe we plus one with 69 and 68. That's why I mean, all of you got sick at the end. So, but look at this. We have a side with 26, right? So we actually we find numbers that we only need five digits. So every five digits, the sequence repeats, right? And two hundred, two thousand and eight is divisible by divided by five. We have a three is remainder. So here we can continue all the way to 69. And for last one, which is, you know, have a three numbers continuous sequence. But we might as well in the last sequence, in the last, you know, group, instead of 69, we put 68. So we and then we have after 68, we actually have a five number eight. I mean, after eight, we have a four more digits. Eight, five, one, seven. Actually, we can do 72 minutes here. So we have enough to cover. We have to extend up to 2008, right? So we have two. Actually, when you start with 23, we have two numbers and we have a one, two, three, four, five numbers here. If each gives a two, because in the last step we replaced by 69, we replace 68 by 69. Then we would have a ten numbers in total. But we got to be careful because if we start with 34, the 69 is right in the middle. And if we replace by 68 here, we do not have enough numbers. We only have three digits. It does not go all the way to 2008. We only can end up here, which is 2007 digits. So for the choice 34, we actually have only one. Why is the rest we have a two, two, two, and that's why we have nine numbers. So again, I think this is a good question. Pretty long one. You can, you know, after we figure out that it's cyclic pattern, most likely we're going to think that have a ten numbers. But again, requires a little bit of, you know, work and realize that actually we don't have enough in this case to extend up all the way to 2008. So we have only one number and the answer is nine. This is one of those problems where you kind of have to just batch it. Yeah, we can say, you know, it's great to try to find a neat solution. That's what most math kangaroos are like. But in the end, you have to check for every single one. Does it work? Just go through it. You know, see what happens. Exactly. So the modular arithmetic kind of. OK, thank you. There's one more question. I think we have time to get through and I really like it. And I think it's good to review. It's a bit of geometry over here. Not geometry, some drawing. I think it's a good question. So let's go for it. The points a0, a1, a2 lie on a line such that a0, a1 is 1 and the point an is the midpoint of the segment an plus 1, an plus 2 for every non-negative integer n. What is the length of the segment a0, a11? And for this one, I recommend you just draw the line with a0, a1 to see the patterns, which you can figure out. So here we do have numbers, we have points, and we have length. This is in the interest of time, I'm just drawing, and I hope you are doing the same thing. You've been doing the same thing. I start with A naught here, and then I go to A1, right? That's one. But now, because one is midpoint of, I'm sorry, A naught and A1, and then An is the midpoint of the segment, An plus 1 and An. So that means A2, let's see, for any non-negative, OK, so non-negative, that means A naught is non-negative. I'm sorry, it's positive. So it means A naught is midpoint of A1 and A2, so A2 has to be on the other side of A naught, right? This is A2. And again, now A1 is midpoint of A2 and A3, so you can flip it all the way here, and A3 goes on this side. And then now A3 is midpoint of, A2 is midpoint of 3 and 4, so A4 goes back to this side, right? And now we clearly see the pattern. All the odd ones, if you start with A naught in the middle here at the origin, then all the odd ones is on the right-hand side, and all the even ones is on the left-hand side, right? So at this point, we can just quickly write out A naught, A1 is 1, and then A1, A2 is 2, but I would write minus 2 and explain why. And then A2, A3 is 4. We know it's double the other one. And A3, A4, I would write as a minus 8 instead of 8. And the reason we do so is that because if we mean later on, I don't know if we have a steady vector or not, but if you do A naught, A1, you consider the vector, and A1, A2, it points the other direction, then you could have this vector. And if you do like that, if you add A naught and A1, you would have A, sorry, if you add, sorry, this one. If you write it this way, then we add A naught and A2, you get A naught, A1, plus A1, A2. You're actually going to get A naught, A2, which is 1 plus minus 2 is minus 2. So I got the direction correctly because A naught, A1, A naught, A2, the magnitude is 1, but it's on the left, so it's minus 1. And the same thing, I add all of them here from A naught to A4. I got A naught, A4, which is 1 minus 2 plus 4 minus 8, which is minus 5. And that I get both the magnitude and the direction. So everything has a minus on the left. Everything has a positive on the right. So that means when I have A naught, A11, it is the odd number, so I know that A11 is going to be here. And in order to calculate A naught, A1, I just have to add everything. And I know the pattern, so I can add it this way. It's going to be A naught, A11. It's going to be the sum of all of this, A naught, A1, plus A1, A2, and plus all the way to A10, A11. Earlier, if you remember, we talked about since you can consider it as a geometric sequence, but I wouldn't bother about it. It's too much to pull in the full formula with the general letters. So I just do it this way, 1 minus 2. And remember, we wanted to remember by hat power of 2. So it's going to be 1 minus 2 plus 4 minus 8 plus 16 minus 34, I think 32, and then just plus 64 minus 128 plus 512 minus 1024 and plus 2048, which is 2 to the 11th, I think, yeah. So this is all you need to do, and then you add up all this thing. I think it should be quick, because if you do this, you're going to have a, if I do it from left to right, I would have, what, 1024. And then I minus this thing, which is 128, 128. And then this 2 is going to give me plus 32, plus 62. This one gives me minus 8, plus 6. No, 1022. And this gives me 1022, I think. Is it what I made a mistake somewhere? Did I make a mistake somewhere, David? What do you get? I get 6. I get e. I got e, right? Yeah, I think I got e when I was doing here. So e, 11. So we start with, it should be 2 to the 11th, which is. Which is that, yeah. Which is that, right? So yeah, I think you could have some mistake. We can do this. We can just go, it's 1 plus 2 plus 8 plus 32. Oh, wait, no, no, no. I see. Because it's 128, then it's 256, then it's 512. Oh, OK, thank you. 256, thank you. Yeah, I was thinking the numbers were big. I didn't do that because I just, I grouped every 2. What you can do with it is. Yeah, you can group exactly. Yes, you can do that exactly. So like. You know that 4 minus 2, which you have, is just equal to 2. Yeah, so you can prove that. That's what I was grouping, yeah. So 122, and then this minus this, yeah. So minus 550, OK. Yeah, you can group it different ways. And then, and minus 22, and then plus 64, I think. Plus 64, plus 64, plus 32. Anyway, sorry. The answer is 683. You can double check at home. But that's the whole idea. So for this one, it's an example. You have a geometric sequence, but don't bother about it. You can get away with just grouping plus and minus, rather than pulling the general formula. And the important way is that recognize the pattern and write it in a way so that you can carry on the calculation easily. Plus and minus, recognize that plus and minus, and odd and even. I think that would help you get it correctly. OK, so thank you, everyone. I mean, the takeaway for this question, a lot of exploration. Make sure you know the rule. Try a few examples. Sometimes, we can see a lot of the time, actually, algebra helps, number theory helps, trial and error. And sometimes, you really have to do the hard work and construct the sequence. So next week, we're going to be doing counting, right? And see you next week.

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