 Hi, I'm Zor. Welcome to Unisor Education. We will solve a couple of logical problems today. This is part of the course called MEST Plus and Problems presented on Unisor.com. Totally free website, by the way. No advertisement, no strings, even signing is optional. And on that particular website you can find a prerequisite course for this one, which is called MEST for Teens, and there is a Physics for Teens and there is a Relativity course. So everything is free, as I was saying. So I have relatively completed, I believe, the MEST for Teens course. This is theoretical background for regular school mathematics and, well, maybe a little bit higher, maybe up to the first year of college. And this course, MEST Plus and Problems, is basically the continuation of that course. And it presents certain problems which you will unlikely meet in regular schools. These are problems to force you to think about something. It's not like to check whether you have correctly got the theory part, like some theorem or something like this. These are problems which you really have to think about. And what I suggest to you is, first I will explain the problem itself and then I will go to solution of this problem. So I suggest you just to pause the video right after I present the problem and think about this yourself. The most important part is for you to think about it. And that's the whole purpose, actually. MEST is the great gem for your brain. And unlikely you will meet anything close to these problems in the real life. But at the same time, the whole course of mathematics is, again, very unlikely you will find something in regular life except like calculations or something like this. But it's a great tool to develop your analytical mind, develop your creativity. So thinking is the purpose of this particular course, MEST Plus and Problems. Okay, so let's go to the problems. As I said, there are two problems. One is easy and another is based on the same thing but a little bit more general. So the first problem is consider a traveler who came to a different country and he would like basically to arrange his staying there. So he's staying in a hotel, pays with a credit card, but he does not have a local currency. However, he has a silver chain. Silver chain has seven links. One, two, three, six, seven. And he has agreed with the restaurant owners where he is going to have seven dinners for seven days that he will pay one link from this chain for every dinner he has. All right, I mean obviously some artificial situation. So the restaurant owner agreed but he said that he liked these links to be basically like cold links, like rings. He doesn't want to cut them basically. But then he has agreed, okay, I can agree to accept one link which is cut. So my question is, how can you arrange cutting only one link out of these seven to be able to pay for seven dinners with one link every time? Okay, so that's where you probably have to pause the video and think about it. And I will give you the solution. The solution is this. So if you will cut this link, you will be able to pay for the first dinner with this link. So these are connected. So this link is cut, so it's disconnected from these two. So for the first dinner, that would be the first. Now, the second dinner, what he can do, he can give two and take one back. So that would be the second dinner. Now, for the third dinner, he can give these two and this one. For the fourth, he will take back these three but he will give these four. For the fifth, he will leave these four and add this one. That would be the fifth dinner. For the sixth, he will take one back but add two of these to these four. And for the seventh, he gives all of them. So as a result, only one link is cut. And we individually pay for each dinner with one ring extra. So that's simple. Now, here is the second problem. Let's assume you have the same chain with n links. And you are making only k cuts. So k links can be cut. And our task is how can we cut it in such a way to basically maximize the number of days you can pay individually one link at a time. So this is much more general problem. And again, pause the video and think about it and I will go into solutions. Okay, so let's assume that we have this n links and we have made k cuts. Well, first of all, immediately following from it is that from the days from number one to number k, we have individual links which we can pay one at a time. So the first k days are covered out of n. Now, what's the next step? Well, using these k links which have been cut, we can pay only with the k days. Now, there is a day number k plus one. So what do we do at day number k plus one? Well, we have to somehow make these cuts in such a way to be able to pay with the number k plus one without any additional cut. Well, what does it mean? If we don't cut anymore, we cannot pay anymore with individual links, right? So our next choice is to have a link, a piece of a chain, a segment of a chain which has k plus one links together. So the first cut of these k should really cut in such a way that you will have one unbroken segment of chain with k plus one links. Let me give you an example. Let's say you have two links cut. So for the first day you give this one, for the second day you give this one. For the third day, which is k plus one, k is equal to two in this case, you need to have three links in one segment, and then you will give these three links and take back two which have been accumulated at the restaurant. So you need three links here together. So now you can do three, then you can do four, you can do five, and what's next? Next is six, and you don't have out of these anything which is six. So six supposed to be the next linked together segment of chain. So one, two, three, four, five, six, all of these should be connected. So you have three of these and six of these. What does it give? Well, it gives you six and five, and six plus one would be seven, plus one would be eight, plus three would be nine if you take back these two, ten would be all of these, three, one, and six, and eleven would be all of these. So for the twelves you need one, two, three, blah, blah, blah, twelve linked together. And that's basically as much as you can do because with two cuts you can have three segments uninterrupted. With K cuts you can have K plus one segments, but no more than that. So with two segments this is what you can do. You can do everything from one to eleven plus, well, to twenty-three. So up to twenty-three days you can have a chain which contains twenty-three links, and you can pay one link at a time for twenty-three days, cutting only two. So with cutting only one you can have maximum seven. With cutting two links you can have up to twenty-three. So what I'm asking is with cutting K links how big the chain can be so you can actually do this payment every link at a time. So you've got the idea. And what's important is that after the first K cuts we can pay for K days and for K plus first we need this one. Okay, now how many days you can pay now? Well, you can pay with K and K plus one. So up to two K plus one you can pay a day at a time. For two K plus two you have no choice basically. So the next segment uninterrupted piece of chain should have two K plus two. So this and this is K, this is K plus one, and this is two K plus two. No, sorry, this one is two K plus two. So if K is two, two K four plus two six exactly six. The same thing as we have here. So having first segment as K plus one link then you have a cut. Then the next segment should be two K plus two then another cut. So what's next? So let's just count K plus one plus one plus two K plus two plus one. How many are all together? It's three K, three K plus one, two, three, four, something like this. Okay, so this is K plus one, which is three. This is one, this is two K, okay. So next one would be, so what can we actually pay right now? In this case we can pay up to eleven, which is four K plus four, is that right? Here it is. We have K, we have K cuts, right? K individual ones. We have K plus one and two K plus one. These have already become two. So it's K plus one plus two K plus two and plus K individual one. One, two, three, four plus three. So up to four K plus three we are covering. So the next segment should have four K plus four. Yes, correct. Okay, now let's check about this twelve. If K is equal to two, two times four plus four, this is twelve. Okay, so that's five. So we are good. So now let's think about next. Let's have the case when you have three cuts. So for three cuts we have K plus one, two K plus two, four K plus four and K, which all together is what? Three, four, seven, eight. Eight plus one, two, four, seven. So this is all together eight K plus seven, which means this one, the next one, is supposed to be eight K plus eight. So you see basically the law is very simple here. It's a geometric progression. You have K plus one as the first member of this geometric progression. And then every time we double this value. So if you have K cuts, you have K plus one segments, right? Cuts are in the middle of the chain, right? If you have K cuts, you have one, two, three in this case, one more segments around these cuts. So we have K plus one segments, which means we have K plus one members of this geometric progression plus K individual links. And that's the total number. So all we have to do right now is to basically count how many links we have if this total number of links is equal to sum of geometric progression and K individual links together. Okay. Now, so what is this geometric progression? The first A is equal to K plus one. The factor of geometric progression is equal to two, right? So this is the first member and this is the multiplier. How many members do we have? Well, if you have K plus one, so we have total number of participants in this geometric progression, which we have to summarize, is K plus one. Since we have K cuts, it will be K plus one members. All right. Now, I do not remember the formula. I'm very bad with formulas, but I do know how to derive formula for geometric progression. So let me just do it right now. So you have A, AQ, AQ square, blah, blah, blah. And the last one would be if it's K plus one members, it should be K plus one, right? This is Q to the first and this is K to the K, sorry, to the K. And that would be K plus one, right? So that's what it is. No, not A to the power of K. It's AQ to the power of K. Okay. How to summarize them? If this is S, you do S times Q equals to what? If you multiply by Q, it would be AQ plus AQ cube, I mean, square. AQ K plus AQ K plus one, right? So if you will summarize them, you will have S plus SQ, sorry, not summarize, subtract SQ minus S. From this, you subtract this. This would be canceling out. You will have AQ K plus one minus A. So S is equal to AQ K plus one minus. Well, I'll take out of parenthesis and this will be Q minus one, right? If you will take S out of parenthesis, it will be Q minus one. This is the formula in my case. Now I have to basically substitute, so it would be Q minus one would be one. So S is equal to K plus one times two to the power K plus one minus one divided by one. Now this is my sum of geometric progression. So these are only segments, plus K individual one, individual links which I have cut in the very beginning. So I will have this plus K, which is K plus one, two to the power of K plus one minus K plus one. That's this one. And plus K equals to K plus one, two to the power of K plus one minus K minus one plus K, so that's minus one. So this is the number of, this is the length of the chain basically, which I can use to pay for this many days If I make K cuts, if I cut K links in the positions which leaves my first segment as K plus one, then cut. Then the second would be K is plus two and cut. And then four K plus K plus four and cut, etc. And the last one would be whatever it is. So this is how we can cut this chain which contains this number of links to be able to pay for that many dinners one link at a time. So that's basically the answer. Now let's check just in case if the formula is correct. Now in case of seven links, we have two together, one cut, and then four together. So one, two, three, four, five, six, seven. Okay, now link is only one cut, so K is equal to one. Okay, let's put the K into this formula. K plus one is two times two square, that's four, that's eight minus one, seven. Right, seven links. Okay, when we had two cuts, we had three here, we had six here, and we have twelve here. So it's three plus one plus six plus one plus twelve, which is ten, eleven, twenty-three. Okay, if K is equal to two, two plus one, that's three times two to the power of three, which is eight minus one, twenty-four minus one, twenty-three. So formula seems to be correct, and we have derived this formula using whatever the logic we were just discussing. Okay, so basically, let me just say it again, you will never have this type of a problem in the real life. However, again, the seeking solutions to these type of problems really is very, very important for developing your own creativity, which you can use somewhere else in real practical problems. And that's the reason for this particular course, Mass Plus and Problems. Now, within this course, I have a certain number of parts. This is the part which is called logic, and this is the number four lecture in this part logic. There is arithmetic, there is algebra, there is geometry, so you go to the website, and basically in any order, the previous course, the mass for teens, is logically arranged with menus and sub-menus, etc., etc., so you have to really go sequentially because everything is related to whatever was before. In this case, I'm using all that theoretical knowledge which have been accumulated in the mass for teens course, and here I just present problems absolutely unrelated to each other, so you can solve them in any sequence you want. Okay, that's it for today. Thank you very much and good luck.