So, in this series of 10 webinars, we're going to cover topics that are popular in math kangaroo exams. So, basically, the first lesson, we do problem-solving strategies, and then we move to algebra, number theory, sequences and patterns, combinatorics, which is basically counting, logical reasoning, and geometry. Oh, by the way, I forgot, Andrew is our teaching assistant, Andrew, please quickly say hi to everyone. Hi, I'm Andrew, and I'm the teacher for this webinar. Yes, so feel free to send questions through the chat, and Andrew and I would be happy to answer your questions. And then, so, what are the steps in solving a problem, right? So, the first step, obviously, is to understand the problem, and this is especially for math kangaroos when, you know, the problem, sometimes the problem statement itself is not so straightforward, right? So, do spend a little bit of time to understand the problem, and then plan how to solve the problem. Basically, think about what concept is being tested here, what strategy you can use to solve the problems, and then carry out the plan. And at home, in exam as well, we have a time, definitely look back, check, and reflect to see if the answer makes sense, whether you can sort of dive deeper into the questions modified it. But you see the sort of, the challenge here is to, and especially in the exam, you see you have 30 questions in just 75 minutes, so a lot of the time the question do not test on heavy computation, but it tests on like how well you know the subject, not like how much you know ahead of the curriculum, right? Check your fluency. So, a lot of the time, to be able to successfully tackle those problem, we just have to pick the right strategies, and that's the focus of the first webinar. We're going to go through the common problem-solving strategies. So, the first one is a draw a diagram, so diagrams help us visualize the problem, right? And not just for geometry problems, but a lot of problems that have numbers only. If you can draw a diagram, usually that's really, really helpful. And then you are probably familiar with some well-known diagrams, like Venn diagrams, tree diagrams, like this, and branches like this, graphs and networks that study connection between pairs. But really, any problems, just draw pictures, as you can see in like problem-solving, that's really, really helpful. Then trial and error, so sometimes we don't know the exact answer, we have a feeling that it is like, you know, approach me where the answer is, so start with a guess, and then re-find it, like find the right direction to go, and check and revise the guess until you get to the correct answer. And then because exams are, you know, MCQ, multiple choice questions, so a lot of the time we can actually eliminate the obviously wrong choices and narrow down the number of options and only focus on those remaining ones. Parity, this is actually a harder, a pretty hard strategy, so parity is the property of an integer being odd and even. So a lot of the time, this strategy is sort of kind of implicit, it's hidden, and we need to recognize it ourselves, and for a lot of problems it can get really hard, and until we recognize it, it's really like we just don't know how to solve it. Then the next strategy is the prior pattern, this one is a very, very popular, very common strategy that you probably have seen a lot in lower grades. So basically, we examine the problem carefully for the first few cases, and then find the patterns, then we can extend the pattern to the general case and prove the formulas. And the next one is solve a small version of the same problem. This strategy is closely related to find a pattern, right, so many times the problems that involve very big numbers, big parameters, and complex situations, then try to simplify it, use smaller numbers, smaller parameters, and try to simplify the condition, but at the same time still keep the structure of the original problems, and see if you can solve it, and then find a pattern and generalize it back to the original problem. And make and organize this, it's also a very common strategy, like data, study structures is always better than disorganized data, right? And then this strategy is very helpful when you have to deal with casework method. Then consider the extremes. We have one problem like this that we examined in this lesson, and what it means is that you have a parameter in the problem, we push it to the extreme, to the minimum, and to the maximum to see what happens if those ends, right? And then symmetry is, yeah, we also have quite a number of problem of symmetry when the parameters, the variables in the problems play the similar role, and then we can use that to simplify a lot of computation. So and there are more, there are more strategies that are specific to the topics at hand, but these are more sort of like common problem strategies that apply across the board. So now we're going to do problem solving, and I think I probably won't like specify which strategy you need to use. Use whatever strategies that you want to use to solve the problem, and then we, when we go to the solution, we're going to, you know, mention those strategies. Okay, so let's start with the first one. A hare and a tortoise competed in a 5-kilometer race along a straight line. The hare is five times as fast as the tortoise. The hare mistakenly started perpendicular to the route. After a while, he realized his mistake, turned, and ran straight to the finish point. He arrived at the same time as the tortoise. What is the distance between the hare's turning point and the finish point? So the race is five kilometers, kind of in a straight line, so usually I like to sort of underline important information, the hare is five times as far as the tortoise, and then when he started, he didn't go in a straight line, but he went perpendicular, and then after that he realized he maybe came back, and the important point here is that they arrived at the same time, right? the times the distance divided by speech, right? Because one is five times larger. And you probably haven't studied on it, but like for this question, obviously it would be very useful to start drawing diagram, right? Just a picture to illustrate the situation. So I'm not doing like solving the question, I'm just drawing because I think that's probably what you've been doing. So if you go in a straight line, then the hair of the tortoise go this way, right? And the hair, it goes perpendicular. So it's going all the way up. And because it's a speed, it's five times the speed of the tortoise. So, I mean, this section is probably longer than this. So the hair goes here. And then after it relies, it's this section, it went all the way back here, right? And then we can call this point is the finish point. And we know that T and F is five, right? And then we are asked to find what is the distance. And we can also say that this one is the velocity is V, this is gonna be five V. So. So basically, yeah, it's a number of questions, but you turn it to a diagram and sort of summarize all the information in the picture itself. And then this is going to be 90 degrees, so we know that we have to write a bunch of equations, right? So I would just like slowly start writing things down. If you have, you can, oh, great. Yeah, I think great. We got the correct answer. So you just continue working. If you got the answers, that's great. If not, just continue working. So th plus hf is going to be equal to phi times tf, right? Because they arrive at the same time and this skit is phi times, this skit is going to be phi times tf. And usually I don't like to put numbers here, but this is just kind of simple problem. So we can just put the number of phi times phi, tf phi is 25 over here. So that is good. So we have two unknowns, tf and hf. We need to find hf, but like, apparently if you can find hf, then we can find th, right? So it looks like we have two variables and then we have one equation, so we need one more. And that equation, it comes from geometry, right? And, you know, like obviously the only one thing that we can use here, which is the Pythagorean theorem. So hf square is going to be equal to th square plus tf square, right? tf square. So that becomes hf square, the hypotenuse, minus th square. That's our second equation, equal to tf square, which is 25. Is that correct? Yes, which is 25. So we have one equation where we have the sum of two numbers and the difference between the square. So, I mean, at this point, if you remember Pythagorean triples, then you can get the answer, but I think it would be helpful if we recall the identity, you know, the difference of square is going to be the difference, the product of the difference, th, right? See, th, and then times the product of, times the sum. It's going to be hf plus th, right? So remember, this is a very useful identity. h is equal to 25, and we know that this one is 25, so that one's going to be 1. And so we have two numbers. The sum is 25. The difference is 1. So one number is going to be 13, and the other number is 12, right? So hf is going to be 13, and th is 12. And this is, as you see, the math calculus, I usually do not test you, like, very odd numbers, kind of familiar numbers, so let me just circle it, but it's good to remember a couple of familiar Pythagorean triples, right? You have a 3, 4, 5, and then, I mean, the 3, 4, 5, and 6, 8, 10 also come up often, but these are just a scale version of the other one. And then you have this 13, 12, yeah, 5, 12, 13 is also a good one to remember. OK, I think that's another one, it's like 8, 15, 17, but it's kind of big, so maybe you don't need to use it. But at least these three triples, you know, it's good to remember by heart, and this identity also very helpful. OK, good, so let's go to the next question. Oh, okay, so I have a question, which variable does t represent? Okay, so sorry, I sort of deleted it, so I put it back here, so t is, so this is the tortoise, start here straight, right, so that is going to be five, and the hair goes perpendicular, and then go back to the finish line, right, so that's five, this is 12, and this is 13, and tf is not variable, tf is a known quantity, it's five, the length of the race, the straight line, does it answer the question? Okay, good. On a certain day, each of 40 trains travel between two of the tiles M, N, O, P, and Q, 10 trains travel either to or from M, 10 trains travel either to or from N, 10 trains travel either to or from O, and 10 trains travel either to or from P, how many trains travel to or from Q? So it's a mouthful, question number 14, but I think it's kind of not so easy, and we have five tiles with 40 trains, so you know, the structure are the same, but like we know that from two and from M there are 10, and same for N, and O, and P, so take a moment, yeah, give yourself a few minutes, try to explore it, first make sure you truly understand what the question means, and then maybe you can, you know, see what strategy you can use. This is one of the questions when you say, after you saw it, say, oh, it's kind of easy, I should have realized it from the beginning. At the same time, the thinking process is with the most important thing, most interesting thing. It takes a while to sort of unwind what the question means. So when I first started, I didn't know how to do it right away. But one strategy you can use that now there are five cities, five towns. Maybe you can try to understand the situation better by just considering two towns or three towns to see if you can find the relationship between the total number of trains and the trains that travel between either two or from each town. So there's nothing special about five, which is a bigger number. So do two, do three to see if you can find that relationship. That is to solve a smaller version of the same problem. Okay, so for example, right, I have three cities only, like M and O. I'm not, you know, solving it, I'm just, you know, get started by what I meant by consider a specific case. That is exactly what I do when I solve it, because, you know, I just don't know how to solve it right away. So M and M, suppose that, so let's say train two in front, so suppose that I just, you know, choose a smaller number, maybe like one trains that go from M to N, two trains that go from N to M, something like that, right? And then M to O maybe also one, O to M maybe three, could be two, just keep it small and simple, O to N maybe one, and N to O maybe four, something like that, right? So that, so how many trains we have in total? We have one, one, one, which is three, and then three is a six, four, so I have 12 trains, right? You can even make like smaller number, but we have total of 12 trains. And then for M, now the question that there are 10 trains travel either to or from M, so now if you look at either or from M, that number is really big, right? Because you have to count one plus two, and then one plus three, all of them are to and from M, so that's going to be one, one, this would be seven, okay? So I sort of stop here and give you a little bit more time, because the fun is in solving the problem, not the answer itself. So these numbers are so much bigger than the total number of trains, so how are they related, right? Yeah, and the relationship, if you look more carefully, instead of, you know, just looking at this time, I say here we add one and two because these are two separate trains, right? But then the number from M, you already add one and two, and then the number of N, again you add one and two. So actually each train is counted twice. Does it make sense? And then you can double check, right, because the total number here is 24, and actually we only have 12 different trains, because each of them is counted twice. So once you realize, I say, well, I wish I had, you know, realized it right from the beginning. But usually, you know, it doesn't come that way. We really have to explore it, and the process is not so straightforward. So it's really, really useful strategy. So once we realize that each train is counted one or twice, then we would have 40 times 2 is 80 trains, right, if we stick to the way it's counted. And from M, N, O, P, N, O, P, we have total 10 plus 10 plus 10 plus 10, then we have 40 of them. So that means 2, and from Q, it's going to be 80 minus 40 is equal to 40, right? And then this question, actually, I don't have time to look through it, but if you can sort of solve it, you're going to see that if you do O, C, you can, you know, go home and do it for yourself, but then you can solve a bunch of equations, or maybe you kind of do an argument or whatever, but try to, basically, in the end, all the trains are just between each city and Q, and there's no other trains between, you know, these four cities. That would be the configuration, but yeah, that you can do at home. There's emphasis here in this strategy. Okay, any questions? I'm going to go to the next one. A grocer has 12 different integer weights from 1 kilogram to 12 kilograms. She splits them into three groups of four weights each. The total weight of the first group is 41 kilograms, and of the second is 26 kilograms. Which of the following weights is in the same group as the weight of the 9 kilograms? So we have 12 weights from 1 to 12. That was straightforward. Now we split into three groups, and each has four weights, which is, you know, which goes with which, right? And the first group is 41. The second group is 26. Now we're going to see which one goes with 9. The question only asks for one, but I mean, presumably, we can find all three of them, because there's nothing special between them. So we started by, you know, sort of squeeze out as much information as possible from what's given to us, right? So we have the sum of the first group, the sum of the first group, the sum of the second group, and then from that, right, you know, we don't know what to do next, but at least we can find out the sum of the third group, right? And how do we do that? If we add everything together, I just started not solving, but just, this one is the arithmetic sequence, right? So either you can add 1 to 3, 1 to 12, 2 to 11, or if you remember the formula is going to be just, you know, n times n plus 1, the simplest arithmetic sequence is 2, which is 12 times 13 divided by 2, which is 78, right, is it, 13 times 6 is 78, yes, okay? So the first step is just, you know, extract as much information as we can. So 78, and then subtract away 41 and 26, which is 67. So you know right away the third one is going to be 11, okay? So 41, 26, 11, this is really big, this is really small. So what should we do? Yeah, I think I'll give you a little bit more time. And I can give one more hint, this is the strategy that one of the strategies that we discussed earlier is called consider extreme values, right? Because you know, the weights are given, and you have to make, you know, give four of them that, you have four of them that makes up 41, and four of them go about 11, like some big, some small, so maybe we consider those two extreme cases to see what numbers you can put in there. Oh, I have a question about recording, the recording will be available, I think, could be tonight, could be tomorrow, under your Math Kangaroo account. So you can, you can go back and, you know, pause and where you need to. So this is another hint, give you a bit more time, see if you can figure out, okay, sure. Okay, sure, yeah, yeah, sometimes you have a time conflict, so you have to leave and, just leave and you can watch the recording later. So the first group is 41, and how to make 41, these are, these are big, right, because the weights are only from here to here, so if I have to consider the, how to make 41, I, you know, might as well look at the biggest weight, right, it's going to be 12 plus 11 plus 10 plus 9, and that would be 12 plus 9 is 21 times 2, so it's 42, right, that is like one extreme value, 42, and I need to make 41, so that means the only ways I can make 41 out of these 12 weights is going to be 12, 10, I'm sorry, it's going to be 12, 11, 10, and 8. Does it make sense? Yeah, and then we don't know what to do in the middle, because there might be many choices, but if you consider the extreme ones, and I mean, obviously, there will be fewer choices, right, I mean, hopefully, but we have feeling that that's the case, so for 11, we also can consider the smallest four weights, so that's going to be 1, that's 2, that's 3, that's 4, and this sum is 10, and we need 11, and the only way we can make that is going to be 1, 2, 3, and 5, right, and then that means the leftover weights are 4, 6, 7, and 9, right, and they must sum up to 26, and then so you look at the choices 4, 6, 7, 9, what comes with 9 over here, the only choice is 7, right, so again, it's really about the thinking process and picking the right strategy with enough practice, you can solve it in a short amount of time, so definitely, you know, after these lessons, try to, you know, spend some time thinking more about these strategies and do past year exams for more practice, okay, so let's go to the next strategy, this one to see what strategy we need to use to solve these questions. Kanga labeled the vertices of the square base pyramid using 1, 2, 3, 4, 5 ones each, right, the five vertices and five numbers. For each face, Kanga calculated the sum of the numbers on its vertices. Four of these sums are 7, 8, 9, and 10, and what is the sum of the numbers at the vertices of the fifth face? So we have five numbers, five numbers, five vertices, and five faces. We know the sum of the four faces, and we need to find the sum of the fifth one. 2020, pretty recent, first slide is really not an easy question, right? Because if you add up the sums, then the apex here is going to be sum four times, right? And this one is not, and also we do not, we do not know these sums belong to the triangular faces or the base, right? So there are a lot of things that we need to figure out. So think about what strategy we can use. You can, you can definitely, because there are a few numbers, you can definitely try, try an error, put the numbers on these five vertices and try if you can make up these sums, that's one strategy. And the other one is using variable and writing equations, right? So it takes some time. I'm, I'm going to different routes. So I put it here, but definitely if you can do try an error, do it. If you have a question, anything, you can send it through the chat and we'll answer it either through the chat or the poll. If you do try an error, it might make sense to guess the number on the top, right, at the apex, that is sort of the most special number. Maybe do try an error with that. I think that might also work. could have some attempt here, so let's move forward. If you want to share how you solve it or you can do it through the chat, but otherwise let me, I try to use variables, I want to illustrate a strategy called symmetry. So, you know, these are four variables, then there's certain symmetry between B, C, D, E, not exactly the same because the sum would be different, but we see that they sort of have the same rows when A is different. So, based on that, later on when you solve like a bunch of linear equations and if the variables are symmetric, right, on the left-hand side, even though the right-hand sides are different, then one common technique you use to just sum up all the equations, right, and that way if all the variables appear the same number of times, then you can find the sum, right? So, in this case, not quite so, but, you know, it doesn't hurt. There's nothing else I have, so I just have to sum them up, and if I sum them up, what do I have? A is going to appear four times, right, and everything else is going to appear B, C, D three times. Is that correct? Yeah, B plus C plus B plus E appear three times, right? So, it's as best that I have, so I have to use it, and that means it's going to be A, and then, of course, I should group all of them, right, B, because I know the sum of all of them, C plus D plus E, and the sum of them is going to be the sum of A plus three times one plus two plus three plus four plus five. That is 45, right, 15, so A plus 45. So, that is the sum of all faces, this is the sum of five faces, right? So, if I add the four faces that I know, four faces that we know, there's going to be seven plus eight plus nine plus 10, this one is the 34, right? So, that means the sum of the fifth face is A plus 45 minus 34, that is A plus 11, right? So, I mean, this is one situation that I don't know where I'm heading to, but I'm just trying to extract as much information as I can, so I started sort of manipulating the equation in a way that I can use all the information given to me. Okay, so now we get to A plus 11, and this is where the nice thing happens. If it's A plus 11, then I can compare with this five times to see which one it could be, right, because we also don't know what that fifth face is, the triangular face or the base. So, if I put this one equal to A plus 11, right, for example, then B plus C is going to be equal to 11, but we do not have it because the biggest is four plus five is nine, right? So, yeah, I mean, like a lot of nice things, it just comes up after I did all this manipulation. And then, so we know that, and similarly with C, D, D, E, B, they are all the same, so that means we know that the fifth face cannot be one of the four triangular faces, right, and that means the fifth face has to be this one, it's going to be A plus 11, it has to be the base. But now, I mean, the problem is done because we know the sum of A, B, C, D, E, and we know that A plus 11 equal to B plus C plus D, and so if you add A on both sides, then we would have, so this one, you add A's on both sides so that we can have the sum of all numbers, so we have A plus B plus C plus D plus E equal to 2A plus 11, but that sum is, we calculated before, that sum is 15, right, because the sum of all the numbers from 1 to 5 is 15, so from here we know that A is equal to 2, A equal to 2, and the fifth face is A plus 11, so that it has to be 13, right? And now, you know, once you know the number, you can see, like, if you could have go back and choose A, and then, you know, the four numbers at the base. So, I mean, we might think that the one is A peg is 3, right, because in the meter, but actually not, because it's sort of like these are continued numbers, so we might think it's 3, but it turns out to be 2. So, I don't know, it could be faster if you can just try all five cases, or in this case, we use symmetry and be comfortable with manipulating, you know, variables to get to the answer. Okay, good, so let's go to the next strategy. This one also a good question, it's very appropriate for a few minutes. That one, I admit, is kind of a bit difficult. Two of the altitudes of a triangle are 10 centimeters and 11 centimeters. Which of the following cannot be the length of the third altitude, right? So, altitude is the height, right, of the triangles. So, the question is that two of them are 10 and 11, and which point cannot be? So, skyline cannot be, and there are five choices here. This is a good question, I would, you know, give you some time to think about. Yeah, it's great that if you have an answer, you put it to the poll in the chat, so that we sort of know what you're thinking and that would be helpful. Okay, so, you know, let's explore the numbers that was a, you know, different solution, I will talk about different solution, but, you know, the first one, so look at the choices like the option. Yeah. So 5, 6, 7, 10, 11, 100, right? So 10 definitely is a good one because why 10, 11, you know, it's more like an equilateral triangle, right? 10, 11, 12, 10, 11, 10, it's pretty much the same as it took it here, a little bit here and there and I can make it 11 instead of 10, 10, 10, so it looks like good. And I will go to the spectrum from biggest to smallest and we suspect that, oh, it could be the, this one or this one, right? And now, sort of, in a sense, we can eliminate these three choices and focus on these two. It is kind of tempted to think that the answer is E because Y, it is just so, so big compared to 10 and 11. So now we try to say, but how to show that, you know, it's really E or A. So think of a shape of, I'm sorry, why it doesn't move, think of a shape of a, can you think of a shape of a triangle such that the attitudes are 10, 11 and 100, if you cannot think of it, then sort of, oh, maybe this is E, right? But actually, if you draw a very, very tall and slim triangle like this, this is not good enough, but kind of, right? So this attitude can be, you know, sort of infinitely long, you can just make it as long as you want and then these two can be 10 and 11 roughly, but we really can make it as long as one. In that sense, it could even go to a thousand and you can still make that triangle, right? So E is actually one of the possible choices. And then based on our reasoning, the answer, there's no reason that a strange thing happened in the middle. So the answer is actually A. And why is it A? Because if I want to try to construct a triangle in which two attitudes are big and one is small, so that means I have to make a obtuse, a very obtuse triangle like this, right? It's pretty much like an isosceles. So if I put this one is 10, this one is 10, and that one is 11, 11, and this one, can I make it five? Not quite, right? Because when I try to make this one smaller, I sort of have to make it more and more obtuse And if I make it more obtuse than this, length also go down. And so it looks like five is the correct answer and not a hundred, right? But by this time, when we think about why there's some constraint between the length of the attitudes, by this time, we probably realize that the theorem, the triangle theorem that say that the sum of two lengths of a triangle is always bigger than the third, right? So maybe we can use this to prove that A is correct answer and how to, you know, relate attitudes to length. We know that A, right, A times H equal to S, which is, which is the area, right? Two S, because this is a triangle. So basically we have the relationship, but it's going to be the inverse of that for the length. So A is going to be inversely proportional to H. So we have a one over HA plus one over HB, it's going to be bigger than one over HC. So sort of inequality for the attitudes. And that once we have a one over 10 plus one over 11, that must be bigger than one over HC, the third one, but this, this two is at most, this one is going to be smaller than one 10, one 10, so that's going to be smaller than one, smaller than one fifth, right? So that means HC has to be bigger than five. And the answer is going to be A, and everything else is permissible. So that is another way that, like, even if I don't recognize this triangle inequalities just by using trial and error and eliminate the wrong choices in the exam, I can still get the answer. And, you know, that's the whole point of talking about problem solving strategies. Okay. Hope that's helpful. Let's go to the next one. Oh, we, we, I mean, it's a good question. Why did I eliminate B? This is not a rigorous solution, it just helped me get the answer. I eliminate B because, because there's, when I go, I know that the question says that only one is correct answer, right? So when I increase it from here to here, I know that on this side, all of these are permissible. And the answer, the, the correct answer would not happen right in the middle. It's more like you can think of it like continual change. There's no reason to believe that six would, would, would, would be permissible and five is not, you know, that sort of hand-waving argument, but to absolutely know where that number is, and we have to use, so, so we know it's actually the cutoff is going to be 5.5, right? This one, not 5.5, it's going to be 21, 110 divided by 21. That will be the cutoff, right? And for that, you have to know the triangle inequality, but the hand-waving argument allow us to find that A is not the correct answer. Okay. Okay, thank you. That's good. When I have a question, feel free to ask. Okay, good. Okay, good. So the next one, Andrew, yeah, please help with this question. Okay, so Ken wants to add some line segments to the picture shown so that each of the seven points has the same number of line segments. And we're looking for the least number of line segments that Ken wants to draw. Okay, good. So I just underlined his seven points and the same number of connections. Yes, also a very nice question. It doesn't require anything special, but it's a very beautiful strategy. Andrew, feel free to give them everyone hints. All right, so a big diversity problem is this power connection. So we can see that for every single connection, it connects to two line segments, right? So each point will gain one connection for each line segment. So for every one connection, we get two connections to points. Oh, you can have some tags, yes. And we also know there's 7 points, right? So, because this point has 3 connections, we have to have at least each point having 3 connections. So, if we try to think about it, so if we have each point having 3 connections with 7 points, so the points will have 21 connections in total. So that means we're going to need 10.5 connections. But we can't draw 10.5 line segments, because we can't draw half a line segment. So, we actually have to draw. We have to give each point 4 connections each step. So, if we do this, we get 28. You can see that in the original job, we have 1, 2, 3, 4, 5 line segments already. So, if we have 28 over 2, so we have 14 line segments. So, if we subtract the existing 5 from the 14, we can get that we need 9 line segments. So, 9 is our correct answer here. Yeah, thank you, Andrew. I mean, these strategies go odd and even. I will just type up here, because sometimes for questions that are more complicated, I type up the solution. So, at home, you can pause it and look at the solution. So, this is basically what Andrew said about the connections. And so, because it was twice, we have to multiply, you know, add them up and divide it by 2. I just wanted to add that probably you have seen it before. Have you heard about sort of handshake problems, right? Like, for example, you have seven people in a room, right? Seven people. And, like, each person makes exactly five handshakes. Is it possible? Yeah, it's similar to this question. It's not possible, right? Because each handshake is counted twice. So, we have 7 times 5 is an odd number. It's not divisible by 2. So, if someone makes this, you know, observation, you know that it's incorrect. And it applies to our question that we have already three connections here. But if we have exactly three connections at each point, then that's impossible because 3 times 7 is an odd number. So, we must have at least four connections at each point, right? It has to be an odd number and then even number. And then you can, you know, go home and try and connect that to make sure that, yeah, show that, yeah, that works. You need eight more or nine more connections to make it possible. So, these strategies go odd and even. And I just want to add that in exam, you know, you might not be able to think of it right away. So, it's totally, you know, possible that you can just, you know, continue adding, continue adding the connections and, you know, just keep track of that until you have all four connections at each connection point. So, sometimes I got the answer like five. And that means you didn't, some of the connections, some of the connections you didn't have as many connections as the other connection points. So, definitely you can use that, solve the question without using odd and even, but make sure to really keep track of your counting. So, just come back to the beginning. This is the common problem solving strategies that we talk about. And we illustrate those, some of the strategy with the problems in these lessons. And as we go along to, you know, study other topics, there will be strategies that are, you know, specific to each topic. But that is the idea. We cover all the concepts. And at the same time, we also want to emphasize problem solving strategies. And that's what I really advise you when you study at home for the exam. There are only three more months. So, definitely do a lot of passing exam. And when you study on your own, you know, from books, you know, from other sources, always take time to think about the problem solving strategies and try to solve each question in as many different ways as you can.

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