webinar on Mexican Guru webinars for level 9 and 10. And just know it's a webinar format, so everything is recorded so you can watch it again at your own convenience. My name is Phan Kurulava. I'm a teacher at Mexican Guru. And we also have David Dersh as teaching assistant who will help with the chat and also with some of the questions. David, could you just quickly say hi and introduce yourself? So hello, my name is David Deutsch. I've been doing math for as long as I can remember. I took the Math Kangaroo 3-4 test in third grade, and I've been doing it ever since. I'm in 11th grade right now, which is only one year after you guys, but I've done well in this competition, and I've gotten a gold medal in the last four that I've done. So I think I have some experience to add to the table. So yeah, it's great to meet everybody's black screens and white names. I am very happy to meet you guys. Yeah, let's have a good time. Yeah, thank you, David. He would have some of the questions and answer the chat as well. If you have any, so we will launch the poll so that you can put the answers in. And then we also, if you have any questions, you can send to the chat to me and David only. Other students will not see your messages. So feel free to ask anything you like, and we try our best to answer it. So many of you, probably most of you have taken Math Kangaroos before. We have 30 questions for this level, and the test is 75 minutes. And as you can see that it tests both breadth and depth of the material. And Math Kangaroos questions come in all different shapes and size and different packagings. So what we try to do in this course is that look underneath the question and sort of put them into different categories and also focus on different problem-solving strategies that hopefully will help you do better in the test as well as basically look at math in a deeper level. So Math Kangaroo problem topics, and most problem fall into the following categories. Algebra, number theory, combinatorics, logical reasoning, and geometry. So algebra, we have common things like variables, equations, system of equations, functions, and word problems. When we talk about equations, the focus is not like very hardcore higher order equations. I don't think we should, probably we should stop at quadratic equations for this level. But algebra is a language of advanced math, so we want to be very comfortable with algebra because it's, we need to use it to solve problems in all different topics over here, in number theory, in geometry. Then number theory, definitely a big topic with multiple divisors, primes, prime factorizations. If you're comfortable converting decimal and factors, sometimes we do have a modular arithmetic question when, you know, there's constraints like the solutions can only be integers, things like that. So we would have a different separate lesson for each topic. And then combinatorics, basic counting, we do talk about permutations and combinations, a bunch of other sort of very creative counting questions. And Math Kangaroo, compared to other math competitions, do have a good portion of it is logical reasoning questions, logic puzzles, and you probably notice a lot of questions with nice and liars, questions you have to figure out what statements are false, what are true. And then in geometry, I think for this level, more focus on 2D geometry, but we do have a 3D geometry, and that's going to be the topic for our last lesson. And so, because it's competition, so we know that we have to work with, you know, breath and also speed as well. So, apart from mastering all these subject matters, and it's also important to master different problem-solving strategies, right, you have a little limited amount of time you want to figure out the best strategies that help you solve the questions. So, some strategies are very specific to each topic. So, for example, if you do counting, you see that we have a complementary counting, which we count things that we do not want to count, and then we subtract it from the total number of possibilities. But there are some other general strategies that can be applied across the board, and that's what we are looking at in this lesson. And so, it's just a quick overview of, you know, the timeline of this course. The first lesson today is introduction, and mainly we go through different problem-solving strategies, and we have two lessons on algebra. First one, we're going to review the basic algebra, and then the second one, we focus more on word problems. Then we do one question on, one lesson on logical reasoning, one lesson on logical reasoning, the number theories, and there's some, lesson number six is quite interesting, a lot of sequences and patterns. They, in order to solve this kind of question, you pull in knowledge from all different topics. And then one specific lesson on combinatorics, two lessons on 2D geometry, and the last lesson will be review and 3D geometry. Okay, so there's an overview, quickly overview of the common strategies that we are talking today. It's nowhere like an exclusive list, but these are the common ones. So, draw a diagram. Diagram is, may help us visualize a problem better, and you know, some well-known one like band diagrams, two diagrams that you probably see a lot earlier, and you know, graphs and network is also some very helpful diagrams. Then trial and error. We start with a guess, right? It sounds trivial, but actually a lot of the time it's a very useful strategy. We start with a guess, and we check if it works, and even though the guess is wrong, but it helps us learn something about the questions, and then have, and then we revise our guess to get closer to the right answer. And if the question has multiple choices, then one quite useful strategy is eliminate the wrong choices, because sometimes with careful observation, you can quickly find out that some choices are often obviously incorrect, and then you can, you only have to focus on the remaining choices. And use of parity, that is the property of a number of an integer to be odd and even, also very helpful in many questions. Just, we need to recognize it. It's not obvious, but we need to recognize it. Then fire pattern. This is a very, very popular and strategy that you probably see, seen a lot in lower level math and Google problems. So examine the problem carefully to fire patterns, and this is usually done by solving several simple cases, and then we generalize it. And sometimes we need to do, to really prove that our pattern is correct one, but it really helps us when we start with something small. And then another strategy that closely related to fire pattern is the smaller version of the same problems. Sometimes a question has very large parameters, then what we can do is that we may try with a smaller values parameters to see if we can gain some insight into the problem, and then just like stretching the files and we generalize it, and then move it. Make an organized list, also a very popular strategy. Basically, this mainly means that we organize our data in a very structured way. And if we use, using in conjunction with the casework method, yes, very commonly used. So pretty much we would have seven questions that covers it. So usually when we solve problems, we do not know ahead of time which strategy we would pick, right? But since this is, the question is, the lesson is on strategy, so I would just put the title right here. And for each question, but from the next, the next lesson onwards, we are not going to do it this way. So the first one is draw a diagram. Four brothers have different heights. Tobias is shorter than Victor by the same measurement, by which he is taller than Peter. Oscar is shorter than Peter by the same measurement as well. Tobias is 184 centimeters tall, and the average height of all the four brothers is 178 centimeters. How tall is Oscar? I just launched the poll, but don't worry, take your time. Whenever you are ready, you can enter your choice. There's no hurry. So I just underlined some of the probably most important information. So maybe we get started, question number 17, this is 12 points, so let's get started together. So the question suggests that draw a diagram, so let me do it and maybe we can discuss some other ways you can solve it, but draw a diagram, so remember this, I do it vertically, tall bed is shorter than Victor and taller than, by the same amount, okay, so if Tobias is here, then Victor is up here, right, and Peter is down here, and Oscar is shorter than Peter by the same amount, so Oscar is here, and all these four, three segments are the same, I draw to scale so that they are the same, and then because these are segments are the same, then we know that the average is going to be here, right, average of the four of them to be here, and this average is 184, 184, and Peter, oh I know, Oscar is 178, is that correct? Oh, I'm sorry, Tobias and, yeah, I made a mistake here, yeah, I made a mistake, so Tobias is 184, and the average is 178, right, 178, yeah, so we know the difference here is six, so we know that each segment here from here to here is 12, from here here's 12, and this here is 12, and over here is 12 as well, and we know that from here we would know that the height of Oscar, Oscar is going to be equal to 178 minus six, that's 12, which is 160. Great, so those of you who put the answer got them correctly, if you, I just want to quickly show a different way to do that, right, because, you know, you see that doing diagram got it just a question to the answer pretty quickly, but you might as well show a different way, right, we know that Algebra is our friend, most questions like this can be solved by using Algebra, so we have V minus T equal to T minus P, right, and then we also have P minus O, and that also equal to P minus O, right, minus O, and then we would say that T is 184, and the average for them, so that's V plus T plus P plus O divided by 4 is 178, so we have four variables, and how many equations do we have here? We have four, right, because actually these are two equation sizes, put them together, so we have four equations, four variables, so in principle this problem we can solve using Algebra, but it just shows the advantage of drawing diagram that helps us get to the answer much quicker without having to solve this system of equations of four variables. Okay, good, and anytime you have a question, feel free to put into the chat, and then David and I will answer the questions. Next one, what is the largest possible remainder that can be obtained when a two-digit number is divided by the sum of its digits? So we have a two-digit number divided by the sum of its digits, and we are looking for the largest possible remainder. Oh. Okay. That was a good story. I'll come back. Thank you. So you, you will see the host everything was going on right. Yeah, um, since you're sharing the screen the screen. Count out. I see but otherwise I can share the slide because I have it up to. Ah, yeah, it's good. I just wanted to see, could you please make me the host again. Oh, I'm so sorry. Yeah. Oh no no no, that's good. Thank you. That's why you are here. I made some modifications. You are already the host, you're automatically the host of this meeting. Okay, thank you. Okay, good, so let me launch the poll because it looks like no one lost hands for me. Oh, could you please launch the poll, David? I lost everything. Oh, I think I have it back here. I'll also share the problem. OK, so let me do this. OK, so the poll's launched. Let me share the problems real quick. Okay, it's loading. Can everyone see? Just, if anyone cannot see, let me know. Okay, so this is, let me find it. Let me find the right problem real quick. Okay, here we go, this is it. So, all right, so the problem is, what is the largest possible remainder that can be obtained when a two-digit number is divided by the sum of its digits? And we're told it's a trial and error problem, guess and check. And I guess you can see why that would be because you've got five pretty small numbers, right? So you can kind of try and check it. So go ahead, look at the problem, try to figure out what it means, and we'll come back and discuss it in a few minutes. Yes, I think, I think we can get started now. I do this on you do this next one year. Okay, we. So, try an error, I mean, for the interest of time, which is get right into the solutions. We tried several different things but didn't quite work so let's, let's start with the hint from from the problem try an error. We, we are looking for the largest possible remainder right and we are divided, we divide the number by the sum of these digits so we want to some of the digits as big as possible. So also, this is making an organized list which is started from the biggest to the smallest so the biggest would be 19, we have a 99 right and then the second biggest some would be 17 and we have two numbers, 98 and 89. And then we have 16 which is 97. We could do 88 and 79 to be organized like the listing from the biggest to the smallest. And then we just try 99 divided by 18. What is the 99 divided by 18. What is the, the remainder is for a 72 In it right so we remember in this case, remember is nine for the first case. Remember is nine. And then you just try with all of this. I think some of you got the correct answer. The answer is D 16. When we divide 15 right 79 divided by 16 16 times for a 64 and 64 79 minus C for 15 so the answer is C. The answer is C, I think, but that is just, you know, without going into the details. That's how we, we, the method is that we try from the biggest to the smallest in this order. So 79 divided by 16 that give us four as a remainder is 15. I just want to add that initially I when I look at the problem we do like a B and then you divide a B by a plus B looks great. I mean, this method works in many cases, but unfortunately didn't give us a lot of insight to solve this. So try and Actually give us bring the answer in this case. Thank you. Sorry for the hiccup about the power. I had a snowstorm and something was not working well. But now we are back to normal and David, could you please help us with question number three. Alright, so This question is about parody and you'll see if you have done questions like this. It's a very classic use of parody. So for each integer and greater than or equal to two. Wonder why let the symbol that symbol on some sort of brackets and denote the largest prime number not greater than So, for example, for six, it would be five. So, for example, for six, it would be five because that's the largest prime number less than the question is how many positive integers satisfy this equation. k plus one in the brackets plus k plus two in the brackets equals to K plus three brackets. So give me a second to launch relaunch the poll All right. All right. And once again, so let's work on this for a few minutes and then we'll come and discuss it. This is problem 27 So it's very, it's very late in the competition. So you can expect that it's either very long or there's some sort of very neat trick involved. So just be prepared to work a little on this one. It's a small group, so you have the poll, but if you have any questions, any insight, anything, you can also send us a chat message, and one of us here. So far, we have two answers that have been given. Let's just wait a bit more and see what happens. Now, the average length for problems is two minutes and a half, which is basically now, but this is one of the last five-point problems, so there should be some more time. About 45 seconds. Let's just go over it together. We chose the sum of higher number of problems because it's more to discuss, because once you go to this level, we noticed that the number of participants dropped compared to elementary school students, but actually, average is higher. So once you go to this level, you want to look at it at a deeper level and not just doing three-point questions. Alright, so it's been some time. I'm going to start, you know, progressively giving hints and then seeing how much more, you know, I'm seeing how much more people answer it as I go through the solution. So, um, basically, the trick in this problem is it's a very, it's a priority trick. It's a very common priority trick that has to do with primes, right? We have one prime plus another prime that equals a third prime, right? Primes don't have any factors other than themselves. And one of the most important facts about primes that comes out of this is that two is the only even prime, right? So if we had a bunch of big primes in this answer, if k was a really big number, then we'd have odd plus odd equals odd. That's what we need. And that's not possible, right? Because odd plus odd always equals even. The only way for two primes to equal to another bigger prime is if one of the primes, in this case, the smaller one, is two. So we've used parity to show that one of the primes has to be two. This is a very common trick, not just the parity in general, but specifically with primes. You're adding primes. We see this a lot. You need to make sure that one of them is two, otherwise you're in trouble, right? So now let's just look at what this means. We know that one of these primes has to be two. Since two is the smallest prime, logically, it's this one. If k plus two, if k plus two with the brackets was two, then this would also be two or be smaller. Either way, you'd either get a smaller prime than two, or you would wind up with two plus two equals four, which is not a prime, of course. So that's not it either, which means that we know that k plus one in the brackets is a prime. What this means is that k plus one is greater than or equal to two, right? Because two has to be less than or equal to k plus one. That's the definition. It also means that k plus one is less than three, because otherwise three would be the answer, not two, and that wouldn't work either. And that's really nice all of a sudden, because that means that k plus one has to equal two, which means that k equals to one. Therefore, there can be at most one positive integer that satisfies the equation. We don't know if the answer is one yet, because what if one just doesn't work either? So we have to plug it in and check. So let's do that. If k equals one, then one plus one in brackets plus one plus two in brackets would have to be equal to two plus three in brackets. Just plugging in, you get two plus three equals five. And I think we can all agree at this point, at this level, that two plus three equals five, and therefore that the answer is b, one. So this is a tricky problem in the sense that you really need that parity trick. Otherwise, you're going to keep going for really huge k, and there's no way of explaining why there is not an infinite amount of solutions. But at the same time, this specific pattern with primes is something that should really be memorized, because I've seen it crop up in so many math kangaroo problems over the years. It keeps coming back. One of them is always two. So just make sure to remember that. It's extremely important. I see we have a wide distribution of answers, which is normal. And OK, this is a very hard problem. It's problem 27. It's almost at the end. But we do have a majority for we do have the most popular answer being one. So good job. This is a hard problem. And remember this trick. So, yeah. Yeah. Thank you, David. And I think as you were explaining, one of the students corrected the answer to one. So that's great. And somehow I couldn't see your typing. So I type here because of a recording. Yeah, it didn't appear on the screen. Yeah. So this parity is a very nice we call trick, but it's something that comes with experience. And once you figure it out, you have a smooth sail, but you don't figure it out and you get stuck for a long time. It's good to keep in mind. Next one. Fire pattern. Julia creates a procedure for turning a set of three numbers into a new set of three numbers. Each number is replaced by the sum of the other two. For example, 3, 4, 6 becomes 10, 9, 7. How many times must Julia apply this procedure to the set 1, 2, 3 before he first obtains a set containing the number 2013? So what did you do with the procedures? We have a three set of numbers and we turn it into a new set, right? Three set instead of converting a new one and we just do it again and again like a machine applied again and again. And we check that from 3, 4, 6. Yes, we add 3 with 4, 4 with 6 and 3 with 6 and we do get 10, 9, 7. So our problem is that we first make sure that we understand the rule and make sure that it works. And now we want to apply these procedures again and again, starting at 1, 2, 3, and when after how many procedures, how many times we can get a set of three numbers and one of that is 2013. So take so times 24. Sometimes some four points questions could be more tricky than five points, but in general, the higher the number, the more difficult the questions. Not all the way, but a general chance. And we only focus on four and five points questions in this course. Remember, we're talking about pattern. We start with smaller case and see if we can find any patterns. So I'm currently the one sharing screen right now. Do you want me to stop so that you can. I think I already share mine because because I because I'm recording on my computer and somehow it looks better. Let me stop sharing. Yeah. Yeah. I see. OK. Yeah. OK. Thank you. All right. If you find a pattern, try to generalize it. Sometimes you can get the answer without proving that it's the right pattern. But if you have time, you can, especially when you do it at home, by all means, try to prove it for the general case. In the exam, you pretty much can get away with just figuring out that it's the right I would just get started and then see if we can figure out things together. We have 1, 2, 3, right? And the next one, if I apply the rule, 1 plus 2 is 3, and then it would be, the next one is going to be 3, 4, and 5. And then 3, 4 is 7, so I would have 7, 8, and 9, right? And then, and then what do we do, 7, 8, and 9, and the next one would be 15, 16, 17. Okay, so 2 people already got, 3 of you already got the correct answer, for the rest after you see this, can you figure out the pattern? What's in 10? Okay, good, so yes, pretty much everyone already get the answer, so the pattern over here is that if you look at the first number, right, 1, 3, 5, 7, and 15, does it ring a bell? This one is 2 to, raised to 0 minus 1, right, and then 2 squared minus 1, that's 3, and then 2 cubed minus 1, and then 2 to the 4th minus 1, that is, that is the pattern, and then, I mean, when we have time, we probably will come back and try to prove that this is the, this is a general pattern. In the exam, we can pretty much sure that this pattern is correct without having to prove that, so if you go all the way to the nth set, it would have the general formulas 2 to the power n minus 1 as the first number, and then 2 to the nth, and 2 to the n plus 1. And can any one of these numbers be 2013? It also helps that we remember the 10th power is a 2, right, so 2 to the 10 is 2000, 1024, and the next one is 2048, and the gap is very big, and 2013 is between these, so we know that the answer is in, which is great, most of you got the correct answer, yes, so we see how useful it is to, yes, thank you, David, for that, thank you, David, David said that probably it's easier to look at the middle number, 2486, yes, I agree, that would actually be more obvious, 2 to the nth, than what I see, so you see, like, we all figure out the pattern, and then we all have different views, the same relations, so that's great, and by all means, you go home and try to prove that this is general patterns, I don't know if you have a different method, David, I would prove it by induction, that's another, another, so you can look it up, prove by inductions, we do not introduce in this course, but it's something that, also a very powerful method of proofs that you can look it up and use it to prove in this case, but in the exam, you can get away with just figuring out in the pattern without having to prove it, okay, good, so let's go to the next one, solve a pretty much related strategy compared to the other one, solve a smaller version of the same problem, there are 2016 kangaroos, each of them is either grey or red, and at least one of them is grey, and at least one is red, for every kangaroo K, we compute the fraction of the number of kangaroos of the other colour, divided by the number of kangaroos of the same colour as K, including K, by the sum of the fraction of all 2016 kangaroos. So what problems take some time to fully understand what's going on, I will try to underline, compute the fraction of the number of kangaroos of the other color divided by the number of kangaroos of the same color. For each kangaroo we do this and we sum up all the fractions. What do we get? So we have a hint here, right? So a smaller version is the same problem. We put this title here because that's the focus of today's lesson. In general, we do not know which strategy we would use. We have to explore different strategies, but you know, that's a good suggestions. We have 2016, that's many kangaroos. You can try with, you know, a few kangaroos to see if we can, you know, have some insight, if that may help us figure out some patterns, and then we can generalize it to bigger numbers. I mean two kangaroos are too little. You can try maybe three or four or five just to see. And that is also characteristic of mass kangaroo questions. Sometimes it takes a while to understand what the question means, what problem is here. Oh, great. I think almost 50% of the students got the answer, and it's only three minutes, so let's do it together. And all of you who gave the answers got the correct one. So just for the sake of illustrating this problem-solving strategy, I will start with smaller numbers. I start with four kangaroos, and the key here is that we do not know how many of them are red, how many of them are blue, and the question asked can find the sum of the fractions, so we may infer that the answer should not depend on how many of them are actually red, how many are actually gray. So I have four kangaroos. I can assume one is gray and three are red, or two of them are gray and two of them are red. And now I have to figure out what it means. For this gray kangaroo, I have to take the number of kangaroos of the other colors and divide it by the number of kangaroos of the same color. So for this one, it's going to be three over one, right? But for the red, each of the red here, the other color is gray, and the same color is red, so the fraction becomes one over three, one over three, and one over three. And when I sum up, I need three over one times one, right? So it's going to be three, and then this one over three add together plus one, so that's four. Well, in this case, for each green, the number of kangaroos, say, the other color is two, so it's two divide, and the same color is two, divide by two, two, and same for red is two over two, two over two. I sum up all these fractions, and I got the same answer. The sum is four, and that's what it should be, right? The answer should not depend on that. And what is number four here? Four is the number of kangaroos. So in the exam, you could right away infer that you have a 2016 kangaroos, and the answer should be 2016, right, NCSA. But of course, we want to do better than that, and remember algebra is our friend, so if we assume that let's call the number of gray kangaroos is g, and then the red kangaroo is r, right? So the notation is probably self-explanatory. And then, so the fraction, the first fraction when the kangaroo is gray, for each of gray kangaroos, the fraction is actually r over g, right? But then we have to multiply, we have by g, because that's many, that's how many fractions we have. And similarly, for the red kangaroos, the fraction for each of them is g divided by r, but then we got to multiply by r, which is the number of red kangaroos. So in the end, we have a r plus g, which is exactly the total number of kangaroos in each case, 2016, and that's what we do at home. Yes, thank you, David. Yeah, it's, some of, they are very much, very much related. We find the pattern by using a smaller parameter, in this case, a smaller kangaroos. If you can find it right away, the general solutions, that's great, but a lot of the time, sometimes they just don't come naturally, easily, and by solving, using this strategy, we can, you know, find insight and find the pattern first, before we move on to print the general case. Okay, great, and in the end, most of you got the correct answer while I was explaining, so that's great. Okay, the next question is David, David. Yeah, I just would, would like to say another unrelated way of approaching this type of problem is that whenever there's a condition that doesn't matter, like it doesn't matter how many kangaroos of each color there are, right, whenever you have a problem where that's the case, the most, the easiest cheat way of doing it is just pick the numbers, such that it's as easy as possible. Like, if you have something that's true for any quadrilateral, pretend it's a square, for example. So, what, I actually remember doing this in practice, and what I did for this problem is that I just pretended that half of them were gray and half of them are red. Great, because do that, then the fraction is 1008 over 1008. The fraction is always one, so you just have one plus one plus one, 2016 times. So, that's another way to solve this specific type of problem. It's not as useful of a strategy, because this is pretty rare, that something just does not matter at all in the problem, but when it does, just set it to whatever works the best for you. Exactly, and then that's the one thing that we were considering while preparing this, that on another chassis, they put the parameters to the extreme. In this case, you have like half of them kangaroo, half of them gray, half of them red, or maybe all of them are red, but at least one, I'm sorry, in this case we can't, because at least one is red, but you could do like special parameters or extreme parameters, and then from that, but the keys, as David said, that we had to generalize, we had to realize that the answers do not depend on, like, you know, the fractions of R and G in this case. So, yeah, as we go along and we can see that, I mean, that's part of the focus, of course, and every time we solve a problem, we take a, you know, a moment to actually look at, you know, what strategies we are using, are useful, could be multiple strategies. And the next one is, I think, David, right? Yeah, make an organized list. Consider the set of all seven-digit numbers that can be obtained using the digits 1, 2, 3, 4, 5, 7 for each number. List the numbers of the set in increasing order and divide the list exactly in the middle into two parts of the same length. What is the last number of the first half of the list? So I underline here, we list the number in increasing order and divide the list exactly in the middle, two parts of the same length. And we, I think, exactly in the middle, I think that's good enough. And the last number of the first half. So take a few minutes and David will help us with this question. No, it should be pointed out just from the get-go that don't try to list every single number when it says organized list, because there are seven factorial of them, which is 5,040, and you do not have the time or the energy to list 5,000 numbers. But that doesn't mean that you can't take the list and group it into sublists that help you. So don't try to list every single number, but you can still make the list more organized for yourself. Yes, as David said, we cannot list everything, but it's good to list a few and then try to divide them in a group. Okay, as the answers come in, I'm starting to see that a lot of people have figured out how to solve this problem, and also that a lot of people are falling into this one very cruel trap that has been set. But it's clear that the idea is starting to be understood and how to actually organize the list. So let's begin going through this and let's see if people find the trap that some of us have fallen into. So just one second, all right. So we have this very big list of numbers, right? So we have numbers with one, two, three, all the way to seven. We cannot organize all these, because again, there's 5,000. We can't, I mean, write them all down on their own. But we have to put them in increasing order. And what we know about increasing order of digits is that the first digit is the one that matters the most. 2,000 is bigger than 1,999, right, because two is bigger than one. So what we do know is that however this list plays out, it's going to look something like this. You're going to have a bunch of numbers that start with one, then a bunch of numbers that start with two, then a bunch of numbers that start with three. I'm going to write down every single one, even though it might seem like there's no point, because there is a reason to this. Then all the ones that start with five, six, seven, right. So this is our list. It's a bunch of these numbers, then a bunch of these numbers, and so on. We want the middle of the list. Now it's pretty obvious that each of these little columns has the same amount of numbers in them, because it's just all of the orderings of six different digits. So we can do what we always do with lists and just find the middle, which is this. Which tells us that the middle number of the entire set is just the middle number of this one column. Now what a lot of people did here is that they went, oh, between three and four, they put the first number in this column, which is C. But we actually need the middle number. We still haven't found the middle, the full middle yet. We've just found the column where it lies. So we have to keep going. We can't be so hasty, right. So now what do we do? Well, one other trick that's very important in problem solving is if you tried some problem strategy and it kind of worked, but it didn't finish the problem, try it again. So now we're going to do the same thing. We know that you start with the digit four, and all the other digits are just one, two, three, five, six, seven. And so now we do the exact same thing all over again. We have the ones, the twos, the threes, the fives, the sixes, and the sevens. And now when we split this set in half, because we're just going in deeper and deeper, right, we split right over here, which means that the final number in the first half is just the last number that starts with three, which would be three and then seven, six, five, two, one. So the remaining five digits in the biggest order possible. And since all of these numbers actually start with four, because we're in the four column, that would mean that the answer is E. And so the most common answer was C, which shows that clearly most people knew how to do the problem. But there was this little trap, which was just because you found the column that you're inside doesn't mean that you found the actual number. And that's the little trap. So we have to just keep going a little bit more. Thank you, David. That's why sometimes it's good to just write out a few numbers. I mean, we do this big picture here, put them into a list, but it's helpful to write out the actual numbers in each group, right? So this is the smallest number. And this is the biggest number. And we know that there's something in the middle, and it has only six digits left. So that would be the largest number, starting with three, which is this number in the middle. OK, great. So you see how it's important to organize data in an organized manner. Do we have time for the last questions, David? What, 7.31? OK, so I will do one more question, because we lost internet. So I think we lost a couple of minutes. So I think I hope everyone can stay. If you have to leave, you can do video recordings. But we lost a few minutes, so I think it makes sense that we finish the last question and stay a few minutes late. Elimination method. Seven natural numbers A, B, C, D, E, F, and G are written in a row. The sum of all of them equals 2,017. Any two neighboring numbers differ by plus or minus 1. Which of the numbers can be equal to 286? And this elimination, because in a sense, we are lucky that we have the choices over here. So remember, this method we mentioned in the beginning. Is any way we can eliminate some of the obviously wrong choices and focus on the remaining choices. So we have seven numbers, natural numbers. I already got some answer, but let me just give you a little hint. You don't have to do it this way, but it might make the computation a little bit easier. I mean, these are numbers are big 286 in 2017, but if we notice that if you divide 2017 by 7, we would get 2 times 288 and plus 1, right? So that is if we have these seven numbers and if we can subtract, it's, I mean, the official name for that is the change of variables. If you can subtract 288 from each of these numbers, our life will be a little bit easier. You don't have to deal with cumbersome big numbers. So we can subtract 288 from each numbers. If we do that way, then we have a new problems, A, B, C, D, E, F. We've written in the row and the difference, right, between neighboring is the, sorry, I have to put things down here. The difference between neighboring numbers are C plus and minus 1, C plus 1 or minus 1. But now, because we're subtracting it from each other, then we ask ourselves which of the numbers, okay, the sum of them now equals 1 plus 1, the remainder, and we can ask which of the number can be equal, which can be equal to minus 2, because that's the difference between 2 and 1888. So it's an equivalent problems and this method is called change of variables, makes it a little bit easier for us. So, changing variables. So that's what the math competition see, right? We have to go in and have combinations of several different strategies so that we can get the answer in a reasonable amount of time. And then, okay, let me, and then, you know, try an error as well. I start with minus two first to see what I can get. If you increase to minus one, and then zero and one, and go down to zero and one and two, if we do it this way, due to symmetry, when add up together, the sum it actually becomes one, right? The sum is sum equal to one. We add them up together, sum equal to one. And nicely, we have A equal to minus two. And when you know that A equal to minus two, then we use elimination. This is really one of strategies here, but then we can eliminate B. This is the wrong choice. We can eliminate C. We can eliminate D. And then we are only left with A or E. Still hard, but makes our life, I mean, so much easier. Now we simplify a lot. And this one, another strategy that we can use over here is that use symmetry. Use symmetry. If A can be minus two, and then we can, you know, flip it the other way around, and G can also, G can also be minus two, right? So it's another choice, either A and G or all of them, right? So we are left with only two choices. Could it be A and G or all numbers, right? And in order to do so, we can only have to make an example. If we suspect that it's hard to make, it's impossible for all the numbers can be equal to minus two, then we can just put an example. I think some of you already figured out why when I'm doing this, which is great. So I, again, we choose a special case. I will choose the middle one equal to minus two, and then I try to increase on both sides. Minus two is small number. I need to make the sum equal to one, so I increase as much as I want. I am allowed to, to the left, to the right, and to the left over here, I can, I am allowed to make it minus one here, and zero, and one, right? Because the difference has to be one minus one. If I add all of this together, I do not get, I do not get the sum of this is smaller than one minus two. So that means C cannot be D, D cannot be equal to minus two, right? And then we are left with only one choice. The answer is A and G, and oh great, many of you got the answer why I'm doing this, which is great. So I don't know, David, do you have any thing to add on, but I feel like these questions you had to, you know, got to just not elimination, but I know you can call it a few different tricks or strategy to get the answer in a reasonable amount of time in a few minutes. So what I'd like to say, when I first saw this problem, this is, you know, this is just me, but I don't think it's just me, because the first thing that came to mind was elimination and parity. I was also thinking about the parity strategy, and I think that all the people who did answer used parity too, and what I mean by parity is say that the first number is, say that B or F, for example, say that the second number is even, then the first number has to be odd, because difference of one, the third is odd, even, odd, even, odd, and then this sum would be something even. So if 286 was either B, D, or F, then we'd get an even number, but 217 is odd. So right away, goodbye, this, this, and this, and that leaves you with C or E. The reason I think a lot of people did use parity is because A and C are the only answers that anyone put. It's just there, not everyone agreed. So I think a lot of people did use parity to figure out it has to be in an odd position, which is also a very useful trick. That's also how, when I looked at this problem, I got rid of any of them straight up, because any of them doesn't work, because, you know, it can't be the even ones, and then that leaves you with this, which is how I think a lot of people approach this problem. Great, okay, that's great. So you used it, but then after that, how would you continue to the end? How would you limit? Then you would still have to try one, but the beauty is that once you try only A or G, you don't have to worry about it being any of them, you know, so it's a different way. You still have to either try two things or use parity and try one thing. It's still two steps. It's equally long. It's just another way to solve it, and I feel like most people did the first step, and then I think quite a few just guessed between A and C, which is reasonable, right? You're very close now, but that's the point where you can really go ahead and just try to make one of them work and see if it happens. Yes, that's great. Thank you, David, and I think that's what people have as well, because yeah, we have equally, not equally, but yes, the only choices we had is A and C, and that's great, and then yes, thank you so much, and I think I don't want to go too much over time, but I hope, you know, we've got something useful today, and thank you so much, David, for the lively discussion. We hope to see you. You can definitely watch the recording. It's a small class, so we really hope to see you live next week. Thank you, everyone, and see you.

Mexican Guru

Math Kangaroo

advanced problem-solving

Phan Kurulava

David Deutsch

algebra

number theory

combinatorics

logical reasoning

geometry

creative approaches