 Hi, I'm Zor. Welcome to User Education. So today we will solve problems which I would classify as logical problems. So this is the lecture number five in this department of logic. Now, this is part of the course called Maths Plus and Problems presented in Unisor.com. Well, this is a continuation of the main theoretical course which is called Maths for Teens. On the same website. Well, there are other courses as well like Physics for Teens and Relativity for all, etc. So basically my purpose was, after the theoretical course called Maths for Teens, I wanted to present certain problems which are not exactly like illustration for the theory, but rather to force you to think creatively outside of the box, basically to develop your analytical abilities, logical abilities, creativity, and factors which are really very much needed in real practical life. Obviously, none of the problems which I present here are of any kind of practical sense, well, almost none of them. But again, this is just a gym for your brain. This is exercise for your inner powers. Okay, so after, yeah, I didn't really mention that everything on Unisor.com is totally free. There are no advertisements, no strings attached. You don't even have to sign in if you are just studying alone, just for yourself. If you're supervised in your studying, then we need something like relationship between a supervisor and a student, and then there is need for signing. But again, it's just signing, like name and password, and that's it. Okay, so logical problems. And these logical problems are not really very difficult, but they have some kind of a twist which might be interesting. And what's very important is I will present the problem, but before I start explaining how I would solve it, I would suggest you to pause the video and try to solve it yourself. Alternatively, you can go to the website, and every lecture has a textual part, which is basically like a textbook, where I present the problem, not necessarily with solution. I mean, I'm talking about the problems of this course. So in some cases, I just give a hint in the textual part of this lecture. So it might actually make sense for you before you even start listening to the lecture, go to the website and see what the textual part is. There is a problem, a presented problem, and try to solve it yourself. Maybe there is a hint, use the hint. Sometimes I do present solution as well in the textual part, sometimes not. The solutions are for the lectures actually. All right, so I have five problems, so let's start. The first problem. Okay, imagine, let's say a casino, and casino manager is analyzing how many times people win in roulette. For whatever purposes, management purposes. To extract more money from the people. So let's say that he finds out that M1 number of people during, let's say, a day has won one or more times, and two people has won two or more times during the day, etc. And in more case M people actually equals to M. That's the maximum. Nobody won more than M games during the day. Now, that's what he knows from whatever sources, doesn't really matter. Now what he does want to know is how many times casino lost. Again, absolutely artificial conditions because it's much easier for him to know how many times casino lost, just asking, croupier, whoever. But in any case, it's a logical problem, it's mass, so this is the problem and we have to solve it. So again, M1 people greater than one game and two people greater than two, etc. Now obviously M1 is greater than M2 because among people who won one or more, there are definitely people who won two or more, and M2 is greater than three, etc. And M is the smallest number of people who won the maximum number of games, which is M. How can we solve this problem? So again, pause the video, think about it yourself. Now, before doing this actually, one second, I will give you an answer. The answer is M1 plus M2 plus, etc. Now, when I solved the problem and found that's the result and checked it against the answer in the book actually, I was kind of surprised because it looks like the number of people who are in the group of people who won one or more is basically everybody from this group is contained here, everybody on this group is contained there, etc. So that was kind of a surprising answer for me. Nevertheless, it's true. Okay, now you can pause the video and I will explain the solution. Okay, let's think about it this way. If M1 people won one or more games and two people won two or more games, how many people won just one game? Obviously, there's a difference between these two. These people won one game. Now, let's talk about two people, two games. How many people won just exactly two games? Well, if M2 won two or more and M3 won three or more, so the difference between them is number of people who won two games. So this number of people won't get one game. This won two games. Well, obviously I can continue this. Let's say N minus two minus N minus one won N minus two games. N minus one minus N minus one game. And N people exactly N games as we saw. So this is number of people who won one game. This number of people who won two games exactly. So how many people actually, how many games these won N minus one minus N minus two games. This won N minus two minus N minus three times two because each one won two games, right? This one was N minus two minus N minus one times N minus two games. Then N minus one minus NN N minus one games and plus NN times N. So if we will summarize, this is number of games which were won by people and lost by casino. Okay, so what's the sum of this? All right, look at it this way. N one minus N two plus two N one minus two N three plus three N three minus three N four. Look at this. This is N two, this is N three, this is N four, etc. And what will be at the end? So everything goes with the previous one. So this is N minus one times N minus two with a minus. This is N minus one times N minus one with a plus. So it will be N minus one. And minus N times N minus one plus N times N that times NN. So it will be NN. So this is exactly NN one and two and three and four and minus one and N exactly like I told you in the very beginning. That's it. Next problem. Next one is easy. So let's talk about chess, chess game. Well, not exactly chess game, but just one particular aspect of the chess game. So there is a rook. Rook goes vertically or horizontally, as we know, right? Now there are eight by eight squares on the chess board. So my question is, how many rooks can you put onto the chess board in such a way that every rook can go left and right or up and down without hitting another rook? Okay, let's say we can put it on diagonal. It would be eight rooks. So obviously each one can go all the way up and down without disturbing each other. Is it a maximum? Yes, it is a maximum, obviously. Because since you have only eight rows, let's say, and obviously you have to have no more than one rook in a row, that's your maximum. So eight is your maximum. So we know how to position eight rooks so they don't disturb each other. They can go freely up and down without hitting anybody else. Obviously this is the maximum. So this is a very simple problem. But I wasn't really presenting this problem for this problem, just for the next problem. Next is a little bit more interesting. But it's the same story with rooks. And right now we are talking about a maximum of eight rooks putting on the board. And my question is, let's numerate the squares on the chess board. One, two, three, eight, nine, ten, etc. 16 and up to 64. So we have 64 squares and every square has a number. Now my question is, what if you position eight rooks exactly as before? I mean, no rook is preventing another rook to go up and down all the way to the edge of the board. My question is, what's the sum of the numbers which are where the rooks are positioned? Well, in this particular case if we have diagonals, diagonals have number, what, one? Then next one will be, this is nine, so this is ten. And I always add nine, right? So 19, 28, etc. So if you summarize them somehow, you will get some kind of a number. Okay. Now, we can position rooks different. For instance, we can position along this diagonal. Or we can position it somehow this way. In some order, whatever the order we can put in, in such a way that they do not really prevent each other from movement, we can position them. And obviously we can calculate the sum of the numbers. What is the sum of the numbers? Does it depend on how we position our rooks or not? Well, think about this. Pause the video. And I will tell the result of this. Well, the result of this is that the sum will be always the same. Again, kind of unusual answer. You position it differently, but the sum of the numbers will be the same. So let's calculate this number. Okay, here's how we can calculate it. Now, we have eight rooks. So we will have row number and column number. Row number and column number of every rook. So I put index i. Where i is equal to 1, 2, etc., 8. Okay, so we have this. Now, what is the number associated with the square which has coordinates r i and c i? Well, that's kind of easy thing because obviously it's r i minus 1 times 8 plus c i. If the row is equal to 1 and column equals to 1, so it will be 1 minus 1, which is 0 times 8 plus 1. Let's say we have the very last one, 8, 8. So what we will have? 8 minus 1 times 8 plus 8, which is 5664, obviously. So this is the right number. Something like this number. This is the last row is equal to 8 and column is equal to 1. Now, this is 64. This is 57, right? So let's try if r is equal to 8 and c is equal to 1, 56 plus 1, 57. Yes, that's the correct word, though. And obviously the number of row minus 1 should be multiplied by 8, the number of elements in a row, and then plus the column number. So that's fine. Okay, formula we have. Now, let's summarize r i minus 1 times 8 plus c i from 1 to 8. So what we will have? Well, we will have 8 times sum of r i minus 1 plus sum of c i. Now, let's think about it. From the previous problem, we know that every rook has its own row and no two rooks are supposed to be in the same row because otherwise they would disturb each other in their movement. Same thing with column. No rooks should be in the same column. So all the c's must be different. Now, what are the c's value? It's 1 or 2 or 3, etc., or 8. Now, they're all different. They might be from 1 to 8. So this sum is basically independent. It's basically sum of 1 plus 2 plus, etc., plus 8, whatever it is. Now, this is exactly the same. All r i's are supposed to be unique because no two rooks are supposed to be in the same row, which means when you resummarize it, it will be... Well, r i is from 0 to 1, so r i minus 1 is from 0 to 7. So it's 1 plus 0 plus 1 plus, etc., plus 7. So that would be times 8 plus this. And if you will calculate, it will be 260. So again, no matter what's the order of r i's and c i's, they must be different, which means that r i minus 1 should be from 0 to 7. All the different values. In maybe different order, but the sum would be exactly the same. All right, so that's it. Next problem. By the way, this was also kind of unexpected result that no matter how we put all these rooks, the sum would be the same. Okay, now another kind of strange result. You have two cups. One is coffee, another is milk. Absolutely the same cups and the same amount of liquid in both cups. Then what we do is we take a spoon, take the spoon of milk, put it into the coffee, stir it and then take whatever the mixture is and put it back into milk. That's it. Now my question is, where the concentration is greater? Milk in coffee or coffee in milk? Stop the video, pause the video, think about it and I will tell you the result. Well, the concentration would be the same. Now, why is it the same? Well, again, it's logic basically. Think about it this way. Whatever we mix together doesn't really matter. But as a result, the amount of liquid is the same, right? Because we took the spoon here and spoon full back. So the amount remains the same as it was in the beginning. But there is a mixture. But think about it this way. Whatever the amount of milk we took and put into coffee, if the amount of coffee is, I mean amount of mixed coffee and milk is the same, it means exactly the same amount of coffee we put into the milk. So after these two transformations, or maybe it's four, maybe we did it again and again and again, it goes back and forth. The amount in both cups will be the same. The amount of coffee and milk was the same in the very beginning, which means that the amount of coffee and milk should be exactly the same proportion as milk in the coffee. Well, that's your answer. Although in the beginning you think that, okay, we put the pure milk into the coffee, but back into the milk we put the mixture of coffee and milk. So it does seem to be disturbed, the balance. But no, it doesn't disturb, because again, whatever amount of milk is here is exactly the same amount of coffee is there because the level remains the same. Okay, okay. And my last problem. Let's say you have a table, rows, columns, whatever. Now, there are numbers in this table. Some numbers. But what's known is that every number, let's say this one, is a arithmetic average of its neighbors, where it actually shares the border. So we don't have diagonals. No, just up and down or left and right. Now, if you have this, for example, number, that's average of this and this. And if you have, let's say, this number, it's average of these, these and these, these are neighbors. So every number is average of its neighbors. Now, proof that all the numbers must be the same in this table. Otherwise, you cannot really have this thing really arranged in such a way that every number is the average of its neighbors, where it shows the border. Okay, how can we prove it? Again, pause the video and think about it. It's actually very simple. Now, we are talking about table with certain finite number of numbers, which means there is a maximum number. We can start with the minimum or maximum. It doesn't really matter. Let's start with the maximum number. So whatever the maximum number is, it doesn't really matter whether it's in the corner or in the border or in the center of the board. If it's average of its neighbors, but this is the maximum number, how can maximum number be the average of its neighbors? Only in one particular case, if they are all equal to this maximum number. Because whenever you have any number 5, if it's an average of two different numbers, let's say 2 and 8, either one should be less than the other should be greater, or they are both the same. But this is the maximum. That's what we started with maximum, so we cannot have something like this, which means only this case remains. And it doesn't matter whether it's only 2. If it's only the corner of, or if it's 3, let's say 5, you have 3, 2, and what? We have to have 15, right? So it's 10. Same thing. This is average of these 3 numbers. 3 plus 2 plus 10 is 15 divided by 3 is 5. Correct? But if this is maximum, we cannot have this case. So if the maximum number, which is greater than everything else, or equal to everything else, is average of its neighbors, neighbors must be equal. Okay, so if this is the maximum, let's say this is, then its neighbors must be the same. Now, we can start repeating this thing again and again and again. And this thing would grow and grow and grow. To all directions, it must be the same. And gradually we will fill up the whole table. We will prove that basically all the numbers in this table are exactly the same as the one we started with. So if we start with minimum, it would be exactly the same thing. Which means if we start with minimum, and this is minimum, there is nothing less than this minimum, which means the neighbors must be equal to this number. And then we go again and again and again. All right, so these are all the problems I wanted to present to you today. I would suggest you to read the notes for this lecture, go to unizord.com, go to course called Mass Plus and Problems. And there is a logic section, and this is number O5. You will see it and read the notes for this lecture. All the problems are presented there. I'm not sure I present... I don't think I presented the solutions, but I do have some hints in there. So I would suggest you just to repeat the same thing, use the hints and come up with your own decision, solution, analysis, whatever, logic. Okay, that's it for today. Thank you very much and good luck.