TMUA - JZ Set A Paper 2

[00:00:00]This is Mox a paper two by Jay-Z. We did paper one a couple of days ago. So, let's get straight into it. Um, so yeah, we got a cubic equation here. We're being asked about roots. So, we can't really find the roots of a cubically, but we can um think about sketching it and think about where its turning points are and therefore get its roots from there. This is a very standard 2 method.

[00:00:17]Also, the fact that this is missing the x term means differentiating it goes really nicely because then when you set it to zero, you just have all these x's you can factoriize out and you get solutions x= 0 or x= 2 p. So, the coordinates there are if you put in x and z, you get four obviously. So we have a stationary point at 04 and we also have a station point put 2 P in and you do some maths and you get this. So there's the other station point at 2 P and then this. Now this is a positive cubic. So it goes like this.

[00:00:43]So if P were negative and this stationary point were over here somewhere. Well, if P is negative, this whole term is positive. And so add four and it's bigger than four, which makes a lot of sense. So it would have to be up here somewhere and you'd end up like this and then this and then this. Right?

[00:00:56]So if P were negative, then we'd be absolutely fine to have one route. Like we would we just would it would be great. Unfortunately, that's not one of the things here. Um so we're going to move on. What if P was positive? Well, then you'd have your station over here with a height minus 4 B cub + 4. And what we would need is we would need this thing here to be positive, right?

[00:01:15]Because if it were negative, then the graph would come up down like this. And then you'd have your second and third roots here before going up again. So we need that. Solve it very easily. You get this. And so P minus one is sufficient as well and so is P minus Z. So therefore sorry P less than zero. So I think therefore putting these together anything P less than one I think is necessary and sufficient of course that's not here and we don't want it to be. We want sufficient but not necessary. Now look at this one here that is within the bounds that we've got but also just takes away things like minus2 4 which we also know will work.

[00:01:48]So therefore this one is sufficient. P can be between mass one and one but anything further past mass one than that and therefore it's not necessary good and now this one of course just expand this out. very happy with my binomial expansion personally. So I didn't have a hard time with this. Also writing this as like a power half then cubing it makes this. You guys should be pretty used to that as well I think. And then divide individually this by all of these terms and you get this. And then we can just differentiate. And now we're supposed to just put in four. Again we should be very good at this I think.

[00:02:18]Just root it and then cube it. Uh just square it and then square and then sorry root it and then five. If there's a negative there just flip it over as well. Um, and then there's probably easier ways of doing this than what I did, but eventually you end up with - 48 over 32, which is this at least. There's some birds [clears throat] outside being unbelievably annoying. They really need to show up. Anyway, um, yeah, this question here is quite cool. Um, so we're solving this inequality. Now, he's correct here. We can only square non negative square root negative stuff. So therefore, x has to be bigger to three for this side to make sense. Now, he's just squared both sides. And I think a lot of people would look at that and say, "Ah, squaring both sides is problematic for inequalities because like minus one is less than um sorry, minus 5 is less than two, but if you square both sides, you get 25 is less than four." And that doesn't work. Now, the issue here is though that this is actually fine because if x is bigger than 3 or equal to three, then this side is positive and this side is also positive. So therefore, you're just squaring two positive things because a root of anything is positive, right? I mean this side is just definitionally positive. Um this is actually fine to do right and then of course moving around and moving around find assume that this is all fine and in fact the whole thing ends up being completely fine just because this step is allowed. They probably should have said why this step is allowed on the proper proof but this step is allowed because both sides are positive based on this condition that they've already got just there. Okay. If positive integer n has the property n square is 12 um then n is 12. So okay 36 is that divisible by 12? Yes. Then 6 is divisible by 12. Six is not divisible by 12. 12 is divisible by six, but not the other way around. So there's a counter example. Um 64 is not divisible by 12.

[00:03:58]So this is not a number that discretion is talking about because this positive integer 8 doesn't have the property that n squares to 12. So we don't care about this. Um 18 squares to this, which I know is divisible by 12. Um because of I I just thought about the prime factors, right? There's enough um twos and threes in here that when you square them, you're going to get them in to cancel from the 12. 18 is not divisible by 12 though, so that's also a kind of example. Um now [clears throat] 24 um is divisible by 12. So I don't even need to do the squaring step, right? I can just skip it because I know it's not going to break this rule over here. And so we just end up with one and three being our two. Right? Another proof to look at here. So for every positive integer n if n^ 2 + 2 is prime, then n is a multiple of three. Um so okay he's saying that we have some integer n such that this is prime but n is not a multiple of three then the number is remainder one or two when divided by three so we're saying this number it's a proof of contradiction right because he's setting up the thing like this um but if n is not a multiple of three it's either in this form or in this form right and when you square either of those things I mean when you square this you're just going to put a bunch of threes and then a plus one at the end that's clearly still one more than a multiple of three and when you square this you're going to have this but four is just 3+ one, put the three into this bracket and then you have a plus one at the end. Um, so yeah, this is just saying that um when you square, you're going to have remainder one either time when you when you divide by three. You can also do that using modular arithmetic arithmetic if you want. Anyway, if the remainder is one for n um then n^2 + 2 of course leaves remainder zero because you're just adding two to it. So therefore, you're getting to the next multiple of three.

[00:05:35]So therefore 3 divides n + 2 square + 2.

[00:05:37]I agree with this. So this is just some multiple of three. It follows that this is composite except what if k was just one, right? You're allowed n square + 2= 3 * 1 and 3 is prime, right? So it's not particularly composite and so therefore n= 1 breaks this and the first line is this mistake here on part seven um which is answer h. Good. Um question number six then. So this just looked I don't know whether I'll be teaching too much of this at school, but this just looked like a ven diagram question to me. So I set up a ven diagram. Um, and then I what you want is you want to like minimize this section. Hardback but not illustrated and English. So you want to minimize this section. So what I decided to do was just put as many stuff as I could elsewhere. So there are 70 illustrated books. So you can put 70 right here and just nothing elsewhere in the eye circle. Um, and now I've used up 70 of the hardback books and also 70 of the English books. So I've only got 70 of the hardback books left. Um, I've got 100 English books left and I've got, now this is important as well, I've got 130 school books left because I'm assuming they're all one of these things at least. Um, so okay, well I can't just put a 70 here and a 100 here because that then would total to to to way too big a number, right? Um, so well, how much too much have I got? Well, 70 plus 100 is 170. So I've got 40 too many. So I think I can put 40 here and then wash off the rest here. And this all works out. And I think 40 therefore is going to be the answer.

[00:07:03]Question seven. So which of these are necessarily true? So we just sit here and we think of counter examples. When we're doing inequalities, we think of negatives and we think of things between 0 and one is often the thing that we should do. Um so here a is one b is two just clearly breaks this because you end up with a negative bigger than a positive which doesn't work. Um here likewise if you go for both negatives because 3^ -2 is 9 3^ - 1 is zero. sorry is is is one which makes it uh three then nine isn't bigger than three so that goes away as well. If a and b are negative then this is less than this.

[00:07:36]Again we can mess around with c and d here and make um them both negative as well and then we end up with a much bigger thing over here than over here.

[00:07:42]And actually this one here just shows that this one is clearly also going to be true. Essentially um because A and C uh sorry yeah because A and C are like quote unquote bigger than B and D if they're all negative like they're further away from zero when you multiply them together and get positive. This is going to be a bigger number than B and D are going to manage. So therefore that one's that one's always going to be true and we can move on. Yeah. Cool. this question then um so we can firstly just put in a1 equals one in and just figure out what the others are going to be. You notice a pattern pretty quickly. It's just going to be a half, a third, a quarter and so on. And so basically what we can say here is that a n is just can just be written as um 1 / n because it's 1 over one is the first one. One over two, one over three, one over4. I'm just assuming the pattern continues. This is Timura. I'm not going to sit here and prove that kind of thing. and the next one. And then of course this is just a one over n plus one. Now, you can times these things together if you want and think to yourself, well, okay, this is a half plus that makes I think like 2/3 does it? Um, that's quite small. That's very small. This is not going to get bigger than one. And I actually just underlined one and moved on when I was trying to do this nice and quickly. Um, to get the complete answer here, you might recognize it if I write like this.

[00:08:56]You can partial fractions this if in case you've seen partial fractions before and then it becomes a method differences. This is a very further maths method, but for those who have seen this, it's really nice, right?

[00:09:06]because you get one over one. I mean, sorry, I skipped over this, but just verify this in your head. When you put this over there, this over there, the times these together, you get n plus one minus n, which is just one. And then this all works fine. Now, when you put in one, you get 1 - 1 over one over two.

[00:09:20]And then the next one is 1 over 2 - 1 over 3 and so on. And then everything just cancels except for this initial one. And so, yeah, the answer is definitely one. Um, so yeah, cool.

[00:09:29]Anyway, question number nine then. So, we should know instantly this is just a Vshape with a cusp of minus K over two, right? Like this. I'll just say k is positive for now and it looks like this.

[00:09:38]It doesn't we'll deal with it later. So yeah, it looks like this. And now this is a quadratic. It's U-shaped. It's positive x^2. And I want the vshape to lie below the curve all the time. So I want the curve essentially to lie in this space here. And I can think about this in two different ways. I can say firstly I want the curve to definitely be above this line here. So even if it looks a bit like this, um I definitely want it to if I say this whole line is 2x plus k without the mod because that's what it is. I want the curve to be above that first and then afterwards I'll make sure the curve is above this one which is minus 2x mass a because of course the mod just times the whole thing by mass one to make it go backwards and I can do that separately and then I'll just compare the solutions. So I'll ask firstly that I want firstly I'll just look for the intersections between the U- shape and this line here but I don't want it to have any solutions. So I'll move some stuff over and then I'll say I want the discriminant of this to be less than zero to make sure that I don't get any uh solutions. So I'll just deal with this. I get this. I complete the square and I got that. So okay, that's one that's my solution set to make sure that the U shape is somewhere above this line extended out. Then I'll do the exact same thing with this line extended out.

[00:10:46]So that's mass 2x mask a exact same solving. I still want the discriminant to be less than zero because I don't want any solutions here. Um and I get this. And now I just need to compare these two solution sets. Now I can kind of quite clearly see if I just draw a line here. 1 plus 23 will just pretend it's there and this will pretend it's there. 1 plus 2 5 is clearly bigger than 1 plus 23. Um I'm essentially doing I should have put circles here. I'm just doing that GCSE thing where you have circles and lines. And now I just want the intersection between the two solutions which is just this initial solution here because the other two things like 1 - 2 5 is less than one 23 clearly because you're taking away more here from one. So it's fine. Um, so yeah, we're just looking for this here, I think, to be our answer. And uh, and yeah, we're done. The other one turned out to be irrelevant because yeah, I thought where the question was going was you'd end up with like here and here as your other solution set and then you'd be looking for this space here where it was kind of overlapping, but anyway, doesn't matter. Number 10. Uh, so yeah, this question is really cool actually.

[00:11:41]You need to I mean firstly draw this is not hard. This is y= - x + 4. So it looks like this. Now this one's quite interesting to draw. What I decided to do was to think about it in two different ways. Firstly, if y um is bigger than x^2, then this mod is doing nothing, right? So, we just remove it and get this. Now, if y is bigger than x^2, then I mean we can rearrange this.

[00:12:00]This is y= x^2. So, what I'm saying here is I'm going to specifically look at this area and say if we're in this area where y is bigger than x^2, then we're this, which is just y= x + 2, so we're that which is entirely in that area we were talking about. So, that's good. Um, except we're less than nothing. So actually we're we're in this shaded colored region here which I won't bother to color in but we're in that region there. So if we're above the red line then we're below this black line. So we're in this region here is essentially what this says. Now if we're the other way around then this mod is going to times the whole thing by bar one and we end up with this which we rearrange. And so if we're below the red line then we're above x mass 2 which is here and so we're in this area here. So either we're above the red line and we're in this area or we're below the red line which is means we're in this area. So in total this space just represents all of this area. It's just those two areas come together. And so if we're both this thing and this thing that means we're in the space still below this diagonal line. Remember so in this little shape here all the way down all the way down and up to there. So it's just this between above and below the red line up to here as well. Now so that's the region we're talking about. So for every xy y is less than six. So the highest y in this region is the y where this uh u shape hits the curve there. Now remember that's this u shape. So I'm just going to look for the intersection between that and this which is remember was just y= x^2 sorry mass x + 4 was just this.

[00:13:28]So let's look for the intersection here.

[00:13:30]We get these two things here. So x is mass 3 and if x is mass 3 y is 7. And so up here there's a y-coordinate of not quite seven because we're not allowed to be on the surface but just less than seven does work. So um I don't think that's true. I think we can be slightly bigger. So I should have crossed it out but no I don't think that one was true.

[00:13:47]Now what about y bigger than minus2?

[00:13:49]Well yeah if we see on our region here every coordinate has y bigger than minus2 because that's the point back here. So that one's definitely true and x less than two. So again um I mean I've actually done the work already because I looked for the intersection between this curve and the line to find this at minus 3. But also this is the two here right this is the two and that's the furthest to the right we can go. So therefore yeah x is less than or equal to two um everywhere. In fact, it should be x less than two. No.

[00:14:15]Um, why is x allowed to be two? We're not allowed to be on the region. Uh, oh, maybe the answer is two only. Okay.

[00:14:25]Well, you can look in uh Jay-Z's solutions to see if I'm right about that. But perhaps this should be strictly less than two for that to be correct. Um, which is annoying, but whatever. Cool.

[00:14:36]Good. Next one. Uh so yeah when we're integrating two things added together we can just integrate them separately. And now if we just firstly check sufficiency if k is a positive integer or an integer multiple of pi just make k pi right and we can see quite clearly the integral between 0 and pi of cos is zero because those areas cancel and cos cub is essentially the same as co because 1 cubed is still one and mass one is still cubed is still mass one. So we're still just going to do this traveling motion that we do before. 2x just of course squashes the graph in. But because the integral of cos between 0 and 2 pi is still zero because cancels cancels cancels cancels. And because cos cubed is essentially just going to look the same as this. This is still going to be zero between 0 and pi. So it's just 0 plus 0. And so this is definitely sufficient. And it's not going to make a difference if you go to 2 pi because of course the integral of co between 0 and 2 pi as discussed is still zero and it's that's not going to matter if you times it by two and make it sh uh squash it in a bit. That's still going to work. So yeah, this is definitely sufficient. now necessary. I basically just stared at the graph and was like, is there anything even remotely obvious about how to find a point which would make this zero without just using a model of pi?

[00:15:45]And I couldn't see anything at all. And so I decided that it was that it was probably necessary because, you know, if I could find a counter example where the whole thing equals zero without using a model of pi, then it wouldn't be necessary. But that just doesn't seem obvious to find at all. And I know I'm not supposed to actually integrate this because this is actually quite hard to integrate even for someone who's got um integrating skills much more much higher than what Tamura requires which I think most of you probably do. And so I just went for this is this is necessary because there were no obvious counter examples and usually you're you're trying to find quite obvious counter examples. Um so yeah, if you don't have time to do it, just back yourself. You you know you didn't miss anything obvious just go with it and you should be fine. Now there's a bunch of ways to think about this question here. um which is that you could there's there's a bunch of ways to rearrange this for y firstly you could think about like this you could then actually raise both sides to power one over y and then add the one to think about this you could also use the change of base law Taylor Swift's law to write it like this and all of these things are quite helpful ways to write this which allow you to answer this I think um for y bigger than x sorry y bigger than zero for every x in the domain I mean this allows you to think well okay I can actually just force y to be mass one because something's F mass - 4 means this is a quarter and x is 4 point sorry it's 5 over 4 makes that work. So yeah this this can't be true because I've just found an example that that that makes that completely work fine. Um for the next one y is strictly decreasing. Well okay if I were to instead force y to be one which is making this bigger then that would make x five because 5 - 1 is four and so x has got bigger whilst y has got bigger. So this is an increasing function and like it turns out when you sketch this graph in a thing in like a graphing software which I recommend you do it actually has an asmtote which we'll get to. So it does actually end up being decreasing for all the parts where it actually exists. But because it has an asmtote a bit it looks a bit like tan does. Um it does end up breaking the decreasing rule because I've increased x and increased y. So it can't be a decreasing function. So that goes away and that goes away. Um now here for every number y exists that is a value of x such that this happens well if y was zero we can see from this rearrangement that if y is zero that's going to be problematic. Um and actually it's going to cause the asmtote as well. Um the asmtote uh there's lots of ways to see that this probably has an asmtote but here is here's quite a nice way of saying it. If y is zero um then we've got some number which is just just a number right over something equals zero.

[00:18:13]And that just that's just a classic asmtote, right? This is a classic asmtote. We can rearrange log 4 over y equals this. And then we're dividing by zero. It's going to make an asmtote. So we're going to be fine here. And circle the answer and we'll be good to go. Only obvious thing to do on this question at all is to write this as 1 - cos^ squ.

[00:18:29]I'm going to code everything in c's as well because I can't be bothered to write this obviously. Expand out. Um factoriize out minus two because then I could easily factoriize both these things and cancel them. Um and now okay, I still want the maximum value of this.

[00:18:42]Um, I think just because I've done lots of integrating, I thought of this nice little method of just expanding this out and then forcing C + 5 to exist on the numerator because now I can cut it here um, and just cancel those two and write it something like this. And this is really nice because if I'm looking to maximize this, I'm clearly looking to maximize this. And if I'm maximizing this, I'm going to be minimizing the denominator. And the minimum cause can be as mass one. So this makes mass 1 plus 5 which is mass 4 uh sorry which is plus4 and then we get mass 2 plus 4 is two and that will be our maximum value I think. Um so yeah quite a nice little method there just there for this bit. Um which is worth bearing in mind. It comes in handy sometimes when you're doing questions like this. Good to the next one we're can see a minimum thing. So we're going to differentiate this straight away. Um when we get this set it to zero. Now the second quadrant is this one right? So either we want x to be negative and y to be positive or um I'm sorry that's all we want x to be negative y to be positive right so here x is plus or minus a clearly now um if uh a was positive then that's not something we want um assuming that a itself is yeah it doesn't really matter one of these will be fine right one of them will be negative one of them positive regardless of whether a itself is positive or negative um if I put in a we get this and if I put in a minus a we get this Now notice this one. If we focus on this one for a second, if a is positive, then this is positive because something cub plus something is that's all positive, right? So this either goes positive positive which is over here. So that's not correct. Or if a were negative, this is definitely negative because a negative cub is negative plus negative is negative. So it's either a positive positive which is here or it's a negative negative which is here. Neither of those things are things that we want. So, we're going to focus on this one and we're going to say that a needs to be positive because that would make this x coordinate negative.

[00:20:35]So, we'd be over here somewhere. And if a is positive, we want this to also be positive, right? Because we want it to be up here. So, we're looking for this to be bigger than zero. We can factoriize out the a and times by three, I guess, and stuff like that. And we find a is either zero or um this stuff.

[00:20:50]Now, we know it can't be minus the thing because we're only looking in this set anyway. But it's okay. This is some kind of cubic graph. It's actually a negative cubic. So it's going down like this and it has solutions at zero and plus or minus this route. So it looks like this then through zero then down like this.

[00:21:05]So when is that bigger than zero? Well, it's when uh we're less than the first negative route when we're over here and when we're between the two roots over here. Now this one doesn't matter because again we have to be bigger than zero for any of this to make sense in our um system anyway. So we're going to be this as our solution only. And uh and yeah, that'll be that. This is a nice question. I'm going to take this triangle which is equilateral after all on the base and just put it there. So this is A, B and C. And then I'm going to make the net of this. I'm going to pull this down like this and this down and the front one down and make this like this. Um, okay. So we've just written out the net and X is either here or here on the midpoint. Now I'm going to put it over there and you'll see why.

[00:21:43]Because Y is just here. And now basically I just want to if I put in some distances they've told me I just want to know the distance from here to here because this distance is equivalent to doing the walk through this side or or down along the edge of this side I should say and then just straight through like this. Right? Whereas if I put the X on this side that will be equivalent to doing this walk which we can clearly tell is going to be longer and the net tells us why it would be longer. It's just on the wrong side.

[00:22:09]Right? So we're going to put it here and we just need to find this distance. Now, this isn't too hard to do because what we can say is, okay, firstly, let's just take this 12 and let's move it six units to the left. So, we're finding this blue distance. Move that 12 six units to left parallel to itself. So, just parallelly move it two six units to the left, which makes this still 12, but it makes this distance to where the base of that 12 is going to land 9 because it's moved six to the left. And now this angle is still 60 um because I've just moved it across.

[00:22:41]So, it's going to be 60 here. And now I can just do cosine rule um to find my x, which is pretty boring, but um I got bit panicked here, but 1 plus 1 + 7 is nine.

[00:22:50]So, this decimal definitely has a three in it. So, we're fine. And uh and yeah, we're good to go. Question 16 then. So, we've got a safe with three levers. This is the rule we have. If A is right and B is left or C is right, then the safe is open, which these must be true. if the safe is open. Now, this immediately can't be true because this safe could just be open all of the time, right?

[00:23:13]They could have just given us a random example of it being open, but it could just be open all of the time, and therefore, we don't know anything upon walking up to it about it being open, right? It could just be open all the time. If the safe is closed, then lever A is left. Now this can't be true either because there are cases in which the lever is on the right where the safe is closed. If A is right and these two are in the wrong place then the safe is closed. So if the safe is closed then lever is on the left is not something we can say for sure. It could still be on the right. Likewise we'll get rid of this for the same reason. Now here if the save is closed then either A is on the left or B and C are in the wrong positions for it to be on the right essentially is what they're saying is that sounds better. We'll go with that.

[00:23:56]and uh we'll underline it or say that's probably definitely going to be good to go. Cool. Question number 17. So the first thing I like to do sometimes when I look at this question is see if I can think of any really obvious examples to ascertain whether or not this statement is true or not because then I could eliminate um you know that the this statement is true version and I could look really hard for where it is actually not true. And now this doesn't end up being too hard to find. um if you use n is four uh four doesn't divide one factorial, two factorial or three factorial, but four isn't prime. Now four is actually the very first thing you it's possible to think of um because aside from I guess one one divides one factor else that's pretty obviously gonna uh work I guess uh sorry is is not relevant I guess because it does divide one factor in fact no one doesn't have anything less than it so you can't even use one you can't use two because two is prime so it's doesn't contradict the same and likewise three but four contradicts it right so I know this isn't true and I'm just going to look very carefully for where this isn't true so that's just where I decide to Um okay so assume n does not provide k factorial good fine contradiction so n is a composite number where a and b could be the same or they could be one lower than the other they times together make n fine a is less than b so we're just doing this particular case where like a is two and b is four and you're multiplying to make eight whatever um but then yeah they they a and b definitely both exist in the set one or two to n minus one right because they're both less than n and so therefore they're going to be in the product of n minus one factorial And now in case two, a equals b. So n= a squ. Since a is bigger than 2, a and 2 a are both this except they're not quite sorry, they are both this, right? They are both this.

[00:25:42]Now this is where they messed up. I think if a is bigger is two, we have a less than a squ. Um well okay sure 2 is less than two 2. But here 2 * 2 is not less than 2 * 2. So if a is two and I was looking out for two the entire time, right? Because I know I had four as my color example and two * two gets me the four that this whole thing had set up.

[00:26:05]So yeah, seven is the line that this is uh incorrect on. And uh and yeah, it's just because of this one little statement there, one little line there, which led them to then closing the bracket early here, which means that they are then making this statement about how they both appear in this thing when they don't because one of them is just n itself and uh and yeah, cool.

[00:26:23]Anyway, question number 18. So this is quite a cool question. Um firstly A, B and C are all positive which means all of these logs are positive. Now this is quite nice because it forces you to think very clearly about when logs are positive or not. Um like when the answer to a log is positive. So if we think about the classic log graph, it looks like I think this is just log base 10.

[00:26:44]Um a log is positive when the thing that you put into it is bigger than one. Um, and if it's less than one and bigger than zero obviously because you can't put negatives into logs, then it's less than zero. Except there's actually another case you can use here, which is that if you're doing log of a decimal itself, like log of 0.5, then the log goes backwards and you require um the input to also be between 0 and one. So if the log to base is between 0 and one, the input would also have to be zero between 0 and one to make it a positive thing. If it's bigger, then it would be a negative thing.

[00:27:19]according to this graph. Um so okay that's pretty cool. So if we want all of these logs to be positive um then we either need a so you know think about these one by one either we have um two things that are bigger than one because this is just representing any log graph with a base bigger than one and we want the input to be bigger than one as well to make it positive or both the base and the input are between zero and one but we we know all that our numbers are bigger than zero anyway so it's fine. So yeah looking at all these logs if we want them all to be positive we either want a and c and b and c and a and b that means all of them to be bigger than one or we want them all to be less than one and obviously bigger than zero as it said. Now the thing is if you write these all in terms of their powers so a to the c= b and b to the a= c and so on um then well okay if we just do these one by one if a b and c is bigger than one. Now if c is bigger than one then a to the c is bigger than a so b is bigger than a but then here if a is bigger than one then that means this is bigger than b and if b is bigger than one that means that's bigger than c. So now we have a chain that says b is bigger than a but c is bigger than b but a is bigger than c.

[00:28:28]That doesn't make any sense. B bigger than A, but C is bigger than B. So that makes the order largest to smallest go C be A. But then it says A is bigger than C. So this is a contradiction. So we can't have this one. But then if we try and have this one, and we have to be a bit careful here, but if we try and have this one where they're all between 0 and one, then that's the number between 0 and one. And when you raise it to a power between 0 and one, you actually make it bigger. Like 0.5^ 0.5 is the root. It actually makes it bigger. Makes it like 0.7 or whatever. So this is bigger than one or sorry bigger than A.

[00:28:58]Uh this is bigger than B for the same reason and this is bigger than C. And we're actually in the exact same situation I just talked about. And so therefore this is a contradiction as well. And so therefore there are no solutions uh for any of this which is quite interesting and a very difficult question I thought. Um yeah this question is basically just a step question. Um I think it's quite a challenging one to put into new to paper. But yeah we have this step function here. You're never supposed to integrate functions like this mod functions step functions anything like that. um because there are just no rules for integrating. You just have to draw them and figure out the areas by hand.

[00:29:31]So if x is between 0 and one then this is just one * 2 to the 1. That's how this this this ceiling function works.

[00:29:39]It's just one time 2 to the 1. You just uh any input between 0 and one you just round to one and just put one in. Any input between one and two you round to two and put two in. So it becomes 2 * 2^ 2 which is eight. And I can just work out these areas uh like rectangles as I go along, right? And so we're getting all the way to 1920, which is going to become 20 * 2. And then I'm just going to add up all of these areas by hand to get the whole value of the integral.

[00:30:05]Which means it essentially becomes a sum like this, right? Of just n 2n where the first one is just um 1 * 2 1 which is this. Then the next one is 2 * 2 which is this and so on all the way up to 20 which is 20 * 2 20 here. So we're just going to evaluate this. Now I can write it out like this. I just said this out loud that it's just this thing here. Um but how do we actually add this together? And yeah, this is this is I think quite challenging for a to paper I would have said but um what we can do is and if you've ever seen the proof for the sum of a geometric series this is quite similar to this. We're going to times the whole thing by two. I mean I can call the whole thing s but sure times the whole thing by two to get 2s.

[00:30:44]And when you times the whole thing by two I'm just going to add two to every single power in each term. Um, and then I'm going to do this one minus this one.

[00:30:52]And the reason I can do this one minus this one is because if we think about this very carefully now, this has no counterpart because there's no 2^ 21.

[00:30:57]That just goes there. But then the previous one would be 19 * 2, which when you minus 20 * 2 just gives you minus one lot of 2. So 19 lots of 2. Um, minus 20 lots of 2 is just one lot of 2. And you just keep doing that all the way down until you get this mass. This is mass one lot of 2 to the two. just that one and then we just have this on its own which is just minus 2 ^ 1 and now if we factoriize out this negative from all of those terms we get this and now this is just a geometric series with first time two times by two every time and there's 20 terms so I can just use the geometric formula for 20 terms which is that I believe one mass 2 is mass one so the mass cancels with this mass expand it out and you get this and then we can expand out some more and group together put that two in that power I guess and then we can take this away from there and we end up with this as our answer quite core question. Now here final question we can say this is root of this over root of this and root of this over root of this and then cross multiply those roots and then because root a root b is just root ab we can just say it's this right we can just put those roots together like this and just cross multiply now when you multiply these out this works out quite nicely right difference of two squares and then of course we can replace them with their causes but when we saw this on his paper one as well when you root a square you end up with the mod because roots by definition are positive we can't just write cos here because cos x could be negative. So we have to put mod cos x here and mod cos 2x here. Now I drew this in power. I kind of regretted doing it but it's kind of funny so I didn't really care. I'm going to say the cos graph essentially um sorry the mod cosgraph essentially looks like this because co looks like this essentially doesn't it? It's basically just a U- shape. I know there's some curviness going on but it doesn't matter. And then when you mod it you just put any negative bits on the top and it looks a bit like this. Now, these are supposed to be cusps and this kind of curves the other way, but it doesn't matter cuz all I'm going to be doing, of course, is counting intersections. So, it really doesn't matter. So, let's just say this is cos x between 0 and 360. Except what I'm actually going to do is copy that over to make this disgusting graph, which is essentially cos 2x between 0 and 360 here.

[00:33:05]Because of course, if that's cos x between 360 there, then if I just double it out and then put 360 here, I've essentially just squashed the graph up, right? So this is cos 2x modded up to 360 and cos x will just look a bit like this. It's just one lot of those things.

[00:33:19]So I've just stretched out this section like this. Now all this should be symmetrical but again I'm just counting solutions so it doesn't matter. Um it this is up to 360. Counting up to 360. I have one solution here. 2 3 4 5 6. Now this one doesn't count because if you notice P is less than. So what I'm going to say is there are six solutions if P were 360. Right? There's 1 2 3 4 5 6.

[00:33:43]Now if P were 361, P has to be an integer. So the next one could be 361.

[00:33:48]Then there would be a seventh solution, which is going to be relevant in a second. So if P was 361, there would be seven. But if P was 360, there would just be six. Now therefore doubling it.

[00:33:57]If P was 720, there would be 12. Um, and I've got to 12 solutions so far. I need 16 of them. So if I get another three solutions, that will go into 15. Now another three solutions would just come from adding 180 to this where I pick up the the next one on on the edge and then these two. So up to 900 there'll be 15 solutions. I'm essentially getting to this point on the graph now after two copies of this and a half copy. And now to get the 16th solution as we were discussing up here I just need P to be 901 just so that I can grab this solution here. Um and then I can use uh the 16 get the 16th and we're done here.

[00:34:35]Thank you again for Jay-Z for writing