solving strategies, and today's topic is algebra. We make sure we cover all the basics so that in next class we're going to dive into word problems. So algebra is a big topic, but I just have a few slides summarized what we need to know for math kangaroo competitions. Definitely we want to know system of equations, and the common techniques for solving equations or system of equations are isolation of variable, right, substitution, one in terms of the other, eliminations, keep our eyes on the ball, right, whichever we are after, so we can eliminate others to get to the answer. And if we have a linear system of equations, then we are, you know, in good shape. Usually the number of equations is equal to the number of variables, and then we can use the linear combination of the equation to solve for the variables. For example, we have three variables here with three equations, then we notice that in the third equation there are only y and z, then we might try eliminating x in the first equation, right, and we do so by multiplying equation one by three, equation two by two, subtract one from the other, and we get an equation number four that has y and z only, and now from three and four we can solve for y and z, right. For example, we can just add them together to eliminate z and then get to y. So all these manipulations are called linear combinations when we multiply both sides of one equation by either constant, we don't want to multiply by any variable, right, because then it won't be linear anymore. So as long as they are linear equations, pretty much we are sure that we can solve for the variables, no problem here. But once, you know, we are out of the realm of linear equations, things get much more complicated once we are in the non-linear domain, right. So for quadratic equations, there are formulas for the roots, but most of the time I think we can just get away with completing the square method because the coefficients are either integral or very nice fractions. So for example, have an x square plus six x equal to maybe 20, and just completing the square, then we add nine to both sides and have an x plus three square equals 29, right, from here you can get two roots. Definitely, I don't know, most of the time I don't see a lot of questions with cubic quadratic-quantic equations. Most of the time you, I think probably you can start with quadratic equations. But if you, you know, encounter linear equations then substitution, elimination are still very useful tools to solve for non-linear equations. And algebraic factorizations, right, it's very useful tool. If we can factorize an expression, then we can find the roots, right. For example, here we have a cubic equation, we notice that x cube minus two x is x times x square minus two, which is the same as this term, right. So we can factorize it into two terms and two factors, and from here we can easily find the three roots of this cubic equation. And also we don't use a lot of this to solve higher order equations. There's some common used identities that make use of algebraic factorizations, definitely you want to know this formula, n square minus one is n plus one times n minus one, right, when b equal to one. And there are more of these identities, I'm sure you encounter it, but at least enough to know some of the basic ones. And then arithmetic sequence and geometric sequence, I think you also expected to know, we will have one lecture dedicated to all kind of sequences and patterns, and we come back to these sequences. So when we solve linear or non-linear equations, any algebra problems, very important that we know how to choose the variables effectively, right, we choose them wisely. Often there are, you encounter what problems, there are many different unknowns, and we want to find the relationship between them, we want to decide, you know, how to give variables to which unknowns, maybe you don't want to give variables to all the unknowns because they are related in certain ways. So a lot of time we want to choose fewer variables than the unknowns, if they are related. And then we're making use of symmetry, this is something that we're going to work in several problems in today's lessons. So if we carefully observe the structure of the equation, especially when we have, you know, many unknowns, not one or two, but many, then look careful as a structure of the equation to see if anything special is something that we can make use of to get to the answer quicker. And definitely inequality, this is a very important topic, especially in equations that have constraints, right. For example, here we have two unknowns, 3x plus 5y equal to 20. So only one equation, but we have the constraint that they are positive integers, and then we can use inequality to put upper bound and lower bound on y, and from here we can only need to solve for three cases of y, and from there we can find x. So we're going to go through all of this in today's lesson. Okay, let's get started with problem solving. Four positive numbers are given. We choose three of them, calculate their arithmetic mean, and then add the fourth number. This can be done in four different ways. The results are 17, 21, 23, and 29 respectively. What is the largest of the four given numbers? So we choose three of them, we can choose three of them, and arithmetic mean, right. We have a geometric mean, but most times when we say mean, it means the average, the arithmetic mean. So four numbers, in this case I think we just have to give four variables, one for each unknown and write down the equations. So you can write the first one if you put ABCD as three, four unknowns and the arithmetic mean of the first three is this and plus fourth one equal to 17. I got a few answers, so just continue this way. I have D, so I can start with maybe C. We can do it together and see where you can get. And then, this is not very good, but A plus B, no, but C plus D over 3, B equal to 23. Not very important, but I like to get them into certain order. So D, C, D, and then we have A, last one, 3 plus A equals 29, right? So if we look at these four equations, there's some certain structure on the left, on the right is symmetry, right? They pretty much look the same. We can exchange the rows. Of course, they are not the same on the right-hand side, but at least we can, you know, make the symmetries on the left-hand side. And one way to do that makes use of this property that we should add all of them together. We add them together, have a third of A, third of A, third of A, so I would have one A here, and then one whole A here on the right. Therefore, I would have two of each variable, A plus B plus C plus D equals to here, and, you know, adding 21, 29, 50, 17, 23 is 40. So that gives me 90, and then from here, I have A plus B plus C plus D is 90 over 2, which is 45, right? Yeah, I got, yeah, we, as I'm doing it, some of you got the correct answer. So once we got this, we start substituting this into these equations, right? If you look at this, which one you think would have the highest, the largest of all of them? It's probably A, right? Because it puts the most weight on it. The most weight is in the coefficient, in this number here, outside of the arithmetic mean. So this one is big, so we only focus on these equations, what we need to look for the biggest number. So if we put in here a bit more manipulation, I put a third of A there so that I can make use of the sum. So that last equation is equivalent to A plus B plus C plus D divided by 3, and then I have a two-third of A left. A equal to 29, and that sum divided by 3 is 15 plus two-thirds of A equal to 29. So two-third of A is 14, and from there, I have A is 21. So this question number 22, please state it in a five-points question. You know, it involves some calculation in the keys to recognize these structures so that, and then do the computation quickly, right, so that we can get this answer quickly. Yeah, thank you, David, for putting it here and adding multiply and symmetric equations. When you see symmetric like that, adding is a very common manipulation. Okay, good, let's go to the next one. I'm sorry, David, you have with this question, yeah? All right, so I chose to actually do this question because I remember doing it in the actual computation. And when you see how to do it, you know, this is, it's really satisfying, I'll just tell you that. Oh, okay, that's how it's done. And I remember in the real competition a couple years ago, this specific problem, so that's why I wanted to choose it. So the numbers A, B, and C satisfy A plus B plus C equals 0, and A times B times C equals 78. You know, there's a lot of things that come to your head when you see sums and multiplications. Just don't overthink it. What is the value of A plus B times B plus C times C plus A? So this is also a five-point problem, so just, you know, think about it and try to find the manipulation you do here. Let me just launch that. All right, so it seems most of us have answered. I'm going to start going slowly through the explanation and give anyone who still wants to answer a chance because it's really, it's really a snap problem. That's what I also saw with the patterns of the answers choices, you know, there were immediate answers and then a pause because it's kind of about whether you see the trick or not. And again, I remember doing this problem and I had a great time with it. And so the general idea is this is really a problem that exemplifies wishful thinking more than anything else. You have these two things and you know that you want to turn it into this somehow. And you know, your most fervent wish is that somehow this simplifies into something related to these two things, right? We want some relation here. And it's actually funny because we also have what looks like a bunch of symmetric equations. If you think about it, you have A plus B plus C, that's symmetric. You have ABC, that's symmetric. You have sums of pairs, which is also symmetric. And what we did last time we had a symmetric problem was we turned A plus B and B plus C and C plus A into A plus B plus C. That's essentially what we did for the other average. So what you realize if you look at this long hard enough and think about what you know and what you want is that A plus B is just A plus B plus C minus C, right? Why am I doing this? Well, because we know that A plus B plus C is zero, right? So it's just negative C. So actually you have negative C as your first term. Similarly, B plus C is just negative A because A plus B plus C minus A, right? And this is negative B. So the answer is negative C times negative A times negative B, which is just negative ABC, which is negative 78. We had a couple of 78s, which makes sense. You see, oh, it's A times C times C, 78, bam. But actually, and this is a bit cruel, I was mad at them when I took this. The answer is none of these, just because you have the negative product. So you actually have a negative answer. It's not 78, it's negative 78. Negative 78. But yeah, that's how you solve this. Really the lesson is, it's wishful thinking, because I know it kind of looks like magic if one didn't have the idea to realize that A plus B is A plus B plus C minus C and so on. But what you're really doing to find that is you really want to turn what you want into what you know. And you really want to compare it to the two equations you've already gotten. That's really what leads you into finding the correct solution. Yes, great. Thank you, David. Yeah, usually they don't test you on very heavy computation, like you have cubic equations or higher orders, and maybe it's time to think of some other trick to simplify it. Next one. Kanga is writing 12 one-digit numbers in the cells of a three by four grid, so that the sum of every row is the same and the sum of every column is the same. Kanga has already written some of the numbers as shown. What number should she write in the shortest square? Yeah, so we can read it quickly, but I usually read it a second time, and sum of every row is the same, and sum of every column is the same, but these two sums may not be the same, right? So when we read it too quickly, we may assume all of them are the same, but no, the columns are the same and the row is the same. These might be different. So we did symmetry, we did substitution. Over here, this question, um, we have five, six, uh, five unknowns, right? So this question is to test, you know, what is, what is the right choice of variables? We have five unknowns. We have to need, only need to find one, not all of them. We can, in principle, put all five unknowns, right, and write out all the equations we can write. Is there a different way to do that? Or you can just put unknowns and go and see how it goes. Okay, so for a minute, let's do it together. Yes, as I said, we have five unknowns where you can put, write down three equations, three sums for the columns, four sums for the, I'm sorry, three sums for the rows, four sums for the columns, equate those columns and those rows, and figure it out. That definitely we guarantee to get the answer, it's just a little bit too long, and we'll see what else can we do over here. Let's start with one variables. In principle, you can call this as one variable, but when I look at this, for example, I go with A, I mean, usually we use X, but sort of a little bit confusing with multiplication, I use A. If we go A, it's hard to write the equation because both the rows and the columns, we are missing other numbers. So my first thought would be put A, the variables we are after, but maybe not the best choice. So I choose this one as A, because over here, I already know the other two numbers. So if this one is A, then I can find out the sum of this column in terms of A, right, which is A plus eight. So if this one is A plus eight, and I move to the next column, then I can immediately fill in this square, the cell, which is A plus one. And then the nice thing that I move to the next column, again, I can fill it in this A plus four. And that is really nice because we sort of translate the, you know, the constraint that the sum of the rows are the same and the columns are the same into, you know, the variables. Here they are related, they are not independent, right? They depend on each other, that's why I don't need all five different unknowns. With just one variable, I can represent others in terms of it. Now I don't know what to put here because there are two missing cells, right? So one thing I can do is that now I turn into the row. So this row, I know, I add everything together, I would have A plus A plus 12. Yeah, if I make a mistake, please correct me, David. So A plus 12. And now that's nice because this one also A plus 12. So I have all the A plus six, and I know right away this one would be six, right? Because that's also A plus 12. And this one I know is A plus 12. So that's A plus eight. So I need four here, right? And once you do that, we know that the sum here is equal to six, 12. And from here, you can also figure out that A equal to four. That means you find out all the missing numbers, not just the one in the shaded. So yeah, this is an example of how we can get away with writing a bunch of equations. And I mean, there's tons of problems like this in Maths with Kangaroo, but just in general, any words problem, it's good to translate the relationship, how they are interrelated to each other so that we don't have to deal with like so many equations and so many unknowns. Okay, good. In the end, yeah, we got actually 75% got the correct answer, which is great. Okay, go to the next one, another questions with this kind of diagram. We have the numbers from one to six are placed in the circle at intersections of three rings. The positions of number six is shown. The sums of the numbers on each ring are the same. What number is placed in the circle with the question mark? So we have a lot of constraints here. The numbers are from one to six only. We only have a six, so just one, two, three, four, five to put in here, such that the sums of the numbers on each ring are the same. We are asked to find out this number. We have only one number known, so probably we need to put variables for each of the circles here, right? It's hard to calculate one in terms of the other. and not too many equations so you can Let's try first to write down the equation to see what we can learn. Sometimes the direction is clear, sometimes there's no clear direction which is how to start from somewhere, and hopefully by examining the equations, I can lead us to some good insight. So, okay, so let me write, so David will put it A, B, then I put A, B, and C, D. And remember to make use of the constraints the numbers are from. Without those constraints, we can see that we have too many variables and not enough equations. We cannot solve for that, but that's when the constraints come into play. The numbers have to be from 1 to 5, because 6 are already used. OK, looks great. I think we've got some good answers here. So let's do it together. We have more interesting questions down the road. So I just named variable a, b, c, d. And the question said, the sums of the numbers on each ring are the same. So I can start by writing. We have three rings. So the first one is a plus b plus c plus d, right? And equal to the second one is 6. This one over here, I write it as 6 plus c plus question mark plus d. And the last one is this one, b plus a, b plus question mark plus a plus 6, right? So now we have only two equations. What can we do? We can extract from here to simplify it, right? Simplification is one manipulation as well. So if we consider only these two, then we have a plus b equal to 6 plus question mark, right? And if we consider these two, 6 plus question mark, and that would be 6. Yeah, if we do this one, then we have a plus b equal to c plus d, right? We consider two. And that's great, because we can actually have six numbers, but we break them up into three pairs, right? a plus b, c plus d, and 6 plus question mark. And that's golden, because we have this constraint. And we know that the sum of all is what? 1 plus 2 plus 3 plus 4 plus 5 plus 6, which is 7 times 3. That's 21. So that means each pair has to be equal to 21 divided by 3, which is 7, right? All of these have to be equal to 7. And from here, we know that the question mark has to be 1. And a and b, you can make two pairs, like 3 and 4 and 2 and 5, and we can double check. Probably it doesn't matter where you put them as long. You can exchange the, I think we can exchange the orders of a and b and c and d, because as long they are 3 and 5, and 2 and 4 and 2 and 5, they satisfy the equations, then we meet all the conditions. So that's where 60% of you got the correct answer. Again, it's also a case of symmetries and constraints. Always, when we don't have enough equations, think about the constraints. Exactly, yeah, the symmetries that Dave said in the, he typed up on the screen, because of, you know, even before writing down the equations, you might just examine the symmetries and sort of make a guess. And also, you can double check with writing the equations. OK, thank you. So the next one, David. Let me just start the poll right now before I forget. All right, all right, awesome. So this is another problem. I wasn't doing 9 and 10 back in, back, OK, wow, back then, but I remember doing this as a practice problem a while ago. So I also have, I have good memories of this problem. After one of the numbers from 1 to n was eliminated, the mean of the remaining numbers was 4.75. What number was eliminated? So it's a very short problem statement, but think about it. There's a lot in it, right? You have the numbers 1 to n. You have no idea what n is. And you take away a number, leaving you with a certain mean, which means you actually have two unknowns, really, in this problem. So I'll leave you to it, and let's see what you come up with. This is problem 27. We're approaching the end. And I do remember this as being a trickier one than average. So do your best. Try to find whatever solution works best, and we'll reconvene in just a moment. The nice thing about this question is that we can, once you find out the answer, you can easily check, right, if it's correct or not. So yeah, we already have a few answers, so we can double check if your answer is correct by doing the calculation. If you really want it, I suggest you don't. You could just say that the number you eliminate is each of these things, and then see what happens. And that gives you a good idea. It doesn't solve the problem, because maybe the answer is impossible to determine, because there's an infinity. But if you get that only one of a, b, c, or d works, you could just do that, and in competition it works. But let's try to solve it for real just this time. Oh, maybe we can give them the formula, David, like if you have, I mean, we talk about, I mean, if you need the formula, we talk about the arithmetic sequence, right? This is the simplest arithmetic sequence, and if you sum up, I think you probably remember by heart by now, but if not, this is the formula, n times n plus 1 divided by 2, and n times is always even, so. Yeah, the famous Gaussian sum. Now, what I'll actually talk about in my explanation is that we do not need to rely on this formula as much. Oh, you don't need to, okay. Yeah, I mean, that's a way you don't need to rely on this formula to do this, yeah. It's very good to know, and it's very good to have memorized because it comes up commonly, but it won't actually be used as much as it could possibly be used. I'll give about 34, 30 more seconds. This is a five-point problem, so there should be more time, of course, but nonetheless, two minutes and a half, so, you know, in a bit, it's been a while, and we'll start going through the solution together. Also, pay special attention to 4.75. Yeah. Did we talk about, no, we didn't talk about that, right, but you might try to remember about terminating decimal and repeating decimal because we need to define the mean of the remaining numbers, so, yeah. That's an interesting way of looking at it. All right, so let's begin. There are a couple tricks that go into this problem. I will tell you because, again, I did this problem. I remember my experience when I solved it, and the first thing that I was thinking was, okay, 4.75 looks a lot like five, right? It's very close to five, and the reason that's important is because the mean of the numbers one through nine is just five, right, because, you know, one, nine, the middle is five, right? One plus nine over two is five, two plus eight over two is five, and, in fact, really using this idea, this is the reason that the Gaussian sum works as it does because you take the first and last numbers and the second, second to last numbers, so, anyway, the point is that when you notice that the median is about five, what you immediately start thinking is that maybe the n has to be close to nine just because if n was one billion, the mean would be way bigger than four, and if n was three, the mean could never be four because none of these numbers are as big as four, right? So, we want to, we want a constraint on n. We want to figure out just about where n can be, and as has been talked about in one of the first slides, a common way to do that is using inequalities. So, let's look at what it means to take the numbers from one all the way to n and take something away, right? Now, normally, if we were using the Gaussian sum and we did algebra, say we took away the number k, we'd get that. Let me actually type here. If we were trying to do algebra on n and k, we'd have n times n plus one over two minus k would be the sum of all the numbers, and the mean would be that divided by n minus one, but that's a little ugly. So, what we're going to do instead is we're just going to look at the upper and lower bounds of the mean. We're going to say if we take away the number one, then we're left with just the numbers two to n, and their mean is obviously n plus two over two, right? Because you have the first and last number, and then you divide by two. You don't have to take the sum of all of this and then divide by n minus one and do quadratics. You just need the first plus the last, and then you take the average. On the other hand, if we took away the number n, you would have a mean of just n over two, because you have one plus n minus one over two, right? n over two, which tells you, and this is very important, that for it tells you that for any n that we pick, the mean is in between n over two and n plus two over two, right? So again, you know, you're using inequalities. It may seem kind of magic, but the idea is I already know immediately, just by looking at the problem, that n is somewhere around nine. I want to figure out exactly what n can possibly be, and so we just multiply by two. Here we get that n is less than or equal to 9.5, which is two times four from 75, is less than or equal to n plus two, which immediately tells us that n is eight or nine, right? So now that we've done this, now that we have figured out what n has to be, and we didn't even do that much algebra, really. In fact, when I did this problem, I sort of just looked and eyeballed and went sevens too small, tens too big, and nine. But with this basic idea of bounding the mean, now we just have to basically try both. So if we try both of those cases, here's what we get. If n is eight, and we take away the number k, we have one, we have a sum of one plus two plus three, all the way to eight, minus k. So that's the sum of our seven numbers, divided by seven. And that gives us the mean, which is 4.75. If we try and solve this, we can go ahead and we find that 36, I believe it is, minus k equals seven times 4.75, at which point it's pretty obvious. Seven times 4.75 is not an integer. So then k will not be an integer, which means n cannot actually be eight. We need to eliminate an integer k. So then instead, we try nine. And very similar strategy, we have all this, we divide by eight, equals 4.75. We find that 45 minus k equals 38, just by multiplying by eight on both sides and expanding, which means that k equals seven b, which is the answer. So there are two steps to this problem. The first one is to try and find n, because n could be anything, from zero to infinity, and we don't want to worry about that. But because n is only eight or nine, we can literally just try both cases and see what happens. And we get seven. Thank you, David. That's very neat, by using this very smart, very neat to use these inequalities to bow and to become eight and nine. Great, thank you. I think let's do, yeah, I think this question probably should take maybe three minutes and we have time for another one. Let's do it quickly. By symbols BC and ABC, we indicate respectively a two and three digit numbers with C as the digits of ones, B as digits of tens, and A as digits of hundreds. Let letters x, y, z indicate different digits and let the sum of the numbers x, x, y, y, z, z be equal to the three digit number z, y, x. Then the letter x represents a digit. So basically, very long questions, but if we are familiar with notations, there's nothing new over here. This z, y, x is just a three digit number. And these are, they didn't say two digits, but could be equal to one. I mean, actually, it could be just zero, but at least we know that these are digits. In general, we can use like AB, we call it the generalized form of two digit numbers. ABCE is called the generalized form. And now, so what we do is that we add up all of them. I just type it here. So we add up x, x plus y, y plus z, z equal to z, y, x. And x is a number, so it's like maybe 11, y, y, 22, something like that. And z, y, x is three digits. This is important. It's a three digit number. That means z cannot be equal to zero, but x, y, z, they are from zero to nine inclusively, but z has to be different from zero. Z cannot be equal to zero. So these are the constraints. Again, only one equation, the constraints are here. So what did we do just now? We have to use lots of time inequalities to be our friends, right? But try as much as possible how to simplify this equation first. Simplify as much as we can before we use the constraints. It's good, yeah, we already got some answer and some correct ones, so let's see what we can do. x here is, they say, it's a tenth digit number, so it's actually equal to 10x plus x, right, which is 11x. It's a different way to write x here. And then if we put everything together, the left-hand side is actually equal to 11 times x plus y plus c, and the right-hand side, we kind of expand it, it's going to be 100z plus 90y plus x, and then we just have to subtract x and y on both sides to simplify as much as we can. On the left-hand side, we have a 10x and 10x, and this is plus, is it 89? 11z, so 100z, so, oh, and 1y, oh, I'm sorry, thank you for correcting me. This one is actually just 10y, so it's 10x plus 1y, and this one I have to subtract 11z, so we have 89z. I think that's it, right? And that's where the constraints come here, because z, this is so big and this is so small, this is at most 99, so z, and z is not zero, so z has to be 1, right? And z equal to 1, we have a 10x plus y equal to 89, so the only answer is x equal to 8 and y equal to 9, so the answer is d, which actually, yeah, 50 percent of you got, so that's great, yeah. Common strategy, simplify the equation as much as you can before using the constraints. I think we are doing great in time, and hopefully I can do another question. Yeah, very common for integer equations, for equation when the coefficients are integers and the what is it, the variables, yeah, also have to be in integers. So what is the smallest integer n for which the number is a square of an integer? So again, some common manipulations that we discussed earlier. So it feels daunting, right? Because you have to multiply all of this until n. We can try, but the number, you know, grows quickly. What if it's not six, not eight, but it's like 27 is more or less, You might think of one way to factorize this, right? N minus one square, we can factor that. We have a formula earlier. Oh, good, some of you, I think, probably recall the formula and get the correct answer. N square minus one. Remember, this is a difference of two squares where you can write as N minus one. Times N plus one. This might help. Might not, it's not the only way, but it might help you get this question. So try to expand everything to see if you can find any pattern, any general formula. If I do the first one, I would have 1, right, times 3, and then the next one would be 2 times 4. Let's just write down the first few terms just to get an idea of how it looks like. To me, this form looks better, right, minus 1 than the original form because we can see some patterns over here, 1, 3, 2, 4, 3, 5, n minus 1, n minus 2, and if we sort of rearrange the factors, we could have 1, and 2, and 3, all the way to n minus 1 together, and then we can group 3, and 4, and 5, and all the way to n plus 1 together, right. So the first one would be, the first one would be, the first one would be 1 times n minus 1, which is actually n minus 1 factorial, right, and the second factor is from 3 to n plus 1, which is n plus 1 factorial, but we have to divide by, factorial, but we have to divide by 2, right, because we didn't have a 1 and 2 in the first place. So if we do this, we expand it, we would have n minus 1 factorial, and the whole thing square because we have two of them, but then we have two actual factors here, n, and then we have 2 times times, I have to write here because it's the same line, times n, and n plus 1 divided by 2. I want this number to be a square, and this one is already a square, right, so the left over here must also be a square, you have n, n plus 1 divided by 2 must be equal to a square number, and if you look through here, it's a trial and error with 5 answers, 8, 8 times 9 is 32, 72 divided by 2, 36 is a square number. So actually, yeah, okay, good, so some of you got the correct answer. I don't know, sometimes, yeah, none of these, not very popular, sometimes it's the questions that they present that actually none of these is the answer, but sometimes when questions say it's impossible to determine, actually I haven't never seen any questions when they say impossible, then it makes it impossible. Oh, you can speak up, David have a different way to think about it? Yes, not a different way, but at the end there, because I actually noticed that when I looked at the solution proposed originally in the coach's slide, they actually had a mistake here and forgot the n plus 1 entirely, and still got the same solution, which is funny, but if you want to algebraically calculate that it's 8, what we can notice is that since n and n plus 1 are co-prime, either n over 2 and n plus 1 are both squares, or n plus 1 over 2 and n are both squares, because they're relatively prime, so it can't be that they share a factor that is double overall, right, they need to both be squares on their own, and allows you to just assume, okay, if n plus 1 is a square, well, then n plus 1 is 9, and n is 8, and look, it worked out, so yeah, you can actually do it that way just using number theory. Okay, great, yeah, we're definitely going to touch up on co-primes when we get to number theory lessons, yeah, that's very good observations, not many of them can make n square, there's something special about n and n plus 1 to make this a square, right, so again, summarized, I think we could best to go back to our introductory slides, you can review it, of the techniques that we used today, isolation, substantial elimination, symmetries, choice of variables, these are the things and some common manipulations here, some common identities here, that is something that gets used again and again, again, we, I don't think that Mathematica would test you heavily on solving like complex equations, it's really about like how to use a variable wisely, how to, you know, identify something, some special features in the system equations to get this answer quickly, so today we covered the basics, and next week would be, we have a lot of interesting questions, some word questions that use Algebra, so thank you everyone for participating live, and we look forward to seeing you next week, yeah, thank you David.

algebra

problem-solving

Math Kangaroo

variables

substitution

elimination

symmetry

linear equations

algebraic identities

sequences