 This is going to be a pretty easy day today, because we're going to end up, you know, all said and done having lost about 10% of this course due to illness, you know, me being out of town, snow even though there's no snow. There's the weirdest thing ever, really. What's that? I don't know. No, no, no, no. Let's see. Okay, so here's what we're going to do. I decided to give you a little break, and this is the last homework you turned in with the exam. It's been a while since I took the exam. I just gave you a completion grade on this, so most of you got, you know, high scores on this. If you did it, you got 100%. So if you did all, at least you attempted the problems, you got 100%. Most people got 100%. So I'll pass that back in the class today. The next section is a little tricky. Some of you may have experience with this. It's called linear congruences. I am going to try to break this up over two days at least. This is kind of a long section, and, you know, it's actually pretty important. So what we're going to do today is I'm just going to kind of introduce you to the computational component of the section. And there's going to be really, I'm going to state a theorem. I'm not going to prove the theorem. Proving the theorem would take me about 25, 30 minutes. I'm just not going to do that. So I'm just going to tell you what it is. And then we'll do a few examples. And that's going to be it, I think, for today. And then we'll finish up on Tuesday. Now I should mention that you have an exam coming up in two weeks from today. So it's Thursday now. So I can't really give you an assignment on this section and have it do on Tuesday. I could give you an assignment due next Thursday, but the problem is I still think you might want a little more time. So what I'm going to do is, I think what we're going to do is this. The next assignment I think will just be due at the exam in two weeks. Okay? That way you just have, you have enough time to work on it and you won't be rushed. Okay. And I think, not super huge. I mean, maybe a little longer. Not disproportionate to the extra time. No, no, no. I mean, you okay? Yeah. Okay. No, that's okay. Okay. So that's kind of what we're at. So I guess we'll just go into the next section here, section 4.4. Let's see. Yeah, yeah. Yeah, we're definitely not doing that. No. Okay. No, we're not, we're not going to go into that. Okay. So section 4.4 is a section called linear congruences. And so basically what this section is about is solving equations except we're solving the modulo some positive integer. Okay. So I'm just going to, again, I'm going to kind of go at a leisurely pace here. There's not, I'm purposefully not doing a lot today just because I want to get you comfortable with this before we move into some trickier stuff. So first example, let's consider the, okay. So yeah, the book, I'm not sure if their terminology is actually right. But let me see. Just a second. Yeah. Okay. It is. Okay. 2x is congruent to 1 mod 2. Okay. So the question is as follows. Are there any values of x making this congruence true? Is there an x such that 2 divides 2x minus 1? Yes. Yeah. That's right. So can I plug in a value so that this congruence is actually correct? Correct. X can be any integer now. Okay. So if you think about what congruence means, okay. What does it, what does this really mean? Really saying i.e. such that 2 divides 2x minus 1, right? No, no, no. But that's not the question, but that's not the question I asked. I asked, is there a value of x that makes it true? No. No, because then an odd number would be an even number. Okay. Oh, okay. Yeah, yeah. So the answer is no. And I'll just put in parentheses again. Since if this was the case, 2x minus 1 would be both even and odd, right? You guys, does this make sense? Right? Because this is what 2x congruent to 1 by 2 just is equivalent to this, right? So you just have to remind yourself what the congruence means, right? a is congruent to b by n if n divides a minus b, right? The left-hand guy minus the right-hand guy. So just remember that as we go through this. That's what it means. Okay. What about this? Okay. So I'm going to not write as much this time, but same question. Let's see. 3x congruent to 9 mod 6. So in other words, okay. So first thing I'm asking is just, is there a value of x that will make this true? 5. Okay. You were really loud there. Okay. You're excited about that. Well, yes. That does work, I guess, right? Yes. Very good. It's nice to see someone that's excited to be here. Okay. Okay. Yeah. Don't get me excited because I'm going to start breaking stuff. Okay. Okay. So I'll just, you know, just to make sure everyone's on board with this. Because 6 divides 3 times 5 minus 9. That's a times. Okay. Right. Okay. So 5 will work. Can you think of another one that will also work? Okay. Well, let's just see. Okay. Yeah. So I'm not going to write all these down, but there are several values you could use here. If you use negative number, for example, minus 5 would work because the question then becomes, so this becomes minus 15, right? So then the question is does 6 divide minus 15 minus 9, which is minus 24? And the answer is yes. 6 does divide minus 24. So that'll work. Negative 1 will also work, right? Minus 1. Then the question is the 6 divide minus 3 minus 9, which is minus 12. Yes. So minus 1 works. And then there's actually several other values that work. A really simple one that works is x equals 3, right? That's pretty simple, right? It just becomes the 6 divide 0. Yes. Of course it does. So, yeah. Okay. So the answer is that what we're going to be interested in is, well, yes, I mean, that's definitely true. There's going to be infinitely many values that will make this equation true. Let's see. Yeah. So, yeah, you can actually generate an infinite list here pretty easily, but we're not going to be interested in that. What we're going to be interested in is finding, let me, sorry, let me finish this thought first. Okay. Now, yeah, we're not actually going to be interested in finding, enlisting out an infinite number of solutions. So what we're going to be interested in is finding different solutions in module 6. Okay. So for example, okay, let me just say this first. Look at, you don't have to, I'm just trying to draw your attention to this for now. Look at these two solutions. Okay. So those are, sorry. Oops. Okay. Yeah. X equals 5 and X equals 3. Those are the ones I boxed. Sorry, Joe. These are different modulo 6, right? 3 is not congruent. In other words, 3 and 5 are not congruent modulo 6. 6 doesn't divide 3 minus 5 or 5 minus 3. Okay. So what we're interested in now is finding all the different solutions. For example, in this case, modulo 6. And so the question is, you know, we want to be finding solutions that are between 0 and 5. Okay. So we've got two of them here. Is there another one that works? I think someone said that before, but yeah. Sorry. This is a little unorganized. I apologize for that. Okay. Does everybody understand what it means for these to be solutions? You guys all buy that? You plug in 1. The question is the 6 divide 3 minus 9, which is minus 6. Yes, it does. Right? Okay. So are there any other ones? And the answer is no. These are the only ones. And so all of these three numbers now are different modulo 6, of course. There's a theorem we proved that actually mod n, 0, 1, 2, 3, all the way up to n minus 1, these are all different. And so since these are all between 0 and, you know, 0 and then less than 6, of course, these are all different modulo 6. Okay. And again, I want to state that again. I want to make sure everybody understands what I mean different modulo 6. What I mean is it's not the case that 1 and 3 are congruent mod 6. That's not true. It's not the case that 1 and 5 are congruent mod 6. It's not the case that 3 and 5 are congruent mod 6. That's what I mean by, what's that? No. What I mean, what I'm saying is just, this is all I'm saying. I'm just saying that you don't have to write this down. I'm just trying to answer his question. So it's not the case that 1 is congruent to 3 mod 6. In other words, 6 doesn't divide 1 minus 3, right? And the same is true for the other two pairs as well. Yes, exactly. Minus 1 is the same thing as 5. So if we are just looking for solutions that lie between 0 and 5, these are all that there are. And in fact, every solution is going to be congruent to one of these three numbers. So that's the point. Now, I'm going to give you the theorem that tells you how to solve these things in general. Like I said, I'm not going to prove it, but I'm just going to state it, and then we're just going to do a couple problems, just sort of seeing how you go about solving some of these equations. They are. I couldn't see it. I could see it was when I couldn't see it. Yeah. So, yeah, I mean, as it turns out, let's see. So if you start at the smallest solution, which is 1, you notice that you're adding 2 to get 3 and adding 2 to get 5. The adding 2 is not an accident. It's actually coming specifically from something, which I'm going to tell you here in a second. Okay. All right. So here's the theorem. And it's a little, there's a lot of information in this theorem. And I'm definitely not going to do the proof of this. It's just, it's way too messy. But so this is the general result. So the congruence AX congruent to B mod N has a solution, if and only if D divides B, where D is the greatest common divisor of A and N. Okay. So that's part of the theorem. And this part actually comes from the section on, it's called diophantine equations, which we skipped. So, yeah, even to prove some of this, we haven't done. So the section's necessary for the proof. So this tells you exactly when a congruence of this form has a solution, if and only if the GCD of A and N divides B. So that's easy. You can always, you can just look at any congruence and say, okay, I just have to check this. And then of course, if it does, then the question is, what are the solutions? What do they look like? So if D divides B, then the congruence has exactly D solutions. The book uses this terminology, which is appropriate. But you may see in the book, they'll say, find all the mutually incongruous solutions mod N. All that saying is just that they're all different modulo N. So for example, mod 2, you wouldn't say 2 and 4 are incongruent mod 2 because they're the same mod 2, right? 2 and 4 is congruent to 4 mod 2. So that second one doesn't really count in some sense. Only the ones that are actually all distinct modulo N are the ones that count. Okay. So, and the last part of the theorem is if X0 is any solution, then you can get all the rest just from the first one. So X0, X0 plus N over D, there's another solution. X0 plus 2N over D is another solution. X0 plus 3N over D is another solution. And we keep on going. And our final solution is X0 plus D minus 1 times N over D. Sorry, let me finish this. This gives us all solutions. So what does this mean? This means that these, so it's saying two things. One, these are all mutually incongruent mod N. In other words, no two of these solutions are congruent mod N. And the other one is that every solution to this congruence is congruent to one of these guys, okay? So you might say, oh, well, but yeah, but didn't he just say that there's an infinite number of solutions? Well, in general, yes. But again, we're only interested in the solutions modulo N. And modulo N, there's only, there are only N different, right? Numbers mod N, 0 through N minus 1. Everything's congruent to one of those. So there's never going to be an infinite number of solutions from that point of view, right? Because we're only picking from a finite set of numbers here. Okay. All right. So let me give you a corollary. This actually will probably be useful later, maybe in your homework. I'm going to give you the full assignment today, but you're probably not going to get too far on it until next week. The corollary is if the GCD of A and N is one, then the congruence AX congruent to B mod N has a unique solution, modulo N. And that comes right from the theorem, really. So I want to just point out a couple of things. This is pretty simple, but I just want to point it out. Okay, so up here, notice the first part of the theorem. The first part of the theorem says that there's a solution to the congruence if and only if the GCD of A and N divides B. In this case, it doesn't matter what B is, because the GCD of A and N is one, it automatically divides B no matter what B is, because it's one, one divides everything. So we know there's a solution. How many solutions are there? D solutions. D is the GCD. The GCD is one. That means there's only one solution. So that's where this comes from. Okay. B. Yeah. Right. So the way to think about it is you look at the GCD of the outer two guys, and so for the solution to exist, it has to divide the inner guy, the B, right? Yeah. Okay. So then this gives you, you know, a very simple way of generating all of them. I mean, all you have to do is check, does the GCD divide B? If it does, you just have to find one solution, and then once you've got the one, then you've got everything. And that gives it to you. Okay. So what I'm going to do is, let's see. Yeah. I'm going to do two problems for you. And, you know, again, these aren't overly tricky right now. It's going to get a little more, it is going to get a little more complicated next week, unfortunately. But okay. Does everybody have this now? Yeah. Okay. So let's look at. Mm-hmm. Mm-hmm. You say you have to find one of the solutions. Mm-hmm. But the solution, they're just n over d, 2n over d, 3n over d. I mean, you just use n over d. No, but you have to add these to x0. Oh. Which is the original solution. I'm minimal. You need to find one, yeah. So you, in fact, it doesn't even have to be the minimum, but you have to find one first. Because it's, from there is where everything else sort of takes up. It doesn't have to be the minimum. No. No, but. Do you use one other than the minimum? Okay. Maybe I'll show you here in a second. Let's see. Yeah. So it doesn't, okay. You'll see. You'll see how it goes. Okay. But yeah, it actually can be anything. So maybe I'll do this on purpose. Well, I'll dress it in a second. But let me just go ahead and get this started. No, it's a good question. But actually what ends up happening though, is that if you don't start with the smallest one, if you start with a bigger one, then what's going to end up happening is that you're going to end up with your, once you use this algorithm, you're going to, often you're going to end up with something that's bigger than n. But when you take it modulo n, it becomes small again. So you can kind of recover the smallest one just by using the fact that, you know, for example, say you get your solutions mod six, you've got, you know, I don't know. They're mod n, yes. Everything's mod n. So, okay. Yeah, it's no, it's modulo n. Yeah, modulo n. Okay. So 18x is congruent to 30 mod 42. We're going to try to find all the solutions mod 42. Okay. So here's what you're going to do. And again, these problems, actually for a change, you're going to get a little break with proofs. You're not actually going to be writing a lot of proofs in this section. You're going to be solving some things. So let's just remember what the theorem says. So what I would suggest that you do first is this. You just have your notes out or, you know, just go through what I'm going to do here. So this is, this serves as a, right? This is, 18 is a, 30 is b, 42 is n, according to the notation from the last theorem, right? So the number d then is the GCD, right? The GCD of a and n. So the GCD of 18 and 42. Okay. And I do expect you guys to be able to do this. It is six. Yes. It is six. Okay. So what does the theorem say? The theorem says then that there should be exactly six solutions mod 42. That's what the theorem says. We know that the GCD of a and n divides b. So, sorry, let me, I skipped that. So this is, of course, that's important. This divides 30. So there are exactly six solutions mod 42. Okay. That's what, again, that's what the theorem said. The theorem said that whatever d is, it's that many solutions. Okay. So what we want to do now, our next goal, I'll just write it our first goal here, is to just find a solution. Okay. All right. So in this case, I'm going to show you in the next example how to kind of mess with these equations to make them a little bit easier to solve. But for now, you know, let's just do it the naive way. Just think, okay, let's just start plugging in small values here between zero and 41 and see, hope that, you know, we find one pretty quickly. Well, zero is definitely not going to work. One's definitely not going to work. You can see that pretty easily. Two is not going to work. Three, so you get 54 minus 30. Okay. That's too small. 42 is not going to divide that. What about four? Okay. Well, 72 minus 30, that's 42. 42 divides 42. So four will work. So there's one solution. What's that? That becomes our X zero. That becomes our X zero. Yes. Yes. Okay. Let me just write this over here. So one such solution, excuse me, solution is X zero equals four. So now all the solutions, of course, and I, again, I mean mod 42 are given by, okay. So let me see if I can squeeze all these in here. Okay. And this just follows what I did, well, it just follows the theorem. So four is the first one. And then what does the theorem say? Well, it says to get the next one, I want to add n over d. Right? If you look at your notes, that's the next one, plus n over d. Okay. So four plus, and in this case, it's all on the screen for you. Right? n is 42. D is six. So there's the next one. Okay. And then you just sort of keep going up, right? So then four plus two n over d. So that'll be 84 over six, three. All right. So what do we get? 126, I guess. Over six, four. So we should get 168 over six. And now, of course, you have to remember when you stop. Okay. Remember, there's six solutions. So we just keep on doing this until we have six solutions, and then we're done. Okay. So we got five. We just need one more. Okay. So now it's one, two, ten over six. That's six of them. And of course, again, it's very easy to know when to stop. You just stop when you have as many as you need, and then you're done. Okay. So again, is everybody clear on this? Initial solution? Initial solution plus n over d, plus two n over d, three n over d, four n over d, five n over d. And n is 42, and d is six. That's where these numbers are coming from. Is that clear enough? Okay. So let me just go ahead and write these out in a nicer way here. So what do we get? We get four, eleven, eighteen, twenty-five, thirty-two, and thirty-nine. These are all taken mod forty-two. Okay. Right. So what I'm asking, instead of having like all these big numbers, one, twenty-six, two, ten, you know, all over d, if you got that first one and you got the second one and you saw that the difference was seven, could it just keep adding that? Yeah. So the difference, right, because that's how it's obtained, though. I mean, the difference is that, I mean, what are we doing here? So this is, we start with n over d. Okay. And so n over d was seven, exactly. Right. And then we're adding two n over d. We're adding a multiple of seven every single time to get down to the end. That's what I was asking. Yeah. Yes. Yeah, yeah, yeah. That's, yes. Absolutely. Okay. I think of a reason it should work. No. The reason why it works is because it's right here. Okay. So let me just say this again. Okay. So maybe we'll cut down a little bit of your time. Here's, so you got your first solution. We're adding seven to it. Okay. Then we're adding twice seven. So we're adding fourteen. But that's just another way of saying adding seven to the second guy. Right. And then we're adding three times it, but that's just another way of saying adding seven to the previous one. So you just keep generating them all by adding seven every time. Okay. Okay. And now I also want to address the question that you asked before. So suppose we didn't start with four. Suppose we started with something else. Well, okay. Suppose we started with the biggest one, which is 39 mod 42. What are we going to add to it? We're going to add seven. Right. That's what we're going to add first. So what's that? 46. But mod 42, it's four. You see? So it doesn't matter where you start. It doesn't matter because when you take it, mod 42, you're still going to generate all the solutions. Okay. All right. So let's see what I'm going to do now. Does everybody have this down? You guys have this? Okay. Yeah. The book, they sort of start introducing all these tricks without really telling you where they come from. So, which I think is really bad to say, oh, let's just do this, even though you have no idea why this works. It's actually not very hard, but I'm going to just go ahead and single it out as a lemma because the book uses this fact. Okay. Suppose that alpha, A, B, and N are integers. If alpha divides A, okay, I'll be formal, excuse me, formal here. Alpha divides B and alpha divides N. Then, okay, I'm just going to write it this way. AX is congruent to B mod N if and only if A divided by alpha times X is congruent to B divided by alpha mod N over alpha. Okay. So what is this saying? This is just saying that if you want to solve a congruence, you can actually simplify it by dividing through by any number that divides all three of the numbers. So you can, for example, if you have 10 is congruent to 15 mod 5, then that's equivalent to 2 is congruent to 3 mod 1 because you just divide everything true by 5. And, of course, smaller numbers in general are just easier to deal with. So this is just a nice technique. You know, I think I'm not even going to write the proof here. The proof basically just amounts to just going through the definition of congruence and dividing through by alpha or multiplying through by alpha, however the case may be. There's nothing to it, really. I mean, there really isn't anything to it. I'll tell you what. Let me just put this parenthetically. This is not intended to be approved. This is just to give you an idea of where this comes from. If ax is congruent to b mod n, then what does that say? That says that say ny equals ax minus b for some y. Again, I'm not being formal now. I'm just sketching the idea. Well, let's just divide through by alpha. And we get n over alpha times y equals a over alpha times x minus b over alpha. And then you get exactly this. You get n over alpha divides ax over alpha minus b alpha, because that's exactly what this is right here. So, again, it's... And then going the other way, you just multiply by alpha instead, and you get it the other direction. So, it's very straightforward. But this wasn't listed in one of your... I don't think this is one of the parts of that big theorem about congruences that I proved for you the other day. I didn't say anything about sort of dividing out stuff. I talked about multiplying through by things, but I didn't say anything about dividing. And so, this is just saying that you can divide also. Okay. So, here is the last example. In fact, I think I'm going to stop with this example. What are we at here for now? Example four, I think. This one's a little trickier. Let's solve the congruence, 9x congruent to 21 mod 30. Okay. So, again, let's just label this a, b, n. Yeah. To make sure I have this last lemma that you said, there was a lemma right there. Does that mean that we can divide all of these by three? Yeah. In this case, yes, we can do that. Yeah. And, in fact, we are going to do that here in a second. But, yes, that's exactly right. Okay. So, the first thing we want to check, though, right, don't forget this. Okay. I want to be very clear. You have to check this whole GCD requirement. It doesn't work out. There's no solution. So, don't just go right into trying to find the solution because it may not exist. So, you need to do that first. Okay. So, the GCD of 9 and 30, which is equal to 3, right? And it divides b. Everybody with me here? Okay. So, what does that say? There are, how many solutions are there? There are three. Mod 30. Okay. Well, how do we come up with the first solution? Now, again, if you really want to, you know, torture yourself, you could, well, maybe it's probably not going to be that torturous, but you start with 0, 1, 2, 3, 4, 5, 6, 7, 8. But you may not find one until you hit some fairly high number. Now, I will say this much, though. Okay. How are you, okay, this is important. This will help you also. There are three solutions. And let's just say we're going to take the smallest one, mod 30. We don't know what that is yet, but let's say we have it, right? Well, what's the next solution going to be? It's going to be that plus n over d, right? So, plus 30 over 3, which is 10. And then there's another solution, right? So, we're going to add that again. We're going to add 10 to that solution to get the next one. So, what that tells us, though, is that we probably have a solution, our first solution is going to be somewhat small. So, right? Okay, because whatever that solution is, we add 10. We get another one. We add 10, we get another one. And so, these are all taken mod 30. Yeah, we should be able to find one already. That's fairly small. Now, so that's just sort of common sense. So, you can sort of think about this and go, oh, okay, because it's big enough, our first solution can't be 29, right? Because when I add 10, you know, I'm already done. I can't get to the next one. Okay, so that'll help you have some idea of whether it's feasible to just start plugging and chugging right away. Because you might say, okay, well, I only have to check maybe eight or nine numbers and I'll find one, right? I'm going to do this a little bit differently this time. But certainly you can do that. There's nothing wrong with that. Okay, so we need to find one solution first. Okay, so I'm just going to tell you, once I told you what I did, that you should be able to find one that's not too big. If you plug in nine, you get 81, minus 21 gives you 60, 30 divides that, okay? So it's not really a big deal to just do it, kind of naively. I mean, in this case, it works out. It really doesn't take that much time. So you could certainly just do that. Numbers get a lot bigger though. This is a useful technique to have. It might not always be that simple. Okay, so what we're going to do is just, what we're going to do is try to just solve, in other words, we're going to solve 9x congruent to 21 mod 30 for one value of x. Okay, so as Joe said and as the limit before, and dictates that we can do, we can just divide through by three, okay? Three divides everything here, so that's legal move. Well, I mean, three is the GCD of 9 and 30, but it also divides, I mean, so yeah, I mean, we're really dividing through by the GCD, and we can do that because we already know the GCD divides B, okay? So that's generally, that's the thought process, is you're dividing through by the GCD. If it didn't divide B, of course, we'd be done because there's no solution. I was wondering, well, if I try to do it, if I try to just convert it to a different problem, I'm dividing first, and then it can work out to where the GCD doesn't compute once you do that. Oh, absolutely, yes, definitely, because you get an equivalent equation, so yeah, I mean, it's not going to work after you divide, so it'll work after you divide, there'll be a solution after you divide if and only if the original one did, so, yeah, I mean... Well, the GCD of 3 and 10 is not the same thing as what the GCD of that other thing was. Well, the GCD, yeah, I mean, in this case, right, the GCD is different, the GCD is one in this case, but, of course, it also divides seven. So, I mean, the point is, yes, the GCD is different, but the solution still exists. Yes, yeah, sure, so the point is, if you want to divide through first and then start the algorithm, you can do that, because if it wasn't going to work originally, it still won't work when you do the division. Okay, that's the thing you have to be careful of, yes, because your solutions... So, for example, yeah, in this case, right, if you look at this and you apply the algorithm, okay, let's just view this as a separate problem for now. Okay, well, where do we have to check first? The GCD of A and N, 3 and 10, whether it's a relatively prime, divide seven, well, it does, but what's the GCD? It's only one, so there's only one solution, mod 10. It's only one solution, mod 10, but you can't forget that, yeah, that's what you were saying, that you're solving this mod 30, and so mod 30, of course, now there are three solutions, mod 30. Okay, which, you know, isn't too surprising, really, because you... Okay, I'm getting off on a tangent, but you're asking some good questions. I want to say something about this. What is N over D in this case? N over D is 10. You see that? Okay, so whatever my first solution is, now let's go back mod 10. If I add 10 to that, it's just going to be the same thing, right? If you take a number and you add 10, you don't get anything different, because you're, for example, seven mod 10 and 17 mod 10 are the same, so that's why you're only going to get one in this case. Okay, so now what we're going to do, I'm going to do a little bit of a trick here, and I'll show you why we're doing this here in a second. These are just useful tricks to have. You definitely don't need these tricks for this problem. The numbers are small enough you can just do it like we did the first one. That's fine, but I'm just introducing these to you now. Now let's multiply by 7 to get 21x congruent to 49 mod 10. You'll see why I'm doing this in a second. Okay. Right, not the modulus. Right, just the A and B. Okay, and if you remember, this is an allowable operation. This comes from a theorem that we talked about before. You can multiply through by anything that you want. Preserves the congruence. Okay, now here's why we're doing this. Again, this is way overkill for this problem with the numbers so small, but it's going to be helpful when the numbers get bigger. Okay, and you can actually see what the solution is more or less right away if you reduce everything mod 10. So note that what's 21 mod 10? It's 1, right? And what's 49 mod 10? Okay, right? If you get enough experience with these things, you can do these really quickly, and then it does help you to solve the congruence pretty fast. What's that? So what, we end up with x equal 9? Yeah, mod 10. Yeah, that's right. Okay. It gives you your x of 0. Right, so I'll write this down in the next screen once you have this down. But the whole idea here, why did I multiply through by 7? Because when I multiply through by 7, I know that 21 reduces down to 1. And so then I can just get x for free that way. Okay, this is essentially mod 7. This is 1. This is 9. So now you just have x congruent to 9 mod 7 and you can take your pick now. You can easily solve that. I can choose any multiplier I want in terms of that theorem you gave us, right? Yeah, you can multiply by whatever you want if that's what you're asking. Yeah, it doesn't have to be stupid to reserve the congruence. Yeah, you can multiply by anything. Yeah, I can multiply by 230. Yeah, you could. That would be a bad idea, but you could if you wanted to. But yes, that's right. So, you guys have this down? Okay, so this is just where I'm going with the next slide. I'm going to replace 21 with 1 and 49 with 9 because that's what they are, mod 7. Okay, so let's see what do we have before? 21x congruent to 49. So what we have before? Mod, right, mod, let's see. Let's see, what did I just have? What did I just have? Mod 10, okay. Right, yeah, I said mod 7. I meant mod 10. Okay, so this becomes just 1x congruent to 9 mod 10. Okay, so everybody with me on this? That's what I did on the last page, right? 21 is 1 and 49 is 9, mod 10. Now we've got this and now it's really easy. We can, what's sort of the most natural value of x to choose here? How about 9, right? Okay, right? So now we have a number. And so this is important though. This is very important. We are not trying to solve this congruence, okay? So as Mr. Jones pointed out, it's different, right? We need to solve the congruence modulo 30, okay? So don't forget, right? The n over d, you're not taking it with respect to mod 10. You're taking it with respect to the original equation, okay? All right. So all the solutions are given by 9, 9 plus n over d. So I think what n was 30 and d was 3, right? And 9 plus 2n over d. But again, of course, once, as Joe pointed out, once you've got the first two, you know how to get everything else from there, right? Okay, so this just becomes... So you see, I'm writing it out this way so you can see what the pattern is, right? Really, if you look at the solutions, right? You start with 9, then you're just adding 10, and this is just the result of adding 10 to that. So you just keep adding 10 every time. So 9, 19, 29. So that gives it to you. Okay, so again, let me clear mod 30. Okay. So I think that was a nice place to stop for the day. So again, a nice introduction to this stuff. I've got your homework. I'm going to pass back also. I'm going to give you homework for this section. Please keep in mind the due date is not until next Thursday. Sorry, no. Two weeks, actually. So let me give that to you, though. We probably... Well, if there's a chance we may do another section, but probably not because we're going to spend the Tuesday to review. But we still have all of it next week, so we'll see how that leaves us. Okay. 4, 4. So here's what I want you to do in 4, 4. 1 through 6. And, okay, sorry. Should have put 1 through 7. Sorry. 7. Okay, sorry. You need number 11, too. Okay, there it is. It's a little bit out of order. Okay. And keep in mind, because it's two weeks now, it is possible I may add to this on Tuesday, for example. That may happen. You still have until the next Thursday to get it done. Okay. So we'll just stop there for today.