 Hi, I'm Zor. Welcome to Unisor Education. Today we will solve a couple of problems which I will qualify as arithmetic problems, but because they are about numbers. Now, this lecture is part of the course called MESS and MESS Plus and Problems presented on Unisor.com. By the way, if you found this lecture somewhere else, like on YouTube, etc., obviously you are welcome to watch whatever you want, but the website actually gives you a little bit more because together with every lecture on the website there is a detailed description of the same material, which basically is like a textbook. So you have both the lecture, visual kind of presentation and textbook style presentation on the same page basically. The website is totally free, no advertisement, no strings attached, so you can use it as much as you can. Now, this course is a subsequent to a prerequisite course called MESS for Teens on the same website, and I do suggest you basically to familiarize yourself with whatever is in that course before you are approaching all these MESS Plus and Problems. Well, primarily because all the problems are related to some theoretical material presented in the MESS for Teens course, the prerequisite course. Also, there is a physics for Teens course on the same website. There is a relativity for all kind of introduction to special relativity course. Well, use it as well. So, a couple of problems today. The first problem is the following. Consider you have a prime number P, which is greater than 24, and I'm dividing this prime number 4 by 24, which means it's represented as this, where is a remainder. Now, what I would like to prove is that R is prime as well. So, P is prime, I divided this by 24 with a remainder, and I'm staging that remainder is supposed to be a prime number. Okay. Now, as usually before I proceed, it makes sense for you to basically pause the video and try to solve this problem yourself. Now, the textual part, the notes for this lecture on the Unisor.com website does not contain the solution, but it contains a hint. So, maybe if you can think about this problem and you don't have a good solution, take a look at the notes for this lecture, it gives you a hint, some kind of a beginning of logical proof. Okay. Now, that might help you. It's definitely much more beneficial if you solve the problem yourself. Now, here is what I suggest as a solution. Well, let's think about this. Let's assume R is not prime, which means it's supposed to be represented as two different numbers. Both of them should be greater than one. Yeah, greater than one. Okay. Now, let's think about what kind of numbers that can be. Now, don't forget that P is supposed to be greater than one. Okay. Now, let's say P is greater than 24 and R is supposed to be less than 24. So, R is a number which is less than 24, represented as a product of something. Now, what are the numbers? Obviously, we need some kind of prime numbers into which R is supposed to be represented. Well, since 24 is 2 times 2 times 2 times 3, none of these numbers can be 2 or 3, because if this is 2 or 3, then P would not be a prime number because 24 is divisible by 2 and 3 and R is divisible by 2 or 3. So, P will not be a prime number. It will be divisible by 2 or 3. Okay. So, it cannot be. So, what's the next prime number? The next prime number is 5. Now, R is not supposed to be just 5 because 5 is a prime number and there is nothing to prove. We're trying to prove that R is a prime number. So, it's supposed to be 5 times something. Well, and that something also should be not equal to one of these, not equal to 2, not equal to 3. Obviously, not equal to 4. So, the minimum will be 5. But 5 times 5 times 5 is 25. R is supposed to be, it's a remainder. So, it's supposed to be less than 24, right? R is supposed to be from 0 to 23. That's the remainder. So, there is no number less than 24 that can be represented as a multiple of at least 2 5s or more because 5 times 5 is already 25. So, it's impossible to represent it as the product of 2 numbers because each one of those numbers supposed to be at least 5. 2 and 3 cannot be because it will be and 4 as well because 4 is also divisible by 2. So, 2, 3 and 4 are not supposed to be among these R's. So, it's 5 and that's why you cannot have non-prime number as a remainder. Okay, that's my first problem. I'd like to mention that the problems which I present here are not typical problems which you might be presented in school because those problems are more related to just to check how well you know the theoretical material. These problems are supposed to force you to think about something which you don't know how to solve. I mean it's not like somebody gives you the recipe and you have to follow the recipe and come up with a solution. No, there is no recipe here. You have to invent it and that's the most important part I think in studying mathematics. It gives you some kind of a way where you can develop your logical abilities, your creativity, your analytical abilities. So, that's the whole purpose I think of studying mathematics. It doesn't have much utility nowadays. I mean you might actually need to add numbers. I mean that's as much as much math as you might actually need. But more than that really requires certain really good thinking. That's why I would like to present this course called Maths Plus and Problems for you to develop this particular kind of non-arthodox thinking. Okay, next problem. Next problem is considering number 10 to the power of n plus 18n minus 1. And it should be, I would like to prove that it's divisible by 27 regardless of n. Well, just let's check if it's divisible if n is equal to 1, that's 28 minus 1, that's 27. If n is equal to 2, it's 136 minus 1, so it's 135, which is 5 times 27. So, for any n, this is just example, but any n is supposed to be divisible by 27. Okay, and again as usually, pause this lecture and try to prove it yourself. Okay, now, how can we prove it? Well, first of all let me start from the easy thing. Divisible by 27 means it's a 3 to the power of 3, which means we have to divide it by 3 two times. So, what I can immediately see, that I can divide it by 9, which is 2 3s. Why? Because 10 to the power of n is 1 and n 0s, right? Minus 1 gives you 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, and digits 9. Obviously, it's divisible by 9, it's 9 times 1, 1, 1, 1, n digits and 18 is 2 times 9 times n. So, this is divisible 10 to the power of n minus 1 and 18 divisible by 9. So, divisible by 9 I have 3 and 3. What remains is that if I will divide it by 9 and what will be as a result, I will have this number plus 2 n. I would like to prove that this is also divisible by 3. That would be my third times. So, by 9 I have already divided by 9. I have remaining 2 n from 18 and this number, which contains n units, that's the second one. So, the sum is supposed to be divisible by 3. How can I prove it? Now, it's a known fact that the number of n minus 1, n minus 1, n minus 1, n minus 2, any number, any number, gives by division of, division by 3, gives a remainder the same as sum of its digits. It's a known fact, but I'll just very briefly prove it. So, this number is supposed to be written as 10 to the power of k times a k plus 10 to the power of k minus 1 times a k minus 1 plus etcetera plus 10 to the power of 1 a 1 plus a 0. So, if you have for instance 3571, it's 10 to the power of 3 plus 10 to the power of 2 5 plus 10 to the power of 1 7 plus 10 to the power of 0, which is 1 times 1. So, 3 5 7 1. This is a decimal representation, right? Okay. Now, what can we do about this? It's equal to 10 to the power of k minus 1 a k plus 10 to the power of k minus 1 minus 1 a k minus 1 plus 10 to the power of 1 minus 1 a 1 plus 10 to the power of 1 minus 1 a 1 plus, plus. I will add back whatever I have subtracted here a k plus a k minus 1 plus etcetera plus a 1 and plus a 0. Now, these are divisible by 3, actually by 9 as well. Why? Because 10 to the power of k is 1 and k 0 minus 1 would be 9 digits plus 10 to the power of k times 9 digits repeated, which is obviously divisible by 9 and by 3. So, which means that the visibility of this and the visibility of this, this is number n. Number n gives exactly the same remainder by division by 9 and by 3 as some of its digits. This is sum of its digits, right? So, it's a digital decimal representation with digits. So, some of the digits must have exactly the same remainder as n if you divide it by 3 or 9 because all these are divisible by 3 or 9, okay? So, their difference must be divisible by 9 as well because this is equal to n minus sum of its digits. So, n and sum of its digits must have exactly the same remainder if you divide it by 9 or by 3. Well, in our case, we need right now to prove that this is divisible by 3. All right, let's think about this way. n is supposed to give exactly the same remainder dividing by 3, 3 times let's say m and some kind of a remainder r as sum of its digits, okay? Now, this, well, this is a number, awesome number and it's supposed to give the same remainder dividing it by 3 as sum of its digits. But in this case, we know sum of its digits. Sum of its digits is n because it's n times 1 repeated, repeated one after another. So, sum of its digits is n. So, if you divide this number by 3, you will have exactly the same remainder as number n. So, 1, 1, 1, blah, blah, blah is equal to 3 times let's say n, but remainder will be the same, okay? Now, if I will put 2 here, that would be 2m and 2r, right? It's 2n. So, sum of this would be equal to 3 times n minus sum of its digits. So, sum of its whatever 2m plus n plus 3r and it's divisible by 3. So, sum of these is divisible by 3. Now, this is a very, I would say, delicate moment in logic. So, first, you basically prove the theorem that the number, any number, by dividing it by 3 gives the same remainder as sum of its digits. And then, this number has, in dividing it by 3, gives some kind of a remainder as the sum of its digits. Sum of its digits is n. Now, this is just the number n and we also know that number n is supposed to give the same remainder r. And now, all you do is just multiply it by 2 and add them together and you will get the visibility by 3. So, that's the third 3. So, 9 was divided before and now we have proven that this is the result after you divided by 9. This is the result of dividing by 9. You still can divide it by 3 and that's the 27. Okay, the last problem. Now, let's assume that we have 100 different numbers. The problem is, out of this set of 100 different numbers, different natural numbers, I can choose a subset, subgroup, whatever, a few numbers, sum of which is divisible by 100. This is supposed to be proven. So, again, we have to prove that out of any group of 100 numbers, I can choose a subgroup, subgroup, subgroup, subgroup, sum of which is divisible by 100. Okay, now, obviously, you can pause the video and think about this process. Again, there are hints in the textual part of this lecture. So, go on the website and you will have notes and notes contain the hint. Okay, so, here is how I suggest to solve it. Let's form partial sums. So, S k is equal to sum of n i from 1 to k. So, S 1 is n 1. S 2 is n 1 plus n 2. S 3 is n 1 plus n 2 plus n 3, etc. Up to S 100s, which is sum of all of these numbers. So, basically, S k represents sum of certain numbers. Okay, now, there are two different cases to consider. Number one, there is one particular sum which is divisible by 100. Well, fine, that's exactly what we wanted to prove. So, if there is such a sum, then whatever constitutes its components is the subgroup sum of which gives the result, which is divisible by 100. So, in this case, everything is just trivial. So, now, assume that non of S k is divisible by 100. Okay, so, non of S k is divisible by 100. Now, this is where the real thing starts. Well, if it's not divisible by 100, it should give some kind of a remainder when you divide it by 100, right? Okay, so, we have 100 different S ks as 1, as 2, etc., as 100. But how many remainders are available if you divide it by 100 and if it's not divisible by 100? Well, the remainder can be either 1 or 2 or 3 or 99. So, there are only 99 different remainders. If you divide it by 100, if you divide it by 100, number which is not divisible by 100, divisible gives you 0, obviously. So, if it's not divisible, it gives you 1, 2, 3 or 99. So, there are 99 different remainders. Now, how many sums do we have? We have 100 sums, as 1, as 2, etc., as 100. Now, there is a principle. If you have, well, I think originally it was about rabbits. If you have 100 rabbits and 99 cages, you have to put at least two rabbits into some cage. Because there are no places to put the hundreds. Only 99 places, but you have 100 rabbits. So, if you would like to put them into cages, then obviously at least two of them must go into the same cage. Which means two different, at least two different sums, let's say, S m and S n, should give exactly the same remainder if you divide it by 100. Okay, then let's consider the following thing, S n minus S m. Well, if two numbers give the same remainder, then their difference would be divisible by 100, right? So, if you have, let's say, 127 and 3,527 remainder 27 in both cases, if you subtract one from another, you'll get 3,4,0,0, obviously. So, if you subtract two numbers which have the same remainder, if divided by 100, you will have the result, the difference would be divisible by 100. So, let's do the following. What constitutes S m? It's all numbers from 1 to n. What constitutes S m? It's not all numbers from 1 to n. So, let's take numbers S, sorry, n, m plus 1, n, m plus 2, etcetera, n, n. What are these numbers? Well, this is exactly what constitutes this sum, because S m is from the first to n's. S m is from the first to n m's. So, if I subtract them, then this group represents this particular sum. So, this is exactly the group of numbers. If I can choose these numbers, some of them would be actually the difference between these S m and S m, and obviously divisible by 100. That's the end of it. So, again, what I'm suggesting to you, go to the website, Unisor.com, to the course mass plus and problems, go to arithmetic part of it, and it's arithmetic 0,5. So, read the problems, read the hints, and try to recreate all the solutions yourself. That would be very, very helpful for you. Okay? That's it for today. Thank you very much, and good luck.