So hi everyone, let's do a warm-up question while waiting for others. So the size of a pentagon are labeled with natural numbers in such a way that the greatest common divisor of any two adjacent sides is one and the greatest common divisor of any two non-adjacent sides is greater than one. Which of these could not have been used as a label? So we have a pentagon um and the GCD of the adjacent of any two adjacent is one while the GCD of any two non-adjacent is greater than one and then which one could not be used. So usually I just draw to make sure we have five sides because it's a pentagon. you Again, for questions, when they do not give us any numbers in the questions here, then you could either try to construct, you know, find numbers and see if that satisfies the conditions, or go through the choices, pick one of those numbers and plug it in to see if you can construct the rest, or, you know, see if there's anything special about one of the numbers. Basically, we thought we had to do a little bit of trial and error in the beginning. So in a sense, it's a question of number theory, right? Because we are talking about GCD over here. And if we look through all these numbers and 6, 14, 15, 10, 11, these are, four of them must have something in common and one is the odd one out, right? And if we look at these, 6 is composite, like 2 times 3, 14 is 2 times 7, 15 is 3 times 5, and 10 is 2 times 5. In a sense, they are kind of composite numbers and each of them can be factored into as a product of two different primes, right? So in a sense, these are kind of similar, right? I'm not sure, sometimes the questions are the same, but at first glance, when I analyze these numbers, they are pretty much similar. But 11 is the one that's very different from all the rest, it's a prime number. So in the exam, I don't have much time, so I probably try this one first, see what it can do with that. And they say that which of these numbers could not have been used as a label, so I will prove by contradiction. I try to put the number 11 here and I see if I can construct, you know, all the four numbers that satisfy these conditions, right? So I put 11 here. So the greatest GCD of any two adjacent sides, that means A and B here. So the GCD, the question says that the GCD of two adjacent sides, so GCD of 11 and A is bigger than one, right? So that means, is that the GCD of, okay, opposite sides, and I could have a C and D here as well. Okay. GCD of, I didn't say, okay, so. Oh, okay. Sorry, it's the GCD of 11 and C has to be bigger than one because it's non-adjacent. So yeah, so these two are non-adjacent, bigger than one, but because 11 is a prime number, so that means C must have a factor of 11 over there, right? So maybe C is going to be some number 11n when n is an integer. And similarly, the GCD of 11 and D, which are the two sides that are non-adjacent, bigger than one, but because 11 is a prime number, so D must also have a factor of 11 in it, right? And is integer. And from here, C and D, 11n and even n, so that means the GCD of C and D is going to be 11 times the GCD of m and n. So they at least have a factor of 11 in common, bigger than one, and that's a contradiction, right? Because the GCD of any two adjacent sides must be bigger than one. So that is a contradiction. It's called proof by contradiction, and that means 11 cannot be used as one of those numbers. So that's what we see a lot in today's lesson when we have to work with like sequencing patterns, and we're going to use quite a bit of number theory and algebra. Okay. So let's get started. So many problems, right? I mean, in mass cancels that involve a sequence of numbers or objects like in a certain configuration according to certain rule, and you've seen the most common sequences, arithmetic sequence, when each number, each term is, you know, is an increase by a common term from the previous one, and the geometric sequence when each term is the previous sum multiplied by a constant, and then we would have formula like this to calculate the sum of the numbers in a geometric sequence. For the arithmetic sequence, most of the time, what you can do is just add the first term and the last term and the second, first, second, last. That's not a lot of calculation, but that's an easy way, or you can use the general formula. But sequences are not just limited to these two common sequences. You could have something like numbers may not be arranged in a row, but could be in other configuration like in a circle, in an array, or a pyramid like what we saw in the first class in a square pyramid in an array like this. And then sometimes, a lot of the time, actually, you have what's called a series of operation and transformation. We are given an initial list of number, and then we operate the same operation on that initial list of number, like it could be adding them together, adding the adjacent numbers together, and then we repeated that operation many, many times until we come up with the final list. And then the common questions involved these kind of sequences could be like five missing numbers in a series of transformations, or five values of certain numbers after maybe like a hundred repeated operations. And then you could be asked whether or not a particular value can be obtained like you are supposed to arrange numbers in a circle according to a certain rule, but whether or not this particular value is allowed, and then whether or not a certain arrangement is possible. And then in lesson five, which is next week, we're going to specifically focus on smallest and biggest values. So in terms of problem-solving strategies, for questions with sequences and patterns, a lot of times we use several smaller versions of a pattern looking at several specific examples. Sometimes we can actually explicitly construct the formula for the sequence, and then we try an error and parity, which is odd and even. And as you see, we're going to use a lot of number theories in Algebra. So let's get into problem-solving. The first one, the first student wrote the number one on the board, the second student wrote the number two, and the third one, and each of the following students wrote a number that was a quotient of the number written just before the last number and the last number. What does the tenth student write? So we talk about quotients, that means basically it's just the divisions, right? So we have, so the first step is that make sure we understand, we understand the rule, right? So we start with one, then two, and the next thing is the quotient of the number written just before the last number, and the last number, that means it's going to be one divided by two, right? So that's what the question means. So I give you 10 minutes to explore this. You can continue, you know, the sequence, there are only 10 numbers. I do sort of two more numbers to make sure that we clearly understand the rule, right? So the next one is going to be 2 divided by a half, so that's going to be 4. And then we take a half divided by 4, so that's 1 over 8. So you can see that these are not, if you see 256 in 1024, this is 2 to the 8th, this is 2 to the 10th. But as you can see, the order of the exponents, this is a power of 2, but they are not increasing, right? They're not in increasing order, so just be careful. 2 to the 8th, and this is 2 to the 10th, could be here, could be here. And after a while, when you see that they actually only involve powers of 2, you may, you know, rewrite it using different notations to see if that will help us to find out the pattern, right? And then the next step is to find the pattern. It's all about pattern, right? So one way to write, to find out the pattern is that after I realize it's a power of two, then I would write a half is two to the power of minus one, and then two divided by minus one is going to be two squared, right? And then if we take this divided by this, you're going to have two to the power of negative three, right? And then this divided by this and two to the power five, and then the next one two to the power of negative eight. Hope I don't make a mistake. Andrew, if I make a mistake, let me know, okay? And then the next one, this one going to be two to the minus 13. If I do like that, it's sort of more or less I can see the pattern. And then this one is two to the power of negative 21. And the next one is going to be 21, and then it's going to be two to the power of 34, I think. Yes, and then the next one going to be two to the power of negative 55. And then just because I already wrote everything in power of two, I also go back here, and for one, I'm going to write two to the power of zeros, and two is two to the power of one, right? So that's one way we can write this way so that we can see the patterns. And yes, so yes, so this is in the second step is to find a pattern. So when I find, okay, I try to type this way in case you want me to come back and there's already everything here. So that'd be one, two, three, four, five, six, seven, eight, nine, 10. So that would be two to the power of 34, right? So answer E. Now after we look at the pattern over here, can anyone, maybe you can answer through the chat, what is the pattern for the powers of two here, the exponents here? What is the pattern? Do you notice anything? If you can, if you notice it, send it to the chat. So it's going to be zero, one. Actually, one, we don't really worry about that. If I, two and three, five, eight, 13, 21, 34, and 55, what's that sequence? It's a simple Nazi sequence, right? Over here, we have a, in our problem, we have an alternative side, right? Here we have a people not in sequence, but of course we have an alternative, you know, plus and minus. Yes, sum of the last two exponent difference. Yes, we, yeah, we, because, because we already decided it's been, so the quotient actually you can just take the sum, the difference. So yeah, or if you realize that in this number, once you realize that people now see like 34, 55, that's a part of the number, you should have to figure out the side here correctly. So that is very typical step. Do a few, do the first few example and find a pattern. Okay, so let's see if we can apply the same strategy and do question number two here. Andrew, could you please do question number two? The first three elements of a certain sequence are one, two, and three, and each element starting with the fourth element is obtained by taking the sum of the two elements before the preceding element and subtracting the preceding element. So the sequence is 1, 2, 3, 0, 0, 5, negative 2, 7, and what is the 2010th element of the sequence? after we find out the patterns it takes quite some time to find out so maybe Andrew you could you could read the numbers and I can type it up here so I have a record okay so we have that they gave us the first seven numbers of the sequence so we have one two three zero five negative two and seven and you can continue a few and then I and now we can add negative two and five and subtract seven we get negative four then we add negative two and seven and then we subtract negative four so we get nine and then from here we add seven and negative four and subtract nine so we get negative six and from here we can find a pattern okay so okay good thank you yeah so yeah so it's the first step to make sure that we verify the rule and then you know do more numbers and then see if we can find the pattern Maybe Andrew can give maybe one more hint. Okay, so something you can do to look at this is to try to separate the sequence into two patterns. So if we look at the odds and we look at the even numbers, we can see that they follow two different patterns. Okay, so if we look at the odd position numbers, we can find that there's a pattern where each odd position number goes up by 2, right? And if we look at the even numbers, we can see that they go down by 2. So because we know that 2010 is an even number, we're going to have to follow the even position, right? So let's try to find a way to express each even number as a term in the sequence. So we can see that the first term is 2, second term is 0, third term is negative 2. So we can find that each term is going to be 4 minus 2n. So 2010 divided by 2 because that's in the even sequence, so it would be 1005 times 2, so we'll do 4 minus 2010, and we'll get negative 2006 as our answer. So negative 2016, right, is it Andrew? Yeah, negative 2006. Yeah, so this one, right? So, yeah, after we kind of find two separate sequence, then as Andrew said, because 2010th element is in even position. So, of course, you can find a general formula, but if we look at that, we would see that basically the number plus its position is always equal to 4, right? If you add everything up, you can have a 4, 4, 4, so that we can get away with that without actually have a proof, because everything here is kind of linear. So since I also had added up to 4, so that would be negative 2016. So kind of a little bit of calculation here, because even after we get the sequence, I mean, these numbers are very similar. So minus 2006 is the answer. So these are some of the sequence, you can actually can calculate the numbers. Let's go for this one, okay. In 2010, balls are arranged in a row and are numbered from 1 to 2021. Each ball is colored in one of four colors, green, red, yellow, or blue. Among any five consecutive balls, there is exactly one red, one blue, and one yellow and one blue ball. And after any red ball, the next ball is yellow. So we know that the balls numbered 2020 and 2002 are green. What color is a ball numbered 2021, right? So for this one, as usual, I think we should summary. Okay, so any five consecutive, we need one red, any five consecutive, we need one red, one yellow, one blue. And then red and yellow, they are always together, like in this order. And 2020, 2020, both are green. So basically, we summarize this. Okay, by the way, when we say that any five consecutive balls are red, yellow, blue, that's what the question says. That means we need to have, you know, two green, right? Just fill in the details. So that to make up five. As I said, exactly one. I always like to draw a graph or whatever, list that thing whenever possible. So imagine we are arranging them in a row, and because we talk about every five consecutive bozo, you know, at least we consider a few multiples of five, and we should see what we can fill in the details. The second one is green, so I put green here, right, and then what would be next? Red and white together, so that means I could, you cannot put it here, it has to be in this five, right, so I could put red and yellow here, right, that would be the first step. And then I'm trying to do case work here, and green here, and red and yellow together, so it could be three, four, or four, five, right, so that's what it could be. So then do small step at a time, so we have two cases, and now in this case, we have a one and five, we need two, one green and another blue, so we could put either green, blue here, or blue, green here, right, so really we are just listing all the cases and make sure that we do not miss anything. And similar here, I have a green, red, yellow, and so I could have a green and blue here, or I'd have a blue, green here, right, I mean, some people can maybe just look at the numbers and find the answer right away, but I sort of like to do it this way, when I can actually see how the sequence looks like, and I make sure that I cover everything, all the cases. So give me a little bit more time to see, let me see which one it is. Remember, we're focusing on, we don't know the colors of other balls down the sequence, but we know the colors of the green. Does it give you any clue over here? Okay, good. Very good. I think you figured it out, right? Because, actually, every, because the NE5 contains color, and then we know that if we look from 1 to 5, they have two greens, one red, yellow, blue. So, but we look from 2 to 6, they must also have one red, one yellow, blue, and that means another green has to be in 6, right? And then we look at from 3 to 7, again, red, yellow, blue, so that means it's green here. And we see that this pattern actually repeats. So, if we figure out the numbers of the balls in the color of the ball in the first 5 position, it's going to just repeat itself in the cycles of 5, like forever like that. And we know that we need to use number theory here because there will be some, you know, divisor, you know, multiples of 5 and remainders, right? So, because we know that everything repeats after 5 balls, so we're going to look for, you know, the remainders of this number when divided by 5. So, this remainder is 2, this remainder is 2, and this remainder is 0. So, we are looking for a sequence that have that property, and this one, so everyone is 2 is good, that means if 2 is green, 202 is also green, but 20 is multiple of 5, and we look over here, there's only this sequence that has a green in the position that's multiple of 5, right? So, that would be repeated at number 20, and this one going to be repeated number 2002. So, this is actually the right sequence, the only one choice. And now we look at 2021, that has a remainder of 1 when divided by 5, so the answer should be B, right? Which is blue, this is going to be the color of the 2021. So, again, explore a few cases and, you know, draw things out, and case work and number theory. I think that's what we'll learn from these questions. Several points are marked on a line and all possible line segments are constructed between pairs of these points. One of the points lie on and is not at the end point of exactly 80 of these segments. Another point lies on and is not at the end point of exactly 90 segments. How many points were marked on the line? So this question, again, it's going to take some time to understand it. So 80 and 90 are very big numbers, so again, we're going to do a smaller version of the same problem. We're going to consider fewer points. So my first step would be to see, maybe I draw, I don't know how many, 3 is too small, maybe 4 and 5, right? We don't want to consider too few points. So for example, this is the point I'm looking at, and I would see how many segments. So for example, this point. This segment doesn't count, right? Because the end point falls right on this point, so it doesn't count. So actually, that's this segment that goes through this point, right? And then another segment here goes through this point, another segment here goes through this point, right? Is that all? Yeah. So in this case, for example, we know that the 3 segments go through that point. And if we look at this point, for example, this point, for example, there are, it's just a different color, this point over here. Then the first segment here doesn't count, right? Because that's the end point. So one segment is here, one segment is here, right? OK, but then I can also use one segment from this point to this point, and another segment from here to here. That's what I would do. Like, I have no idea what the question asked me. There's so many points. So I just work with a few numbers, and then you see what we should do here is that we're going to see how to relate the number of segments to the number of points, kind of, right? I mean, loosely speaking, that's what we are after. So I'll give you some time to explore this. Let me try to type it here. OK. At this point, I don't know exactly what the question I'm asking for, but kind of, right, along that line. I think this is a difficult question, question number 30. So, for example, over here, for this point, I have a 1 point on the left and 3 points on the right, right? For this point I have a 1 and 3. Because we realize that in order a segment to be counted, it has to be, okay, so, okay, sometimes it's actually very useful to describe it in words, right? For a segment to be counted, right, the two end points, okay, it's two end points, it's two end points, have to be on the opposite side, with respect to, to the, to the, to to the point we are considering, right? So if you look at this point over here, there would be, for example, point A, we'll look at point A. Point A, on its left, right, on its left there is one point, and on the right there are three points, right, and we have three segments. Segments, do you see that? For the first point here. But then for this point here, for this point B, point B and B in the middle have two points on its left, and then you have a B, and you have two points on its right. And that's why we have to connect these two here with two here, then we have four segments. So hopefully we see the, we see the solutions here, sort of, kind of, you know, some, some clue here, because now we have to use algebra, right, to sort of generalize it, because I know that if I have, so after I do that, I know that you have an end point on the left, right, end point on the left, and then the point A here, and end point to the right, then the number of segment will be, well, it's just the number of segment that have A as one of the points, that going through point is just m times n, right, and in this case we have one times three or two times two gives us four, right, and then in this case, in this simple counting, we have two points, so another point going to be, you can call it A and B, or x and y, maybe I call A and B, I don't know, x and y, and then we have a B in the middle, and so the number of segments that go through point B is x times y, right, x times y, so now you have two equations, m times n is equal to 80, that's number of points, we don't know where they are, m and n, but we know that the number of segment is 80, and another one is 90, right, and then the total number of points is actually going to be m plus n, m plus n is actually equal to x plus y, and this equal to the total number of points, and minus the points that we are considering here, right, one, so I think from here you could figure out the answer, because this is kind of golden, right, these are integers, when we have a sum and difference, that's not very helpful, but when we have a product of integers, that's very helpful, because we can do the reversible set of factorizations, and there are not many ways you can factor a number, right, so take a few moments and see if we can figure out the values and in x and y. Basically we have two pairs of numbers. For the first pair the product is 80, the second pair the product is 90, and the sum of the two numbers in each pair is equal. The two sums are equal. you know, it's good to find products, right, so we have 80, we can do 1 times 80, 2 times 40, 3, 4 times 20, is it 5 times 16, right, something like that, 8 times 10, and I think that's pretty much it, 8 times 10, so we do that, and 90 is 1 times 90, 2, I think it's kind of big, so 2 times 45, 3 times 30, 4 don't have but 5 times 16, oh no, 5 times 18, okay, and then 6 times 15, and then 9 times 10, and that's pretty much it, right, so if you look at the way we factorize 80 and 90, and the sum are the same, these are not, 80, 41, then actually we have two pairs here, 5 and 16, and 6 and 15, right, 5 and 16, and 6, 15, and the sum is 21, so what would be the answer? And we would have, you know, point A over here, that would be 5 on the left, and 16 on the right, this is point A, and point B, this right, on the right point A, it has a 6 point on the left, and 15 points on the right, and then if you connect 5 here with 16 over here, we have 5 times 16, that's 18, 80 segments, and we have a 6 on the left connected with 15 on the right, we have 19, and that would give us these numbers, and the total number of points is 5 plus 16, but don't forget the black point over here, so that's 22. So I admit this is a difficult question, but it just shows that if we, you know, spend some time exploring it, find the pattern, and generalize it over here, we kind of, you know, could get the answer. It could be long, I know, could be more than five, six minutes, but yeah, it is doable. Okay, so I've written here in case we need to come back and ask any questions. Let's go for the next one, that's a little bit, yeah, less computation. Okay, this one I'm sure you can figure it out, it's less computation. A graph consists of 16 vertices and some edges that connect them, so here, as shown in the picture, an end is now at the vertex labeled A, and in each move, it can walk from one vertex to any neighboring vertex, crawling along a connecting edge. So at which of the vertices labeled P, Q, R, S, T, can the end be after 2019 moves? So we're gonna go, for example, go to T, they say that it goes, a connecting edge, right, so it can go like this, like this, to get to T, or it can go zigzag, I don't know, all the way here, get to P, or like this, and like this, also get to T. We think it's allowed to go on any edge to get from point to T, and again, 2019 is a big number, so we know we have to find a pattern, because things are going to be repeated. Okay, so I will just be quiet here and give you some time to figure out this question. So if you want to draw it, you know that one, two, three, four, five, six, seven, eight, right, so it's an octagon, you could, you know, draw a circle, two points here, two points here, two here, two here, so that's eight, right, so make sure it's eight and not six, or whatever numbers, and, you know, quickly draw something like that. It doesn't have to be perfect, but just make sure it has eight, it has seven, eight sides. Yes, the ant can go where it has to be. It says, no, I have a question. Yes, it can do so. It can go, you know, it can choose to go from A to here and then go back and forth. It can just do it as many times as it wants, and then at some point it's tired and then it wants to go directly to the destination. But yes, it can do so. So, the first thing you might want to do is just, you know, go from A to each vertex to see, you know, because they talk about 2019 moves, then you could see like how many moves does it take for the ant to go to any of these vertices, right, and see if you can see any patterns. first we have to find a pattern or sort of suspect the pattern and then we try to prove it. You could go by the shortest path for example, right? Okay, good. Good, I think you're starting to figure things out. So I'll give you a little bit more time. And if you already figure out the answer, you have a feeling that is the correct answer, you can sort of go a step further and try to really like prove it, why is this the case, right? So for example, if I go from A to 2, right? I just go 1, 2, then I just mark, like in the exam I said, oh it takes me two steps to go to two moves to go to T, so I just mark 2 here. I go to S, right? I go 1, 2, 3, 4, so I mark S is 4, and Q is like 1, this is the shortest one, good one, yeah it's opposite, so 1, 2, 3, 4, 5, right? If I go 1, 2, 3, 4, so 5, and then for P, 1, 2, 3, 4, so P is 4, and R I would go here, 3, 4, 5, 6, okay? Yes, so after you just like, you kind of figure it out that, you guess that there's one odd, one odd, the odd one now is Q because that's odd numbers, right? Everything else is even, and if you're suspected, say like, oh maybe I don't go directly to Q, I go from here, I go to the outer one, 1, 2, 3, 4, 5, 6, 7, I go more but still 7, and you can sort of try a few cases and say that for Q it's always odd, and for everyone else it's always even, and 2019 is like fittingly odd, so in this case kind of we guess that the answer has to be Q, and we're using odd and even. I just quickly show you a proof that this is the case, right? So one way to do that, it's very common to do that once you've done a lot of questions with parities that we can mark them as, you can mark it plus 1, minus 1, or 1 and 0, basically all you wanted to mark is just the difference between odd and even, and if we mark it this way then we can convince ourselves that every move that it takes, right, it has to go from one vertex to the other, every move that it takes it has to change from minus to plus, or plus to minus, right? There's no other way around, you can look at all the pictures that only finite number of moves, and that means this pattern has to be odd and even, and that shows that it has to be Q, and no other ways. It can go, you know, from Q and this and back and forth, you know, many times, is it one, two, make up 2014 divided by two, you know, that many to get to Q, and for everything else it's impossible because it's the alternative between plus and minus, and that's the kind of more formal proof, but if you already realize the pattern in the exam, yeah, just move on, you know, that it has to be Q. Okay, so I think that's actually a very nice question, and not too long for the exam. Okay, so let's go to the next one and find another. Okay, so I think we have time for this one. Which of the numbers n in the set from 1 to 9 have the following property? It is possible to paint the unit squares in a 5 by 5 square in such a way that in each 3 by 3 square exactly n unit squares are painted. Okay, so let's try, let's quickly draw, I mean, for this way there's no way out, we got to draw it, and you just nicely draw a 5 by 5 square, 5 by 5 square, I mean, kind of make it more or less square. by square. And now the question asks that if you want to paint all the three by two square, I have a lot of three by two square here, right, like this one, one of that, another one here, for example. So it's getting an idea of how things look like, no need to fill in any, yeah, like those are three by two square. So we want that in each of three by two square, exactly n units square are painted, right, so n equal to 1, 2, 3, 4, 5. The reason I want to show this question is that we have a strategy called eliminate the wrong choices. This question would be, would take a lot of time if you do not have the multiple choices, but you know, they have mostly for students in the exam, so they do actually put the numbers here, the choices, in a way that you can use that strategy. Eliminate. Sometimes it's not obvious, but like sort of in a way that we can narrow down the options in a sense, right. So n equal to 1, 2, 3, 4, 5, 6, 7, 8, 9. Just pick whatever n that's easiest for you, right. So I'll give you a few minutes to find out what n's are easy to paint, what values of n's are to make it easy to paint the the five by five square, and such that in each n, a three by two square, they're exactly n units that are painted, right. Some are very difficult, some are much easier, so let's explore that for a few minutes. Or if you want, you can type to the chat what values of n you think are the easiest. Oh, yes, I will just make an example of the 3 by 3 square, but you have a lot more. So in the, you have a 1, 2, 3, 4, 5, 6, you have a 9 3 by 3 squares. I was just showing a few examples, but in total we have 9 3 by 3 squares as a, you know, sub-squares of the, these 5 by 5 squares. So focus on easy value of n first. Oh, okay, so let's go with the clue 1 only, right? So n equal to 1, someone showed that you can do 1, right? And actually you know where that 1 is? You have to paint the unit square, because all these square, all these 9 squares, they share the center square. So if we paint n equal to 1, that would be easy, n equal to 1, that could be painted, right? So that is easy value. But actually this is not the easiest value. What is the easiest value? The easiest value is actually n equal to, yeah, so I think I give maybe a minute to and you could figure it out. You know, 2 and 3 are hard, right? I would not do these, look at the n equal to 1, that's a special case, and actually n equal to 9 is the easiest, right? Because you paint all squares are painted, you paint all 25 squares are painted. If 25 squares are painted, then for each 3 by 2 square, all 9 squares are painted, right? So these are actually the easiest. And then the next one is n equal to 9, right, n equal to 1. So if I do that in the example, I could already eliminate A, eliminate B, eliminate C, right? And now I'm left only with D and E, and now I just have a little bit of, you know, logical reasoning here. D is subset of E. D is subset of E, so for example, I check 2 and 7, that's not very helpful because if I can do 2 and 7, it could still, you know, I still cannot eliminate if E is a choice. So you have to want to pick another number that's not 2 and 7 that's easy to paint. And then you can sort of eliminate D and go for E, something like that, or it doesn't work, then you don't know what to do next. But at least, you know, don't try 2 and 7 because it really doesn't help us to eliminate D or E, right? So try another value of n. 2 and 7, which value is that? And I can tell that the hint is right on the board, right, in the two values that you are looking at here. Okay, good. I give you a look like some already figured out. I give another 30 seconds to see the class figure it out. And it will feel like very satisfying once you figure it out. Like this, just these two simple cases help us eliminate A, B, C, D. And another simple case, not so simple, but kind of is going to help us determine between D and E. So if you notice that when n equal to one, that's the one that one central cell painted. So if I reverse the situation, right? When the central cell is not painted and everything else is painted, what is that value for n? Let's say all cells are painted except for the central cell. And that would actually be n equal to eight, yes, the complimented one, right? So you have a one and eight and nine that has a value and then you can confidently chose E in the exam at the correct answer. Right, put it here. So that question is really, this question really to elaborate the strategy, you can narrow down, lots of narrow down the option without having to do a lot of work. And it's a long question, so I just wanted to show you the example when you actually chose to paint n equal to two. If you do n equal to two, then you can do n equal to seven by reversing the painted and the unpainted. n equal to three, that mean you can do six, n equal to four, you can do five. So you can only need to do two, three, four, but it's very long. Likely we won't be able to do that in the exam, right? So, you know, but this strategy works well. So in the end, I just want to sum up that for, in this questions, in this lesson, we talk about like we see a lot many different sort of unconventional sequences and pattern. And these problems are very good place to practice knowledge of algebra, of number theory, and sometime out of the box thinking is kind of a lot of question, kind of fun when we think back about that. And then in terms of problem solving strategies, so smaller number, a version of the same problem, pattern, sometimes we actually can find a formula, trial and errors, some very useful problem solving strategies. And we all did lesson number four, and I really strongly encourage to do past year exam. You probably you have started already, but you haven't tried to do it every day. As you can see, there are a lot of questions. We need to be very quick with arithmetics, right? Like 80, 90, those numbers should be dancing in your head. Like we would not have time to, you know, do that in the exam and to start from scratch. So make sure that you maybe do it every day for 30 minutes, 45 minutes. That's much better than, you know, sit in one sitting two, three hours a week. So do it every day so that you remain sharp and, you know, treat it as something fun, something different from school. So that would be my advice and advice that I had from a lot of other students. You know, in higher grades, they all the way say that, yeah, do past year exam is one of the best way to prepare for that. And while you are doing that, you can review the concept and especially you can think about problem solving strategies.

mathematical problem-solving

patterns and sequences

algebra

number theory

greatest common divisor

arithmetic sequences

Fibonacci sequence

proof by contradiction

parity

problem-solving techniques