 Hi, I'm Zor. Welcome to Inusor Education. We continue a series of problems which I have kind of gathered together under one roof called mass plus and problems. So today we will solve a couple of four actually, so-called arithmetic problems. Well, I call them arithmetic because there are numbers involved. So again, I will continue using these logic arithmetic algebra as a general kind of title of the series of problems related to either of these particular areas of mathematics. They will not be presented in any particular order because basically they are not related to each other. It's not like I'm trying to put some educational material in certain sequence because I'm using something which I have already covered before in the subsequent lecture. No, these are completely separate, so you can just take them in any sequence you would like. But they are gathered together inside this Unisor.com website under the course name mass plus and problems. Okay, so today I will talk about a few arithmetic problems just because there are some numbers involved. So problem number one. Okay, so there are certain number of pieces of paper. Sometimes it's presented as concrete like 10 number of pieces. Now, I will just use any number and I'll just use as an M. So there are N pieces of paper. Okay, so what we did, we took some of these pieces, let's say K1, and each of it cut into N pieces. Then whatever remains, all these pieces of paper, original which I did cut, which I didn't cut, whatever. Again, it's a group of set of pieces of paper. I took again some of them, let's say K2, and each of them cut into again N pieces. And I continue this process a certain number of times. So after that, I will have a lot of pieces of paper, let's say capital N. What I would like to prove is that N minus N has a number. So total number of pieces minus initial number of pieces. So basically the number of pieces which I have incremented. This is divisible by N minus 1. Okay, now as usually I recommend to pause the video and try to solve this problem yourself. By the way, these problems are really simple, no problem, no difficulties in them. And then after you have solved or didn't solve it, continue with watching the lecture, because you might have your own decision, your own solution, you might listen to whatever I'm saying. Plus there is a note for each lecture where I basically, like a textbook, I present the solution. Okay, how do we do it? All right, so let's start from the beginning. We have N pieces of paper. Now if I took K1 of them and start cutting them, remains uncut with the N minus K1. Now the K1, I cut each one of them in N pieces, which means I add K1 times N. So that's basically the result after the first operation. Let's call it N1. Okay, now N2, next step. What did I do? I took whatever number of pieces I have, which is N1, took K2 minus of them, which means I left N1 minus K2 uncut. But I added, since I divided each one of those K2 into N pieces, I added this one. So that's total number of pieces, which is equal to, okay, N1 minus K2 times N minus 1. By the way, I didn't specify this, N minus K1 times N minus 1. If you factor out K1, that's what you will have, right? Is that right? Plus. Sorry. Okay, which is equal to N1, I know where it is. So it's N plus K1 times N minus 1 plus K2 minus 1. N plus K1 plus K2 minus 1. Right? So just look at some kind of a hint. After the first cut, I have this. After the second cut, I have this. Now, it's very easy to prove by induction. I'm not going to do it because it's really very easy and you can do it yourself. That after, let's say, S steps like this, my result would be N plus K plus 1 plus 2 plus KS times N. Because every time I cut, I have to subtract KS, let's say, and add KS times N. And the formula will be continuing. These K will be added to each other. From which follows that NS, which is basically the total number after S steps, which I am just using the letter N, minus N would be divisible by N minus 1, which was supposed to be proven. So again, simple step to prove by induction that this formula would be correct is really simple. I'm not really wasting time. What's important is I'm offering these problems after a substantial course of mathematics has been covered. While I'm not using all the apparatus maybe, which we have learned during the theoretical part, which was called mass proteins. I'm not really using here at least these simple problems, but I still, it's important to go through these problems after you have learned a lot of theories because your mind is already prepared to logic, to creativity, to analytical thinking, etc. So the problem solving is very important, but you have to have the foundation, you have to have the background, theoretical background to be able to basically understand what this problem is all about. Okay, second problem. Okay, let's say one person decides to think about some kind of a number from 0 to 1,000. Another person is supposed to find out what this number is by asking questions which can be answered by the first person only as yes or no. So I cannot really ask what number do you think about because the person cannot answer. The person can answer only yes or no, which means I have to really ask many questions to find out the number. So the problem is, what's the smallest number of questions which can be actually asked about this? Well, I can offer the way which is actually the smallest, but it's not easy to prove that this is the smallest number of questions. That I will probably leave to you to think about it because it's not trivial at all. However, the method itself I will definitely present right now. It's actually used in many computer programs which are related to search algorithms. So in many cases, the way how we search for something is to use something which is called binary division. So what's the binary division in this particular case? Well, you just ask, is the number in the upper half or in the lower part of this range? So let's say, is it greater than 500? Well, there are two answers, yes or no? If yes, I know it's greater than 500, so it's between 541 and 1,000. Again, I divide the whole interval in half. I'm asking if it's greater than 750. Now, if it's n, if I answer no, which means it's less than 500, which means it's, now in this case, it's in 500 or 1 thousand. In this case, it's between 0 and 500. So I divide it by half and I'm asking is it greater than 250? Now, in this case, if yes, it means it's 751, 1,000. If no, that means it's between 541 and 750. You see, in each case, I reduce the interval. So in this case, it would be from 0 to 250. No, wait, wait. From 251 to 500. And this way will be 0 to 250. So on each step, I divide, basically, the interval in half and reduce my choice, obviously, in half. So how many times I have to do it? Well, how many times you can divide 1,000 in two? Well, I'm asking if it's greater than 10 times. I have to do it. Well, how many times you can divide 1,000 in two? Well, 1,000 is 2 to the 10th degree, a little bit less. 2 to the 10th degree is 1024. So, by no more than 10 steps, I will divide down to one single number. Maybe less, but, basically, I mean, it depends. But, basically, it's about 10. Now, what's the more mathematical equivalent of this? Well, the more mathematical equivalent is this. Instead of thinking about this number as a decimal number, think about this as a binary number. And, again, the theoretical part of the course, mass for genes covers what is a binary system. So the number can be represented as 0 and 1. So what I'm basically asking is, in a binary representation of number which you are thinking about, is the first digit 1? Well, if it's yes, then it's 1. If it's not, it's 0. So, basically, I already determined the first binary digit. So, the only thing is I'm asking to use all 10 digits, including leading 0, maybe. So, in exactly 10 questions, I will get a complete binary representation of this number. So, again, 10. Now, again, I don't want to discuss right now the issue why this is the shortest way to get to the number. I would suggest you to do it yourself. And that's a very interesting, actually, part of the whole thing. You can send me an email which is on every web page of theunisor.com. You can send me your thoughts and I will share it with everybody. Okay, so that's my second question. Again, this binary division methodology is really very practical. It's used in many computer programs which are involved in some search algorithms, given 100 numbers from 1 to 9, each one. So, basically, single digit numbers, 100 of them. Now, if I will add them up together, well, what's the minimum? Minimum is all of them are 1s, which is 100. What's the maximum? All of them are 9s. That would be 900. Now, in this particular case, this number, this sum of these numbers is equal to 789. Now, my question is, can or cannot I choose 70 numbers out of these 100 with their sum to be less than or equal to 500? So, you see, this sum is closer to the maximum. So, most numbers are large numbers. That's why I'm asking, is it possible from all those mostly large numbers to choose 70, which is much more than half, but with the sum with sum less than 500? Is it possible or not? Well, it's a very simple thing. The question is very simply to answer. Look, if 70 numbers have a sum of less than 500, then the remaining 30 numbers should have a sum greater than 289, right? Because the sum of both these 70 and these 30 is supposed to be 789. If 70 gives me less than 500, then the rest would give me this. But now, let's think about what's the maximum sum if you have 30 numbers, each one of them is this. Well, let's say 32 times 9. If all of them are 9s, which is 270, which is less than 289, which means we cannot accumulate sum which is greater than 289. Maximum is 270. So, it's impossible to choose 70 numbers. These numbers are too big and many big numbers closer to 9, like 789. Most numbers are 789, basically, right? So, we cannot choose 70 numbers with the total sum of less than 500. Simple, right? Now, the last one is more interesting, let's see. Okay. This is the clock. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Now, we have 12 coins and we put one single coin on each digit. 12 coins put in 12 digits. Now, one move is to move two coins. One coin should be moved from the place where it is by one step clockwise and another coin counterclockwise. So, let's say if I will make this move, so this coin will be here and this coin will be here and these places will be empty. Now, obviously, I can move them around in many different ways, whatever I want. My question is, can I gather all the coins on one particular number? Let's say on 12, it doesn't really matter which one. Can I get them together? All these 12 coins into one particular step. Now, in this case, for instance, next step can be I 7 move here, so I have three coins and 5 move here and I have two three coins here. Now, I can move them this one to here, let's say, and this one to here. It's two, it's one now. Then again, I move this to this, this will be four, and I move this to this, so it will be here, something like this. So, in any case, I'm moving the coins around. Can I get them together, like four in this case? Can I get all 12 together in one particular number? Well, the answer is no. And here is a proof. What I will do is I will calculate something which I called the value of the position. Coins are in some position. There are a certain number of coins for each number. So, what I will do is, let's say, as number one, I have K1 coins, as number two, I have K2, etc. So, what I will do is I will multiply the number itself by number of coins associated with this number. Plus two times K2, plus etc, plus 12 times K12. I call it the value of the position. So, how this value is changing when I make a move? Okay. If I move from this to this one coin, which means it's changing position from number six to number five, I reduce by one number of coins here, and I add one number of coins here. So, the value of position will be changed by minus six plus five. So, it's minus one, right? So, whenever I'm moving counterclockwise, I'm changing to by minus one. Or the only exception is if I move from one to twelve, then it's plus eleven. So, it's either this or this. At the same time, what happens if I move clockwise? Well, obviously, usually I'm adding one, or if I'm moving from twelve to one, I'm subtracting eleven. What is the result? The result of this is, if I'm making two moves, if I'm making two moves, there are different combinations here. If it's this combination, then I do not change the value at all, right? One coin will lose the value, another will subtract the value, another will add to the value where it remains the same. If I'm doing this, so one coin will be plus eleven, another coin will be minus eleven, will result in the value change, and nothing's changing. The value is not changing. Okay, what if I'm doing this? Well, then the value is changing by twelve. So, either I have no change, or value will be subtracted twelve, or if I'm doing these two moves, the value will be added twelve. So, the value is either not changing, or subtracting twelve, or adding twelve. Okay, fine. Now, what if all my coins are in the same place, all twelve of them? Let's say it's on four. What's the value of this position? Well, it's four times twelve. So, no matter where my twelve coins are collected on which number, I will always have this number times twelve. So, the value which I'm looking for should be divisible by twelve. It should be equal to six by twelve, or ten by twelve, or three by twelve, whatever it is. Can it be obtained? Now, what's my initial value? Initial value is one times one. We have one coin on each number, right? So, one times one plus two times one plus twelve times one, which is seventy-eight. Seventy-eight is not multiple of twelve. But on every move, I'm either adding twelve or subtracting twelve or don't change the value at all, which means that I always add twelve or add or subtract twelve or don't change the value. So, it's always multiple of twelve, which is subtracted or added to this number. Now, this is multiple of twelve. This is not multiple of twelve. And I'm adding or subtracting either zero or multiple of twelve on each step. I cannot get from here number, which is multiple of twelve. I can always, so what's the multiple of twelve? It's seventy-two, right? So, six would be always extra. I will not be able to come from a number, which is not multiple of twelve, using steps which are multiple of twelve to get to the number which is multiple of twelve. And that's the proof that this is impossible to do. So, I cannot gather all twelve coins on the same number together. Okay. Now, what I suggest you to do is read notes for this lecture. They are presented on Unisor.com. Go to the course called MESS plus and problems. And this is the problems which are in lecture called Arithmetic 01. So, you choose Arithmetic topic and then Arithmetic 01 will be the first lecture, actually, of this topic. Okay. That's it. Thank you very much and good luck.