 my one online student. Yes, I didn't say anything that was important, so it's okay. Alright, so here's what you should know. First thing you should know, of course, is just the definition of congruence, right? Okay, and since it's short, I'll just go ahead and write this down. a is congruent to b mod n. This means that, I mean, the actual definition we used is that n divides a minus b. That's what I want, okay? There's a theorem, so there's this difference between definition and theorem. Some of the times they're interchangeable. There's a theorem that said that a is congruent to b mod n if and only if a and b have the same remainder upon division by n. But that's not the definition. This is the definition. That's a theorem. Yes. It's sort of understood now. I mean, technically, a and b are integers and n is a natural number, right? We've only talked about the modulus being a positive integer here, but I'm not going to worry about that. I'm not going to get that picky in this exam. Okay, and so most of these, I'm not going to say a whole lot about it. If you have any questions, I'll try to address these, but so when I labeled these theorems, I mean relative to the text. So you should know theorem 4.1. This theorem, of course, you can find this in your textbook. So theorem 4.1 is about what I just said, actually, about the remainder. So a and b are congruent mod n if and only if they have the same remainder upon division by n, right? Okay? And again, I'm not just like usual, right? I'm not going to ask you to prove these, but you should know them. And now you might say, oh, well, how am I going to, if I ask you this, what am I going to say? What's theorem 4.1? No, of course, I'm not going to do that. But I have ways that I can ask the question so that I can tell if you know what it's saying or not, okay? Of course, as you know, sometimes I'll ask true false questions, things like this, okay? So you should be aware of this. Or I may say, well, if a and b are integers, mod 50 and a has a remainder of 20 and b has a remainder of 25, can they be congruent? Mod 50. And be aware that sometimes I might make you think a little bit, okay? I'm not going to rule out tricky type questions to see if you know what you're doing or not. Okay, theorem 4.2. And again, this is one of these problems where, you know, or one of, yeah, I mean, this is something that I, that doesn't have a name specifically. So again, I'm not going to say what is theorem 4.2. I mean, I guess I could do that, but I'm not going to do that. But I can also, but you should be looking at it, because again, I can ask you things that, by example, will tell me if you have an idea of what this is saying or not, okay? Okay, the next one is theorem 4.3. This actually comes into play in solving some of your problems. This is something that's important to know. This has to do with, basically with canceling in a congruence. C A is congruent to C B mod N. Then you can cancel out the C, but you have to divide N by the GCD of C and N. Okay, this is on page 66. We really use most often the corollary to this. We didn't really use this so much in general. Corollary one, which we've talked about, you probably all know, or most of you know, at this point, what this is saying. This just simply says that you can cancel a number from both sides of the congruence if that number is relatively prime to the modulus, right? C A is congruent to C B mod N and C is relatively prime to N. Then you can cancel it to get A is congruent to B mod N, right? That's corollary one. Okay, so again, this is just the basic stuff in this section. So these are the things, you know, what you need to know in order to be able to do the problems. Also, you might ask, someone might ask, okay, well, if I'm going to, you know, solve one of these congruence problems, am I going to have to refer to theorem 4.2 and say, oh, by 4.2 B, we can do this. By 4.2 F, we can do this. No, I'm not going to expect you to do that. What I would expect is for you to show your work and show the steps. I don't expect you to justify it by saying, by this theorem, you can just do it and not say where it's coming from, you know. But I expect you to, I do expect you to write out steps, you know, and just make your work clear. So I mean, you know, for example, if this goes without saying, but if I give you a congruence to solve and you just magically find me, give me the answer without any work, then that's not going to get many points, right? I'm going to want to see that you know what the steps are, even if you don't have to quote the theorem exactly because I don't really expect you to know what the numbering is. Okay? So for example, let me just tell you kind of what I mean by this. Suppose at some point you have, you know, a congruence, a is congruent to 1 mod n, and you want to raise it to a certain power because you're interested in knowing what a to the 50th is. Then if you just raise a to the 50th, then if you just say, a to the 50th is congruent to 1 to the 50th, which is 1 mod n, that's fine. You don't have to say by theorem 4.2f. I think that is actually what it was, but you don't have to do that. Okay? All right. Let's see. So that's about it for 4.2. What I want to do now, though, is I want to do one of your homework problems that I didn't grade, okay? And as you should all know by now, graded homework solutions are posted online. And the solutions to this assignment, which I'm going to pass back at the end of class today, these are all posted. Of course, the one you're turning in on Thursday's solutions are now posted. But we've also talked at a pretty great length about how to get started on these problems. So I think you've, and we're going to talk a little bit more today. So you really should have gotten enough help to have an idea of how these go. Okay. So let's take a look at number 9. I did not, I thought about grading this one. I didn't look through your work, but I'm guessing that there were probably a good number of mistakes on this one. So this one said, suppose p is prime and p is strictly between n and 2n. Show that, and you can compare the solution to what you did in your homework when you get your homework back. Show that 2n choose n is congruent to 0 mod p. So there's a key step here that I could be wrong here, but I'm guessing that a lot of you are missing that you really need to have in order to do this properly. And that is that you need to know that 2n choose n is an integer. If your dad didn't enter in your proof, it's probably not correct. So what we're going to do is what we're, we're just going to say this. So we know that 2n, in fact it's more, it's a natural number actually, but let's call this k. Okay, I don't, the fact that it's a natural number more specifically than just being an integer is just not relevant. So I'm just going to say k is an integer. I did prove this for you in class. I don't know that this entered into the homework assignment problems at all, but there was a point in class where I actually showed you that n choose k is an integer. I used induction to do that. That's been a long time, but that is something that you are going to want to use. Okay, so what do we know? So definition, right? By definition we know what 2n choose n is, right? It's 2n factorial over what? n factorial times n factorial, right? The last part is 2n minus n factorial. So that's where the last n factorial comes from. Okay, and so again we know that this is an integer k, hence if we just multiply both sides by n factorial times n factorial we get 2n factorial equals n factorial times n factorial times k, right? Everyone see where I got that? Just multiplying both sides by n factorial times, n factorial squared is really all this is, right? Okay, so let's see. So what do I want to say now? Note, what do we know about p? Note that this is certainly true, right? One is less than or equal to p, which is less than 2n. So certainly p divides 2n factorial, right? Think about what a factorial is. Think about the definition of a factorial that should be clear that p divides 2n factorial since p is squished between 1 and 2n. What is 2n factorial, right? It's 2n times 2n minus 1 and all the way down to 1. Well, p's one of those numbers, so it's got to divide it. Okay? Does that make sense? Okay, now look at this, go back over here for a second and just look at this equation that we have, right? Well, if p divides 2n factorial and 2n, this is nothing hard, right? If 2n factorial is equal to n factorial times n factorial times k, then p has to divide n factorial times n factorial times k, because it's the same thing as 2n factorial, right? Well, p's a prime now. And so if p divides the product, these are three integers, right? So if p divides the product of these, I mean, there's only two different ones here, but p divides this product, then what do we know? p has to divide one of them, right? p is a prime. This is a corollary I think I gave you at one point. A prime divides a product, a finite product of integers. It has to divide one of the integers, for sure, right? Okay? I could say p divides n factorial or p divides n factorial or p divides k, but there's no reason to be redundant here, right? So it has to divide one of the two. And so the claim is that p does not divide n factorial. And then the only possibility is that p divides k. But remember what k was, k was equal to this. And that's exactly what we're trying to prove, that p divides this. So if p divides k, because it's equal to 2n choose n, p divides 2n choose n, and that's exactly what we want to show. Okay, so the question is why can't p divide n factorial? Well, p is bigger than n, but n factorial could potentially be bigger than p, right? Because when you multiply everything else, you can't just say, well, p's bigger than n, it's too big. No, not necessarily, because n factorial absolutely could be bigger than p. So you can't quite get off this easily here. So the claim is that p does not divide n factorial. Well, let's suppose it did, okay? Well, then what's the, anyone want to tell me what the idea is here? What's wrong with this? Now it does have something to do with the fact that p is bigger than n, right? But it's not as simple as just saying, well, it's too big in the sense that n factorial could be bigger than p. So it's not, you can't quite use that fact. Okay, here, let me write something else out for you. Then just, just write down the definition of n factorial. What is n factorial? 1 times 2 times 3 times all the way down to n, right? That's what n factorial is. So there is a fact that I just quoted that we just used just a few minutes ago. If p divides this product, what can we say about p? It has to divide one of the elements in the product. So we know that, say, p divides i, okay, I'm running out of room here, but i is somewhere between 1 and n, right? You guys by that? Because it's prime, has to divide one of them? Okay. Okay, this I don't normally do this. I'm sorry for the, okay, this is, this is terrible. There we go. Okay, I'm just going to finish it up here. I'm too lazy to get another page. So, but now, okay, p and i are positive integers. p divides i. So what is this? Okay, so let me just write this down again. p divides i, well, that certainly implies that p is less than or equal to i, right? That's certainly true. But what is i? i is less than or equal to n, right? That's just at the very bottom. You see that? So what we can say then is that p, by transitivity, p is less than or equal to n, right? Okay. But look at the assumption p is bigger than n. Do you see that? It's contradiction. So that can't happen. Okay, so sorry, I'm being a little bit informal now, but I'm just going to write it down here, right? It just comes from this, from this assumption. And it's less than p, in other words, p is bigger than n, right? So that's not possible. So since p does not divide in factorial, we must have, remember what I have down here, right? p divides in factorial or k. We just show that it can't divide in factorial because we get a contradiction to one of our hypotheses. That can't happen. So therefore this has to happen. One of them happens, the first one doesn't. Therefore the second one has to be true, right? So p divides k. And remember what k was? 2n choose n. Thus, and that's really it, right? Thus 2n choose n is congruent to 0 mod p. Mod p. Sorry, I'm running out of room here. Okay, right? p divides 2n choose n. And that's exactly what it means for 2n choose n to be congruent to 0 mod p. Okay? And so that's, that will take care of it. And I don't, again, like I said, I didn't look through your, I decided not to grade this one just mainly because I've, the fact that you needed, we hadn't done in a long time and I thought that a lot of you were probably going to forget this fact. But some of you maybe just wanted to cancel out of p and say, okay, 2n choose n. Well, down here the p is not going to occur down here and it's going to occur up here. But the point is you can't just cancel the p out because if you do that maybe you don't have an integer left over. Maybe it's not left over. I mean you want to say that this is equal to p times something. But when you cancel out the p what you have left, you don't know necessarily as an integer. That's why you really need to use this fact in the very beginning. You really, I don't think you can really get around it in a nice natural way. So if you didn't use that and you improved, then you probably didn't prove it correctly. Okay. Does this make sense? Okay. So just to remind you n choose k is always an integer. So that's useful to know. Okay. So for now anyways we'll go ahead and jump ahead to 4.4. We can talk about a couple more problems if there's time. I certainly have something to say about problem 11. Like I said, this one I think you're going to need to know a few things that one thing in particular that I don't know that I've told you in class. Okay. So let's talk about what you need to know here in 4.4. Things you should know how to do. First thing you should know how to do is solve a congruence, just a single congruence in one variable. And we'll look at a couple examples. Certainly we've discussed this on, like I said, I'll do a concrete example here in a few minutes, but what is it that we do? Remember what we start with. You check to see if the GCD of A and N divides B. That's always where you start. If the answer is yes, then they're exactly, call the GCD of A and N, D. They're exactly D many solutions to this mod N. If the GCD of A and N, D does not divide B, then there are no solutions. And so if you remember, the algorithm then is, all you have to do is find one solution using some of the techniques we've talked about. Once you have that, you just keep adding multiples of N over D to that until you get the required number of solutions. Okay. So that's not too hard. I would expect most of you will be able to deal with this. Okay. And I also want to reiterate again about the numbers here. Some of you are again wondering, okay, well, I may have to do with big numbers. Are they going to get massive and such? And the answer is no. I mean, what would I expect you to be able to do here without a calculator? Fairly simple calculations. I mean, it's possible that in the course of a problem, you may have to multiply 103 by 2 or 4. But I would expect that you could have done that when you were 9. So, you know, and I would also expect that you could do it in a matter of a few seconds. So, you know, I'm not going to have you dividing 14,159 by 62 and things like this. So the computations that you do should be on a third-grade level. And there shouldn't be a ton of them. Now, I will say this too. If in the course of solving a congruence, you end up with the number 68,227, probably you're doing something that is not optimal in your solution, okay, to say the least, right? Okay, so that's a hint that if you get there, you might want to think about your strategy and go back over and try something else. Okay, so this is something that you should know how to do. You should know the statement of the Chinese remainder theorem. And you should also know, of course, how to apply it in practice, okay? And of course, I told you what it is, and I've also applied it for you. So you have an idea, you should have an idea of how to do this. It's just an algorithm, right? So you have to solve a bunch of single congruences and sort of put them together to get the simultaneous congruence at the end, right? With the capital N1, capital N2, and that kind of thing. Yes. You mean on the exam? Oh, I see, right now. More or less, yeah. Let's see. There are four points. Well, let me see. So what is, oh, okay. No, I mean, this is, okay, so this is something that you should know. I mean, so for example, I could ask, I could give you a specific problem that will incorporate that in its solution because that sort of tells you kind of what I was just saying, that you check the GCD and if it divides B, then there's that many solutions mod N, right? I mean, I'm not going to ask you what is theorem 4.7. I'm not going to say that, but I could say, you know, here's a congruence. AX is congruent to B mod N. You know, what has to be true in order for that to have a solution? And it is, you know, the GCD of A and N has to divide B. So, yes, you should know it. You should know it. But just like the other ones in 4.2, I could test your knowledge without specifically saying what is this theorem on page whatever, you know? So the content of the theorem you should certainly know. But of course, you're going to know that anyways when you're solving the problem. You already know what it says because you know what you need to do in order to solve it. So, I mean, that's not really going to be an issue. Yes. Okay. Now, I will say for the Chinese remainder theorem, I will tell you this, since it has a name, I could ask you, state the Chinese remainder theorem. Say what it is. Okay. I don't generally do this for theorems. I don't already have a specific name. Of course I could. I mean, there's nothing that would keep me from saying, what is theorem 4.2? I could ask you that. I'm not, I'm just being nice and I'm not going to do that to you. But here's one that you should know exactly the statement of the theorem. Okay. It's not really that bad. I mean, it's kind of long, but it's not, you know. So, you can all do this. I'm positive you can do this. Third thing, know how to solve, we've talked about this too. You also have a homework problem like this. Congruences of the form, the two variables, right? AX plus BY congruent to C mod N. Okay. And really, we've only looked at this kind of special cases where either A and N or B and N are relatively prime. And then we get all these unique solutions for every value of the other variable, right? So, if you remember, if A and N, I'm not going to write this down because it's already in your notes. If A and N are relatively prime, then for every value of Y, this has a unique solution mod N for every value of Y. And so the point is what you do, what's the trick here? You look at, well, at least for problems of this flavor. You look at A and B and you say, okay, well, which of these is relatively prime to N, right? Then whatever that one is, the other one gets moved over to the other side. And then you just solve it like you do one and just one variable, right? So, just kind of like what I did in class. And so really, once you make that recognition, the problem just gets reduced to the techniques that we've already used with one variable. You just, for example, treat Y as a constant. And then you just solve it the same way. Okay, let's see. What else? Okay, I mean, these are the main things that you need to know here, really. So, I mean, this, you know, this exam is going to be similar in format to the other ones. You know, there's certainly going to be some, you know, definitions on the exam. You know, you probably have a couple of true, false type questions. And then, of course, you're going to have to solve some congruences. This will be a little more computational probably than the other exams were. Probably fewer proofs, although you will probably have at least one proof to do on the exam as well. But like I said, the numbers shouldn't get too unwieldy. And if you know what the techniques are, you should be able to deal with these things, really. I would expect the scores will go up somewhat on this test, actually. Okay. So, let's do a few examples here. Okay. I want to get through the basics. And I also want to talk about something that you need to know to do number 11, which I may not have done for you specifically in class. You could probably figure it out yourself, but I'm just going to go ahead and give you this hint. Okay. So, how about something like this? I'm not going to spend a ton of time on these because we've talked about these quite a bit. Okay. I also want to be clear. When you're solving a congruence, when you're solving a congruence, I'm only looking for all of the different solutions, modulo, whatever the modulus is. So, in this case, it's 12. You're not going to be writing down more than 12 numbers. Okay. So, there are only 12 numbers, distinct ones, mod 12. Okay. I don't care about, you know, 75 and all these things. I don't care about those. Okay. So, I may not write much down here because we've talked about these so many times. What's the first thing that you want to check? GCD of 10 and 12, which is, does that divide 11? No solution. Right? You guys all with me on this? I'm not going to write anything else than just no solution here because it's just, if you've been doing, keeping up stuff, you should know that right away. Right? Okay. There are no solutions to this. Okay. What about this? Yes. If you were to give us a problem that had a solution, do you actually want us to write out why? Oh, yes. So, I'll be pretty, I'll try to be clear on that in the directions. This is to say, okay, give me all solutions, modulo, whatever the number is, and if there is no solution, say why. I'll probably say something like that because, yes, you could always just guess and say none. So, yeah, and then saying what you would just say, you know, GCD of 10 and 12 is 2, it doesn't divide 11. That's all you have to say, just that I know. Yeah. Okay. So, let's look at another one, 88x congruent to 12 mod 20. Sorry. Okay. That can be that easy. And so, the way I'm going to write this out is more or less kind of what I'm looking for from you. I just, I want to see that you know all of the things that you need to know in order to know that you're doing it right. So, again, I'd want you to write something like this, you know, the GCD of 88 and 20 or 20 and 88 doesn't matter. So, you need to compute this. This is not really that hard. Something like this I would expect you to be able to do on an exam. All you have to do is think about the factorization of 20. Okay. Well, it's 4 times 5. Okay. 4 definitely divides 88. You can see that. What about 5? Well, 5 doesn't divide 88. It can't, so the GCD can't be any more than 4. Right. So, if it is, it'd have to be 20. It's not. It doesn't divide 88. So, the GCD has to be 4 in this case. And 4 divides 12. So, there are four solutions. Right. And of course, again, there are four solutions, modulo 20. Okay. So, now what you want to do, remember the next step is just find one. That's, that's all you're trying to do here is to find one solution. And it's not hard to just guess values until you can sort of see what the, what the answer is here. I'm going to go through this in a way that will simplify this a bit, but you can just look at it. Yes. I'm not going to get that picky. Yeah. Yeah. Technically, it's, it's four solutions, mod 20, because there's actually an infinite number of solutions, but that's, yeah, I'm not too worried about that. Okay. So, this is what I want to, you know, it's possible the numbers could be a little bit bigger, but again, the computation would not be that hard. It wouldn't be that tedious. Okay. So, what can we do to make the numbers a little bit smaller to begin with? Divide every one by four. Right. Four divides into everything. So, we can reduce everything right away. And remember, I'm going to reiterate this again. Anytime a number divides into all three, you can just divide all three by that and you have exactly the same solution set. You, if you have a number that divides into this number and this number, then you can cancel it only from the first two as long as it's relatively prime to this. In general, what you have to do though is you have to divide, if you want to, in general, if it's not relatively prime, you can still cancel it, but then you have to take this and you have to divide it by the GCD of this and the number that you cancel. Okay. Yes. Yes, then it's if and only if. Yes. And also, that also works if you're canceling out something that is relatively prime to the modulus, then it's also an if and only if. So, one direction always, hold, hang on just one second. If you have a congruent to b mod n, that always implies that ac is congruent to bc mod n. You can always multiply by the same thing on both sides. But to go backwards in general without changing the modulus, what you're canceling has to be relatively prime to the modulus. Yes. Okay. Well, so, I mean, if, what we're talking about canceling though, what we're talking about is having a number that divides in both simultaneously. So, you need, you need, whatever you're going to cancel, you want to, it certainly has to divide both of those things at the same time. So, in other words, you can't just cancel from one side. You need to, it needs to go into both of them. Okay. So, 4 goes into 88, right? 22 times. So, this is 22 congruent to 3 mod 5, right? Okay. I just divided everything by 4 here. Okay. So, and just to test your skills a little bit here, what's 22 mod 5? 2, right? I left, and I keep doing this. Okay. I left off the x. Sorry about that. So, 2x congruent to 3 mod 5. Okay. Well, the numbers now are pretty small. This is not too hard to, yes. Is that enough there? Oh, yeah. Yeah. So, it's pretty easy to see that x is going to be 4 in this case, right? I mean, you can just go through the first 3 in your head, and it's not hard to do. So, we can choose now, or you don't have to write the word choose, but x to be 4. Okay. So, there's our first solution, right? So, the set of all solutions, I'll just write this down here. Now, the solution set is, okay, we're going to start with 4. And so, remember what the algorithm is, right? What do we do? We add n over d. We keep adding n over d until we get all 4 solutions to the system. Okay. The n is the modulus, the d is the gcd, right? So, yes, and you need to go back to the original one, right? Which is 20. 20 is the original modulus, okay? So, what we're going to do, and the gcd was 4, right? So, what is it that we're going to add to this? Okay, let me make sure everyone's okay with this. Remember, here is the original problem. Once you have your one solution, the next solution you get by adding n over d, okay? So, here is, this, the modulus is n, right? And the d, the gcd is d, okay? So, we're going to add n over d, 20 over 4, which is 5. And we just keep adding 5 until we get all 4 solutions. We're done, okay? So, 4 plus 5, right? So, we get 9, 14, and 19. So, everyone clear on this? Understand how I got the rest of these? Okay. Okay. So, that's it. All right. Let's see. What else do I want to say? I want to say something about, and these two problems, and this time I'm spending more time going over your homework problems, especially since you're not going to get a chance to see the solutions posted before the exam. I don't think I talked about number 5 at all, if I remember. Maybe you can remind me if I did. I'm pretty sure I didn't say anything about that. Yes? Right, right. I would like you to do that, yes. I would like, and I may say that in the directions, give me all the solutions between, for example, in this case between 0 and 20, for instance. So, and we addressed this before too. So, the point is, suppose instead you solve it a different way, and you get 14 as your first solution. That's fine. You're just going to keep adding 5 every time. So, you're going to get 19. Well, then what happens when you add 5? You're going to get 24, but mod 20, that's 4. You see that? Then you add 5. You're still going to generate the same solution set. You're just going to have to loop around and reduce, and then you just keep going, okay? And that holds true for any of these values too that you got. I mean, if you got 19 first, right? It's the same thing. We had 5, you get 24, which is 4, and then you get back down here, then here, then here, and you're done. Okay? So, yeah. So just keep that in mind. There's nothing bad that happened. If you happen to get a solution that's not the smallest, you can just reduce, and then you still get them all this way anyways. It's not that hard. Okay, so anyways, yeah, I don't want to repeat myself here, but I don't think I've said anything about 5, really. Is that true? Can somebody tell me that I'm not losing my mind here? Homework problem number 5? I did say something about, okay, yeah, I don't think I did. Okay, so number 5, I'm not going to do the problem for you, but what I'm going to do is give you the idea. So they give you this congruence, a single congruence, and they say, solve this congruence by actually solving four simultaneous congruences, four congruences simultaneously. And so I want to explain kind of why that works, what they're saying, really. Yeah, so what theorem are you talking about? Okay, so theorem 4.9 is a little bit, I mean, it's not, it's stated only in the second form of the problem. The problem says solve this single congruence by solving four of them at the same time. So, oh, so you're talking about theorem 4.9? Okay, well no, this one doesn't really apply because this is a system of linear congruences in two variables. Okay, so, and yeah, so theorem 4.9 is not really what you want, plus it's really kind of, it's a lot simpler than that, really. I'm just going to explain why it works this way, and then I'm going to leave it to you to do the work. But, so let's see, this is, by the way, I want to be very clear about this. This, I'm not solving the problem for you. I'm only showing you why what the book is saying works. That's all I'm doing here, and I'm going to leave you to do it yourself. Okay? Solving, yeah, so the book says solve this by solving these, this system of four congruences simultaneously, that I'm going to leave to you to do. But you can do this actually by the Chinese remainder theorem, for example, once you get the 17 coefficient to go away, right? We talked about that with another problem. You just have to multiply by the appropriate stuff. Okay? So let's suppose, and if you have your book you might want to follow along here, suppose that 17x is congruent to, this is the original problem, congruent to 3 mod 2 times 3 times 5 times 7, right? That was the original problem you're trying to solve. And they're saying do it by solving the system of four simultaneous congruences. Okay? Then, let's use the definition of congruence. So what does this mean? That means that 2 times 3 times 5 times 7 divides 17x minus 3, right? That's just the definition of congruence. So, okay? This, by the way, this really isn't that hard. See if you buy this or not. 2 divides 17x minus 3, then. Certainly. Because 2 times 3 times 5 times 7 times x, say, or y, I should say, is 17x minus 3. Well, then 2 times something, which is an integer, is 17x minus 3. So 2 divides 17x minus 3. But what does this mean in terms of congruences? This means that 17x is congruent to 3 mod 2, right? Okay? I'm not going to do all of these, but you probably can see where I'm going with this. We also know from above that 3 has to divide 17x minus 3, right? Because it's an element in this product that divides 17x minus 3. So 3 certainly divides 17x minus 3. Again, it comes right from up here, this division that I have up here. So, hence, it's the same, same idea, right? 17x is congruent to 3 mod 3. And you can do the same thing with 5 and 7, of course. I'm not going to write that out, but those four, then those four congruences that they're asking you to solve simultaneously, x is a solution to all four of those because it's a solution to this first one. Okay? That's sort of simple, really. I mean, really, this just follows from the fact that if a product numbers divide something, then each one of those guys in the product has to divide it, too. That's it. That's all there is to it. So, from what my point is, from a solution to this congruence, that value of x is automatically a solution to these four congruences that they're giving you in the book. And conversely, I'm going to show you that if you have an x that satisfies all four of those, then it has to satisfy the first one. So they have exactly the same solution set. So if you want to solve the first one, it's the same thing as solving the four, that system of four simultaneous congruences. It gives you the same thing. Let's suppose that, now, in other words, what I'm saying is, suppose x is a solution to these four congruences that I have up on the screen. Well, then what do we know? Let's just, again, let's just use the definition, or it will be done here in a second. The first congruence, of course, I just have this on the left-hand side, right? Two divides 17x minus three. The second congruence, three divides 17x minus three. The fourth congruence is that five divides 17x minus three. And the fifth congruence is seven divides 17x minus three, right? I'm just, again, now I'm just translating the congruences into their corresponding divisibility relations using the definition, right? That's just, for example, the third one. This is just what it means, right? This just means five divides 17x minus three. Just the definition. Okay. Well, I could just do this all at once, but I'm just going to break this into two things, two separate pieces. Well, if two and three both divide the same thing, two and three are relatively prime, so their product divides it, too, right? So from the first two, we get six divides. I'm not going to write this down, but you've seen this before, right? We've used this fact several times. Two numbers are relatively prime, divide the same thing, their product divides it, too. And over here, five and seven are relatively prime, right? So 35 divides 17x minus three. Okay. So this is the first two and then the last two, okay? And now, what can we say about this? Six and thirty-five. Well, do they share any common factors? Six is two times three. Thirty-five is seven times five. There's no, there's certainly nothing in common here. So their product divides 17x minus three. But what is this? Of course, this is just two times three times five times seven divides 17x minus three. And if we translate this back again into this congruence notation, this is just 17x is congruent to three, mod two times three times five times seven. Again, that's just what it means, right? If it's easier to see it, go from here to here, then you should see it, right? The definition is that this divides this minus that, okay? So this is where the idea is coming from. I didn't get to that in the lecture, but so the point is if you want to solve a single congruence, well, there's a certain trade off here, but if the modulus is really big, you can break it down to a bunch of simpler pieces. Of course, doing those pieces may take you longer than doing the original one to begin with, but this is something that you can do for this reason. Okay, and in this case, since the directions are specifically asking you to solve the system of congruences, you should do it that way, okay? How many solutions are there to this congruence? Modulo, this modulo is here, modulo one, whatever it is, two ten, yeah, okay? So how many solutions are there? You can actually, the fact that for me should be able to see right away. One, right? Seventeen and two times three times five times seven, these are certainly relatively prime. I mean, what could divide them both, right? Seventeen's prime, one or seventeen. Seventeen can't divide that for sure, right? Because it would contradict the fundamental theorem of arithmetic. Okay, once you have the prime factorization written out, there's no other prime that can divide the number, right, than the ones that appear in the factorization, okay? Is this okay? Okay. And so the last thing I want to say, and again, there's going to be some questions about probably it's number eleven, problem number eleven. Number twelve, if you've looked at this, number twelve, you can do it in three seconds from looking at number eleven. There's nothing to number twelve. It's easy, trust me. Number eleven is the one that requires a little bit of work. Okay, can I go to the next page now? Anyone need this? No? Okay. Okay, so here's what I'm going to do with number eleven. Number eleven, I'm not going to do for you. I'm going to make you do this one yourself, but I'm going to tell you the things you need to be looking for, and there's also a certain result that I don't think I gave you in class about the least common multiply. I may have said it, but I don't think I proved it for you, and this is something you really need in order to finish the problem. So here's what I'm going to tell you. These two pieces of information should be relevant in your proof. I'm not saying that this is the only proof. So if you do it a different way, maybe not, but this would be the natural way to go about it. So here's a couple things that you need, okay? Probably. Okay, I should say probably. First thing, suppose you know that the GCD of N and M, say, divides some integer x. Okay, I'm not going to be overly careful saying N and M or not both 0 and x is an integer. I'm not going to say that here, just save time, but this is actually something that goes way back. And actually, we did talk about this, I think. Then x is a linear combination of N and M. Okay, so I'm going to remind you of this. There's really not much to this. So let's just let D be the GCD of N and M. So I'm claiming that x is a linear combination of N and M. Okay, this is a fact. A lot of you still want to use this in your homework and oftentimes it has not really been useful. But in this case, it's going to come into play. What do we know about the GCD? It is a linear combination of the two guys, right? D is a linear combination of N and M. We've definitely talked about that before. So we know that D is equal to, say, alpha N plus beta M for some alpha and beta in Z. Okay, so what's our assumption? GCD of N and M divides x. So D divides x. Say D times Z equals x for some integer little Z and capital Z. Okay, well, x is D times Z, but D is alpha N plus beta M, right? You look above. So this is alpha Z. I'm going to rearrange slightly. Alpha Z N plus beta Z times M. And that's it. That's it. So x is a linear combination of N and M, an integer linear combination of N and M. So the point is not only is the GCD of N and M a linear combination of N and M, any integer that the GCD divides is also a linear combination of N and M. And this is a fact that you will probably need to use in your working of number 11, in your solution to number 11. Okay. The second thing, and this I don't, I really don't recall if I told you this or not, something about the least common multiple. We didn't spend a lot of time on the least common multiple, but there's an analogous property for the least common multiple here that you really need to know in order to finish this problem. And that is, let's let M be the least common multiple of, and I'm just going to use A and B in this case. In the homework problem, the A and B are the moduli of the congruences, but it's the same idea. Let M be the least common multiple of A and B. Suppose that A divides an integer X, and again I'm suppressing this now, but what X is, but it's an integer, and B divides X. Then, in fact, analogous to the GCD, then the least common multiple M actually divides X. This is something I may not have told you before, but you're going to need now. So if you remember with the GCD, right, the property of the GCD of A and B, we call the GCD of A and B, D, say. So, the property D has is that if X divides A and X divides B, then X divides D. X divides the least, the GCD. Least common multiple has a property that if A and B divide X, then the least common multiple divides X, so everything kind of gets flipped the opposite direction. Okay, so, and I'm going to talk about this more or less. Okay, so what do we know about M? Well, if M is the least common multiple of A and B, we know that A divides M, and B divides M, and M is, okay, the least positive, right, positive integer with this property. Okay, well, M divides X is what we want to show. We can definitely divide X by M using the division algorithm, for sure, okay. X equals MQ plus R, where, what do we know about R? Do you remember this? Restriction is that 0 is less than or equal to R, which is less than M, right? The remainder has to be smaller than what you're dividing by. It has to be smaller than the divisor. Okay, so let's see. So what is there to note here? Note that if we solve this for R, R is equal to X minus MQ, right? And so we'd be done in a second. What are our assumptions? Excuse me, our assumptions, A divides X, for sure, right? That's just our assumption up here. A divides X, and we also know that A divides M, right? It's right here. Definition, least common multiple. M is a multiple of A, A divides M. So if A divides X and A divides M, you guys buy that A will divide R then, right? You can replace M by A times something and X with A times something, pull the A out, right? A divides R, for sure. And same argument, B divides R. And like I said, we're almost done here. Now, here's the idea, okay? R is the remainder upon division by M, so zero is less than or equal to R, which is less than M. My claim is that R has to be zero, okay? See if you guys are still awake. Can anybody tell me why R has to be zero? Anyone? What if it wasn't zero? What if R wasn't zero? What's that? Okay, well, yeah, I'm not, I mean, you're on the, you got sort of the right idea anyways. Here's the idea, right? So R is certainly a common multiple of A and B. R is a common multiple of A and B. M was the least common multiple of A and B. In other words, it's the smallest positive integer that is a multiple of A and B. But if R wasn't equal to zero, R would be a smaller positive multiple of A and B than M, contradicting the fact that M is the smallest common multiple of A and B. That's why R has to be zero. If it's not, it contradicts the leaseness of M being the smallest positive common multiple of A and B. Okay? If not, R is bigger than zero, and I'm just going to write these out separately. R is less than M, and R is a common multiple of A and B. This contradicts the leaseness of M. So, well, what, what do we know if R is zero? That means that M divides X. That's exactly what we're trying to prove. Right? M divides X. So this is the fact that you're going to need to, to know, there's one other thing I'm going to say about this problem. And then I'm, I am going to leave this up to you to go through the details here. I'm, I'm giving you the pieces really that you need to, to do this. The problem is not that bad. You just need to play around with the definition and the congruences here. But there's one other thing I'm going to say about this, just so you know at the, at the end. I haven't, of course, written the problem down yet, but the last part of the problem is to show that a certain solution is unique, modulo, the least common multiple of M and N. If you have your book out, you can see that. Anybody still need this? Nick, are you good? If you need, if you need a few seconds, that's fine. Alright, so this is the last thing I'm going to say. So you want to, so basically you have, again, this is number 11. I think what do you have? You have x is congruent to a mod N and x, let's see, x is congruent to b mod M, right? I think that was the problem you were trying to solve. I miswrote that, let me know. I'm pretty sure that's right. Okay, so you want to show, and so the first part you have to prove, and again, I'm going to leave you to this, is that a solution exists if and only if there's a two, this is a two-part problem. If and only if the, again, let me know if I, I don't have my book open. The GCD of N and M divides A minus B, pretty sure that's what it was. If and only if. So you, you really need to do two things here. You need to, there's two separate proofs here, subproofs, if you will. Okay, the second part is to prove that if there is a solution, I may be paraphrasing this a little bit. It is unique mod the least common multiple of N and M, right? Oops, common multiple N and M. I think that's the last part. If I'm wrong, please tell me so I don't jack this up. I'm pretty sure this is right. Okay, so what about the second part? How do you prove that? What is this saying? Unique modulo, the least common multiple of N and M. Okay, so what, what it's saying is this. The least common multiple of N and M is some positive integer, something, okay? Just for the sake of making this more concrete, let's, let's suppose the least common multiple was six, okay? Then what this is saying is that there's a unique solution mod six. In other words, of the mutually incongruent number zero, one, two, three, four and five, the solution is exactly one of those numbers. And that's it, not more than one. But how do you actually prove this? How do you actually write this out? Here's the way that you want to prove this, okay? What you want to do is, is you, you, you suppose that, and really this is what I want to see. You suppose that X and Y are, these are just integers now, okay? Solutions to this system, okay? What you, and when I say X and Y, I don't, I don't mean X goes in, that the first congruence has the variable X and the second congruence has the variable Y. That's not what I mean. Being a solution means that it satisfies, it's a simultaneous solution. It satisfies both of them at the same time. So X being a solution means that both of those hold. Y being a solution means both of those hold with the X replaced with Y. That's what I, that's what I mean, right? So what you have to show, if there's only one solution, then what you, what it is you're trying to prove is that X and Y are the same mod N. So you have to show that X is congruent to Y mod, not N. X is congruent to Y mod the least common multiple of N and N. That's what you have to show. That's saying that, in fact, they have to be the same. There's only one of them, okay? So I'll write it down. What you have to show, this is what you're trying to prove is that X is congruent to Y mod the least common multiple of N and N. That is what you're trying to prove. And this is where you're going to use this second fact that I gave you before. If you want to show that, and I'll just go ahead and say this, then I'm going to pass back your homework as we're running out of time. Well, okay. What did the last, on the last page, what did it say? It said that if N divides something and M divides something, then their least common multiple divides it. So what is it that you're trying to prove here? I want you to think about this for just a second. We're almost done. Just try to think about this because I'm not going to write it down. I want you to think about it. You need to show that the least common multiple of N and M divides X minus Y. That's what that congruence is really saying, right? Well, suppose that you know that N divides X minus Y and M divides X minus Y. Then you invoke that last thing. I just told you on the last sheet or on the last page that says that then the least common multiple divides it. So you don't actually have to sort of specifically work with the least common multiple to show that the least common multiple divides X minus Y. All you need to know is that N and M divides X minus Y. And then by the previous page, the least common multiple divides X minus Y. So really your strategy is just to show that N divides X minus Y and M divides X minus Y. And you get it for free by the last page. Your brain's still working here. You understand what I'm saying? Okay. So that's what you're trying to do. All right. So I think I'm going to stop there. I will tell you, I plan to be around tomorrow afternoon. So what I can say is that I will be in from, I'll be in from three to four tomorrow in my office. So you're welcome to come by and ask a couple questions or whatever, whatever you want to talk about. Okay. So and then of course I have my regular hour before class on Thursday. So of course you're welcome. As I said, you're always welcome to come by and chat with me. I'm happy to try to help you out. Okay. So let me pass your homework back. Like I said, solutions are posted online. Okay. So you can review those. I would suggest that you do that, especially for ones that you're missing. And again, the old homeworks are up here. So if you missed one of your old homeworks or tests, please feel free to come by and get it. Okay. So hang on a second. Let me get this. Get rid of this.