 Hi, I'm Zor. Welcome to Unisor Education. Today we will solve a few problems. These are part of the course called mass plus and problems presented in Unisor.com. This is a continuation of the prerequisite course which is called mass routines, where I'm mostly involved in some theoretical issues in mathematics. There are some problems, obviously, over there, but they're kind of illustrative problems to illustrate the theory. In this particular case, in the case of this course, mass plus and problems, I'm trying to present certain problems which really have to kind of force you to think about the problem outside of the box. I mean, you definitely will be using the theory which you have learned before, but it's in a more creative way, so to speak. So the problems which I'm presenting in this course are not exactly standard. Sometimes they are maybe easy, but sometimes maybe more difficult. Again, the main difference of the problems presented in this course is they are not the usual stuck mass problems presented in school. So the purpose is, obviously, to train your mind. These problems are basically like training grounds for your brain. None of them, or very few of them, have any practical application. They are just to ensure that your brain is functioning to its full capacity. That's training. That's gym for your brain. Okay. So today we have three problems. The problem number one is as follows. You have two numbers. Both numbers contain only digits in its digital recording. So X has 1111111111 eight digits. Number one and number two has 111111100 digits. Now, from arithmetic, and by the way, this lecture is problems about arithmetic. The arithmetic O4. That's how it's called. So in arithmetic we are all talking about basic numbers and their elementary operations. Addition, subtraction, division, multiplication. Now, there is a concept of the greatest common factor between two factors, between two numbers. This is the number both are divisible by, and that's the largest number among all the divisible numbers. So I'm just repeating something which you're obviously supposed to know. So like, for example, if you have number, for instance, six and let's say, what, 15, for example, what is their largest or greatest common factor? Well, obviously both are divisible by three and there is no bigger number than three. Both are divisible. So three is their greatest denominator. So my problem is to find the greatest common factor of these two numbers. Well, obviously numbers are small, like six and 15, I just told you. You just basically go through all the factors and see which one of them is largest. Well, in this particular case, considering this is a very large number is impossible. So we have to do something else. So at this particular moment, I suggest you to pause the video and think about how to solve this problem. And I will continue with the solution. Okay, now, let's just implement certain representation of these both numbers in a different way, the way how usually numbers are presented, if digits are given, only digits. Well, what is digit? It's a decimal system of representation of the numbers, which means this digit has way 10 to the power of zero, which is one. This digit has basically the weight of 10 to the first, power of one, 10 to the power of two, three, et cetera, up to 10 to the power of seven, from zero to seven. So I can write X as one times 10 to the power of seven plus one times 10 to the power of six, et cetera, plus one 10 to the power of one, plus one 10 to the power of zero. So this is a more algebraic, if you wish, representation of this number. This is just by digits, and this is the representation using that every digit has certain weight. Now, same thing we can do with Y. Y is one power of 99, plus one times 10 to the power of 98, plus, et cetera, plus at the end we have exactly the same thing. Okay, so that's the beginning. And look at this. The tail is exactly the same, right? So up to the one times 10 to the power of seven, plus, et cetera, one to the power of 10, zero, this part is the common part in both things. So the whole X is actually is here. Now, what's next after that? One to the power of 10 to the eight, one power 15. That's the next group of numbers, right? So I'm talking about representation of Y. This is the tail, this is the next to the tail. Now, how can I represent this? Now, if this is X, this is 10 to the power of eight times X. Am I right? Because 10 to the power of eight, you just take out of the parenthesis and I will have 10 to the power of seven, then six, et cetera, up to zero, which is exactly the same as this. So that's the next group of eight digits in the representation of Y. So what would be next? Well, next would be, as you obviously understand, next would be one 10 to the power of 15 plus eight, 23 plus, et cetera, one 10 to the power of 16. That's the next group of numbers. Which is what? Which is 10 to the power of 16 times X, right? If I will 10 to the power of 16 take out of the parenthesis, then I will have exactly the same thing as X. So as you see, if I will group it by eight, my result would be that Y would be equal to... Now, what will not fit to the group of eight? 23, next would be what? 31, 47, 63, 90, 95, right? Let me check my... I think it's 95. Yes, 95. So the last one would be 95. The last group would be one 10 to the power of 95 plus, et cetera, plus one times 10 to the power of 80... What, 70? 80 minus... 87, yes. So up to 95, we can represent each group as 10 to some power times X. And only starting from 96, we don't have a complete group of eight. So I can say that this would be 10 to the power of 99 plus... I skip one time, so it's obvious. 10 to the power of 98 plus 10 to the power of 97 plus 10 to the power of 96 plus... And then I will have some number, 10 to the power of whatever, 97, in this case, times X, then again 10 to the power of 90 X. So I will have 10 to the power of 87 plus blah, blah, blah, some other 10 times X. So all these groups I combine together, I put X outside, and the multipliers like this, like this, like this, I put inside. So this is the result of representation of number Y in terms of number X. Now let's recall what we have to do. We have to find the common factor, which is the greatest or the largest among all common factors. Now from this, obvious that this number and this number and this number... So if Y and X have a common factor, exactly the same common factor should have this number, correct? Because if this is divisible by something and this is divisible by something, this must be divisible by something, because this is equal to Y minus multiplied by X, so left part will be divisible, that's why the right part must be divisible. Now this can be actually represented as 10 to the power of 96 as 10 to the power of 3 plus 10 to the power of 2 plus 10 to the power of 1 plus 10 to the power of 0, which is 10 to the power of 96 times what? 1111. So we have this number, 1111. So this is this. So our factor, which is a divisor for this and divisor for this, must also be divisor for this. Well, obviously neither X or Y are divisible by 10 to any kind of power because they're all contains only 1, including the very last digit is 1. So they must have the common denominator with this. Let's just take a look at the X. Now Y is obviously divisible by this. Now X is 11111111, 8 1s, right? Which is times 10 to the power of 4 plus 1111. Correct? So it's 1111 times 10 to the power of 4, which is 1111 and 4 0s and 4 1s. So obviously X is divisible by this. And Y is divisible by this. And there is no more, there is no bigger number because everything else is 2s and 5s and obviously X is not divided neither by 2 nor by 5. So this is the largest common denominator. X is divisible by, and Y, which is represented as this and this is represented as this, is divisible by 1111. So that's the largest common factor. Okay, finished on to the first problem. But it's not really difficult as soon as you understand the representation of the number which is given you in digits in the form of 10 to the power of something times this digit. Now the second problem is something which most likely you did study at school but nevertheless I will just repeat it. Here it is. Let's take any number n, any. It has certain representation in digits, in decimal digits like n1, n2, etc. Well, which means basically that n is equal to 10 to the power of k minus 1 times n1 plus 10 to the power of k minus 2, n2, etc. plus 10 to the power of 0, nk. So that's the representation using these digits. So my theorem, which I'm just presenting to you is that remainder, remainder of n divided by 9 is the same as remainder of n1 plus n2 plus, etc. plus nk divided by 9. So the sum of the digits divided by 9 gives the same remainder as the number itself. Now, well, this is kind of a theorem, if you wish. So let's just prove it. Now how can we prove it? Well, actually it's very easy because I can always represent 10 to the power of, let's say, k minus 1 as 10 to the power of k minus 1 minus 1 times n1 plus n1. So this is equal to this, obviously, right? 10 to the power of k minus 1 times n1 minus n1 plus n1. Now, what is this? This is basically 1 with a certain number of 0s minus 1. That's what this is, which is 9, 9, 9, 9, 9, 9. Which is 9 times 1, 1, 1, 1, 1. Whatever number 1 says. Which means it's divisible by 9. So what I would like to say is that if I will represent each number like this in this format, I will have 9 times whatever some big number n1 plus n1, each of them. So I will have n which is equal to 9 times n1 plus n1 plus this one. It's the same thing. It's 1 and many 0s minus 1 plus 1. And minus 1 would be many 9s. Whatever number of 9s is, I'll put it as n2. And again, minus 1 and plus 1, multiply by n2 plus n2 plus etc. And the last one would be 9 times nk plus nk. Okay? This one. 10 nk. No, that's the previous one. Then it would be nk minus 1. k minus 1 plus nk would be by itself. Just 1. So what do we say? We have this is divisible by 9, this is divisible by 9, this is divisible by 9. So everything is divisible by 9 except this, this, this and this. So n would be represented as 9 times n1 plus blah, blah, blah. Plus n1 plus n2 plus etc. Plus nk. So sum of digits and n have to have exactly the same remainder if we divide by 9 because this piece is divisible by 9, right? So if you divide by 9 and you'll have some remainder here, you must have exactly the same remainder here. Because representation of any number if you divide it with certain remainder, so n times, 9 times p plus q, right? q is remainder and p is how it's called quotient, right? And that's a unique representation. So if n is equal to this, then this sum of its digits must be exactly like this, which means p should be the same and q should be the same. That's why remainder is the same. That's the end of the proof. So remainder of the division of the number and sum of its digits, division by 9 is exactly the same. And incidentally, by 3 would also be the same because 9 is always divisible by 3. So in this representation, whenever I represent it as 999 something, it's divisible by 3. Division by 9 and division by 3 have exactly the same property. Some of the digits of the decimal representation should have exactly the same remainder as the number itself. And as a consequence, if remainder is 0, which means if number n is divisible by 9 or by 3, then the sum should be divisible by 9 or by 3. So this is the rule which you can basically apply to find out whether this number is divisible or not divisible by, but let's say by 9 or by 3. You just take the sum of the digits, which is much easier. If you have a big number, you can't really easily divide it by 9 without some kind of either calculator or long division. To add the digits of this number very easy, you can check whether it's divisible by 9 or by 3. Okay, now. And the third problem, which is very much related to this one, it's based on this one, is the following. Let's assume that you have two numbers, n and 2n. It has its own representation, m1, m2, etc. What's the letter? Let's say p. Now, what if remainder of division of the sum of these is the same? So if remainder of the sum of these digits is exactly the same as remainder of these digits, then my theorem is n is divisible by 9. We are talking about division by 9 and remainder of the division by 9. So if the sum of digits of one number, if remainder of the sum of digits divided by 9 is the same as remainder of the division of sum of double that number when I divide it by 9, if remainders are the same, then the number is divisible by 9. So one number, double that number, we have the sum of these digits and we have the sum of these digits. If these two sums dividing in a division by 9 give the same remainder, then n is divisible by 9. Okay. Here is the very simple proof. Now, if n has certain representation as 9, let's say, p plus q, so p is quotient in dividing the n by 9 and q is remainder. 2n, let's put it p1 and q1. 2n is also, now let's take the sum of digits. Sum of digits, which I will put, sigma n i tan, also can be represented as some other 9 times, let's say x plus q1. It must be the same q1, right? And sum of digits of the 2n should also be 9y plus q2. And now I'm saying that these two are the same. So sum of digits divided by 9 gives some remainder of this and sum of the digits of the double number dividing by 9 gives exactly the same. So q1 is equal to q2. So I can put these are the same. Let's put it q and q. q and q. So what follows? Subtract them from 2n, I subtract n. We'll have n is equal to 9 times p2 minus p1. q minus q would be zero. So n is divisible by 9. That's it. That's the end of the proof. So, again, this is an immediate consequence of the previous problem. In the previous problem, I was telling that division of the number by 9 gives exactly the same remainder as the division of the sum of the digits by 9. So in this particular case, since both numbers n and 2n give the same remainder here, it means that the numbers are the same, have the same remainder. And if numbers have the same remainder, then the representation would be this. p1 and p2 quotients are different, but the remainder is the same. And then I just subtract it and that's it. Easy. Okay. It's very important to read the notes for this lecture. Every lecture on Unisor.com for all courses, including this one, have the visual representation and the texture representation. Basically, you can say that the whole website Unisor.com is a visual textbook. So there is a textbook and on each chapter or part of the book, whatever, have a lecture. Now, if you understood more or less, I hope, whatever I was talking about, it's really very important to repeat the same thing when you just read the text. Text might be slightly different from whatever I'm presenting, maybe different letters I'm using for whatever, doesn't really matter. The idea is exactly the same and the textual part is written basically as a textbook. It's nothing like abbreviated or anything. It's really like a textbook and you can read about the same thing, which basically will help you to maybe understand better whatever I was talking about during the lecture. Now, the whole site Unisor.com is totally free, no advertisement. Sign-on is optional only for those who want to study under somebody's supervision to establish the relationship obviously. Otherwise, if you're just self-studying, you don't even have to sign in. Well, that's it. Thank you very much and good luck.