 Hi, I'm Zor. Welcome to Unizor education. We continue solving problems and combinatorics. This is the sixth series of problems in the first part, so I will have another part. Anyway, combinatorics are great problems actually and I do suggest you to first go to unizor.com and have these problems well, try at least to solve these problems just by yourself. Very important. And then you can just, you know, listen to this lecture. Now the notes for this lecture contain obviously the problems and then they follow with an answer, which you can check against yours and then there is an explanation of the logic which basically I'm doing exactly the same thing here. So again, the main source is the problems and the main activity is to try to solve these problems yourself. Alright, now after that is done, so let's go to the problems of this particular lecture. One is Okay, I have a number. Two to the power of two times three to the power of three times four to the power of four times five to the power of five. This is my number. I would like to know how many divisors or factors does it have. Well, let's include number one and and the whole number as divisors just for simplicity purpose. Okay. Well, I did solve a somewhat similar problem before in one of the previous lectures, but in that program, in that problem all the multipliers were prime numbers and then the problem was kind of a relatively simply solved. This is not the case and that's exactly my first most important point right now here is before doing anything like counting how many different divisors the particular number has, we have to represent it as a product of prime numbers well raised to certain powers obviously. Now in this case two and three and five are prime, four is not. So I have to convert it to prime which means I have to convert four into prime numbers. Now four is two by two and four to the fourth degree is two to the eighth degree, right? This is two square and to the fourth we multiply the exponent so it's two to the eighth degree, which means that my number n is equal to two. I have second power and I have the eighth power, right? So all together are ten and this is a representation as a product of prime numbers in certain powers. Now this is now easier. Now if we would like to count how many different divisors this particular number has, well, let's just consider. Every divisor has to have certain number of twos from zero to ten, certain number of threes from zero to three and certain number of fives from zero to five because these are prime numbers. There are no other combinations if my divisor has anything but two, three and five or it has two or three or five in the in the power greater than ten or three or five correspondingly, then it cannot be a divisor. So all I have to do is to choose certain number of twos, certain number of threes, certain number of fives, multiply them and that's a divisor as long as the numbers do not exceed the powers. So how many choices do I have? Well, I have eleven choices for number two to choose, from two to the power of zero to two to the power of ten, right? So I have eleven choices for my first component, then I have from zero to three it's four choices for certain number of threes in my divisor and then the six choices of five. So any power which I can choose from two, for two, for three and for five, then I combine them together raised in these powers. That's a divisor, right? And this is 264. So that's the answer. That's how many divisors this particular number has. Okay, this is easy. And again, just let me repeat, all I have to do is first represent as a product of prime numbers in certain powers and then everything seems to be very simple and for prime numbers I have already basically solved this problem in its general format. So if n is represented as p1 in the power of n1, p2, power n2, etc., pn, well, let's use some other letter pk, let's say power nk, then if p1, p2, and pk are prime numbers and again I'm referring you to the problem which I have already solved in one of the previous lecture. I have from zero to n minus one, which is n1 plus one, two is for the first and two plus one for the second, etc., and k plus one for the last prime number. And that's the product of these is actually the number of different divisors because that's my prime numbers in any power can participate in any divisor. Alright, so that's easy. Now let's go into a little bit more complex problem. And I have actually two problems here, one a particular one and one general. So let me start from one specific particular problem because it's easier to explain. So I have exactly the same number, two to the ten, three to the third, and five to the fifth. Now I'm interested in summarizing all these 264 divisors. What's their sum? Well, that's a little bit more difficult and a little bit more involved problem. Now before actually explaining anything about this particular problem, let me just make a very short excursion back to the geometric progression. Now if you have a geometric progression, which means you have a first member and then every other member is the previous times certain factor, q is a factor. And let's say the very last one is q to the nth degree. And what if I want to summarize them? This is called sum of geometric progression. I don't know about you. I never remember the formula. However, I do remember how to derive it very simply actually. Let's multiply s by q. What would I have? A times q would be a q. A times aq would be a q square. Now the previous one to this would be a times q to the n minus first. And times q would be aq to the nth. And if I multiply q by this, I will have aq and plus first. Right? So if I will subtract from this, I subtract this, I will have s times q minus one equals. Now these are all cancelled out and I have only the last one and the first one. Aq to the n plus one minus a, from which follows this formula. Oh, sorry, minus, yeah, minus a, right, minus one divided by q minus one. Okay? So the formula obviously is true for this factor not being equal to one. Everything else works fine. But for equals to one, it's actually a, a, a, a, a. So the sum is equal to a times n plus one. So it's a separate case. It's not interesting. We're not considering it. Anyway, this is the formula which I have just derived. And I will be using this formula. So if I have certain geometric progression with certain number as the first member, certain factor and certain maximum exponent of the factor participating in the sum, this is the formula. So it's basically a function of three parameters. The first member, the factor, and the maximum. Okay? All right. So let me just write it somewhere up there so I don't forget it anymore and, and, and they don't have to derive it anymore. So s of aq n equals aq n plus first minus one divided by q minus one. Okay, done. Now let's think about this particular problem. So again, I would like to know what's the sum of all these 264 as I calculated in the previous problem, divisors of this number. I suggest a geometric interpretation. Here is my three-dimensional space. Now, three-dimensional because I have three prime numbers, by the way. In a general form, I will have more and I will have more dimensions. But let's consider three because it's easier, right? So what I will do is the following. I will put numbers here from zero to ten. So this is zero. One, two, three, four, five, six, seven, eight, nine, and ten. And they will represent the power of two, which I'm going to use. So the whole x-axis would be my number of twos in the divisor. Now, y-axis would be the number of threes and I have zero, one, two, three. And the z-axis would be the number of, number five, which participates in the divisor. One, two, three, four, five. Now, let's take any, well, we will build actually some kind of parallel pipette on this. Something like this, whatever. So this is a parallel pipette. Every point inside of that parallel pipette with integer coordinates, I'm talking only about points which have integer coordinates. So they have projection, let's say, this is projection on the x-axis, this is projection on the y-axis, and this is projection on the z-axis, so the point is somewhere here. Now, what does it mean? It means that the coordinates are what, five, one, and two. It means I take five twos, one, three, and two fives. And this is one of the divisors. So my point is that every point inside of this parallel pipette with integer coordinates, of course, corresponds to some divisor. And the number of points, by the way, equals to the number of divisors, right? So this is kind of a geometric interpretation of all my divisors. They're all inside this parallel pipette with sides, ten, three, and five. Ten, three, and five. From zero to ten, from zero to three, and from zero to five. And again, I'm talking only about integer coordinates, okay? Now, so I have numbers inside this parallel pipette. So this point is associated with this number. Another point is associated with another number. Any point is associated with some kind of divisor, right? Now I have to summarize them. The way how I will summarize them will be the following. First, I will summarize all the divisors which are lying on this x-line. Then I will try to expand it on the plane. So I will add these rows as well. And when I have the plane, I will try to rise it and fill up completely the parallel pipette. This is my geometric approach, right? Okay, fine. So let's think about what I have if I will just summarize all the points with integer coordinates along the x-axis. Which means they have zero trees and zero fives. So it's only twos which are participating. Two to the zero, two to the first, etc., etc., up to two to the tenths. So what do I have? Well, this is geometric progression, right? The first member is equal to one, right? Two to the power of zero. The factor is equal to two, right? I'm multiplying by two every time. And the last power, the maximum is ten. Let's use this formula. It gives me the sum along this axis. Sum of all eleven divisors which have only two in their prime composition. Not trees and no fives. Now, what do I have? Well, a is equal to one, so I multiply by one. That's okay. So it's two, two to the power of eleven minus one divided by two minus one. Okay, that's my first. Great. So this is sum of all these elements. Now, let's move along the y-axis. What will be the difference between these numbers and the next row from here? Which have the y-coordinate equal to one. Well, it would be two to the zero times three to the one plus two to the first times three to the one. So all of these divisors have three to the first degree, right? And the last one would be this, which is actually sum of all the previous times three to the first degree, right? So all these numbers are greater than all these numbers by factor of three. So the whole sum of these numbers would be three times greater than sum of these. So again, from points in the beginning, now I'm switching to lines now. This line has already been summarized and that's its sum. And now I'm considering the next line in the increasing y-coordinate. And this line would be exactly three times greater in the summation than the previous line. So in terms of lines, I can consider this sum as a beginning, the first member of the geometric progression. And then the next sum would be three times greater. The next sum would be again three times greater, etc., etc., up to the last one. So what I have here is I have a new geometric progression with A, the first member being my sum whatever I have received here, which is two to the 11 minus one divided by two minus one. Q now would be equal to three because every new line, new horizontal in this case line, would be three times greater than the previous. And the maximum is three, the maximum power I'm raising this. Go to this line, go to my formula, and what do I have? Well, I have my previous one, which is this one, A, times three to the power of four, right? And n is equal to three minus one divided by three minus one, Q minus one. That's the result of my summation of all the different divisors which lie on the XY plane. They don't have any five in their prime representation. Fine, great, done. So now we have found the first step was summarize along the line, second summarize along the plane, XY plane, and the third one, obviously, I will summarize in the third dimension. Let's add five. Now, let's consider any divisor which lies on the XY plane. It has number of fives in its representation equal to zero, right? What if I will switch to the next plane, this one, where the coordinate of z coordinate is equal to one. It means I'm adding five into the representation. So every member of this plane, which is on XY with z coordinate equal to zero, and then if I switch it to the next one, if I raise it to the first coordinate on the z axis, it will be five times greater because it means that every member, since every member on this plane with z coordinate equal to one has five in the first degree, right? So the sum of these comparing to some of the next plane one higher, what will be their difference? Well, the plane which is one higher will be five times greater if you summarize all the members of this plane, will be five times greater than the sum of the basic XY plane, right? Since every divisor would be multiplied by five. What would be the next? If I will raise again this plane to the level of z coordinate is equal to two. Well, again, that means that I have five to the second degree, right? Power of two, which means I'm again multiplying every divisor by two, and the sum would be also multiplied by five, and the sum would be multiplied by five again, right? So every time I'm raising this plane higher and higher, sum of the elements on that plane would be five times greater than the previous one. So I have again geometric progression of these sums, the first member being this one, right? So this is my first member. Now, the factor from plane to plane I'm multiplying by five, right? And the maximum is five. So what will I get now? I have to multiply a, right? Which is this times what? Q to the n plus first, five to six minus one and divided by five minus one. So that's the answer. I mean, obviously you can simplify two minus one is one, three minus one is two, and this is four. So obviously it's simplifiable in some way, but this is the answer. So that's the total sum of all the divisors. Good. Now, if you understand that, let me go to the next problem, which is a generalized version of the same problem. Now, I was kind of lucky that in this particular case I had only three prime numbers, and I have used the three-dimensional space and made the picture for you. If I have in my next problem, I have not three but certain number of prime numbers. So I have k prime numbers, and every one of these prime numbers is raised into certain power. And now the question is again, absolutely the same. What's the sum of all its divisors? So, well, instead of using this picture, I will just use plain coordinates, right? So first, I will consider ordered sequences of coordinates again in k-dimensional space. So my coordinates will be the string with coordinates, let's say, which letter should I use? Let's say I1, I2, Ik. Now, this string of integer I1, I2, etc. Ik represents actually in k-dimensional space my one particular divisor, namely this one. And obviously, each a, j should be no greater than n, j, and zero, and integer, right? So these are all the different divisors which this particular number n has. So when I1, I2, etc., Ik are going among all these different numbers or all the different combinations. And obviously, we know that the number of combinations for j1, for instance, for I1 is n1 plus 1 from zero to n1. For I2 would be from zero to n2, etc., etc. So the total number of divisors would be n1 plus 1 times n2 plus 1 times n3 plus 1, etc. But now we're interested in their sum, not in their quantity. So I will consider these sequences of k-integer numbers conditioned this way as representing these divisors, right? Now, let me first summarize all the different divisors which are represented with something zero, zero, zero. So the first coordinate is changing from one to n1, from zero to n1. So for all these divisors, which are basically described by these particular strings, so zero, zero, zero, zero means my divisor p1 to the zero, p2 to the zero, etc., pk to the zero, which is equal to 1, by the way. Now, the first coordinate equals to 1 means p1 to the first, and these are still in the zero, in the power of zero. Excuse me. Okay. Now, if I want to summarize them, what are the results of the summation? Well, the first number with zero is equal to 1, right? Because every power is equal to zero. The second number would be equal to p1, because p1 would be in the first power, and the rest are zero, which is 1. Next one would be p1 square, etc., and the last one would be p1 to the n1. So that's the sum of these. I put the sigma sign here. So some of all these, if you wish, I can actually do it more scientifically. I can put i1 here, where i1 from zero to n1, right? So these are my old divisors, which are along the first coordinate. And again, according to this formula, I will have, what will I have? The first was 1, so it's p1 to the n1 plus 1, right? That's the maximum, minus 1 divided by p1 minus 1. That's the sum. Now, I will add to this, I will add to this another summation, and i2 would be here. Now, how can I summarize this? Well, again, for i1, which is equal to zero, sorry, for i2 is equal to zero, I have already summarized. Now, what's the difference between the sum for i2 equals to zero and when i2 is equal to one? Well, the sum would be exactly p2 times greater, right? Because every member, like for instance, this one, if I multiply it by p2, it gives me this member, right? From zero to one. Because this divisor contains only p1 in the degree in the power of i1, right? And this divisor contains the same p1 in the power i1 and p2 to the first power. So, this one will be smaller than this one, or this one will be greater than this one by a factor of p2. Now, next would be, if I multiply this by p2, I will get this one. So, if I want to summarize them all, I will have to basically take this as my first member of the geometric progression, p2 as a factor, and the maximum, which is n2, would be my maximum power, maximum exponent, which I'm using, right? So, whenever I would like to extend my summation from just the p1 to p1 and p2 divisors, I have to multiply it by correspondingly p2 to the power n2 plus one minus one and divide by p2 minus one. So, I'm basically explaining exactly the same thing as I was explaining in the case of a three-dimensional. And obviously, the formula would be, as you understand, more or less the same. But I would like to prove this formula a little bit more rigorously. How can I do it? Well, the best way is actually by induction. So, let's consider that I have already guessed the formula based on all these considerations. And the formula is this one, pk nk plus one minus one divided by pk minus one. So, this is the formula, which I have kind of guessed. How can I prove it by induction? Well, let's do induction by k, by number of prime numbers which are represented in this particular number n. So, what if I have more numbers? I have k plus one numbers. Well, first of all, obviously, we should check it for k is equal to one. So, if I have only one particular prime number in the distribution of this particular, in the, in the representation of this particular number, then divisors are one, p1, p1 square, etc., pk, p1 to the n1, right? So, these are all the divisors. And obviously, their sum is equal to p1 to the n1 plus one according to this formula minus one divided by p1 minus one, which corresponds to this formula with k is equal to one, right? So, that's obvious. Now, let's consider the formula is correct for k prime numbers. And let's add the k plus first number in the representation. So, now, my representation is this one. Now, all the divisors of this number are all the divisors of the old number, which means all the different combinations of these prime numbers and all the combinations of these prime numbers multiplied by pk plus first and all combinations of these multiplied by pk plus first square, right? And the cube, etc., etc., up to nk plus first, right? So, we are adding more divisors. Now, what would be the sum of these divisors? Well, if I have all the divisors separated in groups, these are only p from one to k. These are p from one to k times pk plus one. This is not a minus. This is from two. Now, this is from one to k. All the different combinations and all the different powers times pk plus one square, right? So, I have divided all my divisors of this number into these groups. This group contains only prime numbers from one to k. This is contains only prime numbers from one to k multiplied by pk plus first to the first power and then to the second, etc. What's the difference between these sums if I will summarize all of these, all of these, all of these? Well, if I will summarize them, it's actually sum of these guys and this is sum of these guys multiplied by let's quote s. This is s multiplied by pk plus one. This is s multiplied by pk plus one square, square, etc., right? So, again, it's a geometric progression. So, I'm basically using exactly the same logic and I can definitely use exactly the same formula. So, my sum of these would be equal to whatever was before here of this and I have this formula, right? This is my a, right? My q would be equal to pk plus one and my n would be equal to nk. So, I have to multiply, according to this formula, I have to multiply the first member, which is this, which is this, times the q, which is, which is this to the power of maximum, which is mk, sorry, that would be k plus one, right? The maximum k plus one minus one and divided by pk plus one minus one, right? The q. And this is exactly the same formula with k plus one rather than k prime number. So, basically, the induction is working here and logic is exactly the same even without this visual representation. I thought visual representation is a parallel tippet in case of three dimensions would be, you know, kind of nice. Okay. And the last problem is very simple one and I will not spend a lot of time on this. So, let's consider you have a circle and you have n points on this circle. Now, question is how many different polygons we can draw if we connect all these n points in some order. So, the polygons are not necessarily convex. So, not necessarily, I have this one. I can also have this, this, this and this. This is also polygon, right? And this is also polygon. So, I can connect this, this, this and this. So, how many different polygons I have? Well, obvious is the first consideration is, well, the order I'm going from one point to another basically determines my polygon. So, number of n points I can place in different order by n factorial different ways, right? Well, that's not the end of it obviously because, you see, if this is, let's say I start from one, two, three, four. One, two, three, four. Right. So, if I will do it in this sequence, from this to this, from this to this, from this to this and then I end the story from four to one. This is exactly the same polygon as if I, for instance, start from this, then go to here, then to here and then to here. Right? This, this and this and this. Which means that there are certain repetitions, the same polygons, even if the sequence is different. So, what's the different, what kind of different sequences produce exactly the same polygon? Well, obviously it's the cyclical permutations. So, if you have a cyclical representation, all of them gives exactly the same polygon, right? From first point to second and then to third to fourth or, and then back to the first to complete the polygon, or from the number two point to number three to number four and then to number one. It's exactly the same sequence of lines, basically the same segments which make the polygon. So, I can actually make this cyclical permutation without changing the polygon itself. So, now the question is how many, for any particular permutation, how many cyclical permutation it has? Because all of them are supposed to be glued together into one polygon. Well, the answer is, well, if you have four, for instance, members, well, obviously, you can start from any one of those from the first or from the second or from the third or from the fourth, right? So, does it mean that it's four different cyclical permutation? Well, not exactly because you can actually go to the opposite direction. You can start from the first and then go to the fourth and then third and then second and then close to one. This is also exactly the same. So, you have two different directions, four, three, two, one, three, two, one, four and two, one, four, three, something like this, right? So, I have actually eight in this case. So, for n points, I can have it moved cyclically one way, let's say clockwise, for instance, or whatever you call it, to do the cyclical permutation. Or I can do it counterclockwise and it will be exactly the same polygon. So, I have two n different cyclical permutation and the answer is I have to divide it by two n. Or if you wish, it's n minus one factorial divided by two because n factorial means from one to n, n would be canceling and from one to n minus one would be remaining and two is still there. So, that's the answer. That's so many polygons I can draw. All right, that's it for today. I do recommend you to go to Unisor.com and review these problems again. Try to solve it again just by yourself and then whatever the solution is provided. Basically, that's it. If you, by the way, if you will register to the site, then you can just take the course, you can take exams, etc., etc. Everything is free and I definitely will welcome you to do this type of things. All right, thanks very much and good luck.