My brother got asked the distance to the moon in his Oxford University interview to study Physics. He said he didn't know, and they told him, "work it out". From his time in air cadets, he knew that aeroplanes travel around 550 mph. The only other fact he knew was that the moon was roughly far enough away from Earth to fit 30 Earths inside the space between. So he applied the information he had. A flight to the opposite side of the World took roughly 24 hours. 24 hours x 550 mph = 13,200 miles. Double it for the circumference of the Earth = 26,400 miles. To turn a circumference into a diameter, divide by pi. I think he rounded down to 3.1 or something to make it easier, but then he times that diameter of Earth by 30 for an answer that was only a few thousand miles outside of the moon's maximum distance from Earth. The interviewers loved him. He didn't just know the answer. He knew how to *apply* information that he already had to be able to *work out* the answer. He got his place at Oxford.
There's a lot of different variations on questions like this. It's called a Fermi approximation, the idea is because your errors are randomly over/under you'll probably end up in the right ballpark
This series on Oxford is incredibly humbling. It's incredible to think college grads are disciplined enough and have the drive to succeed in these admissions. That sort of discipline only really kicked in for me during my masters (a bit too late lol). Kudos to you guys
So interesting seeing you learn these type of “subjective/soft” skills where the success of learning it isn’t as clear - they’re actually a lot more relevant in day to day life
Yeah, for example it hadn't occurred to me before that getting "unstuck" while solving a problem is a skill that you could improve. And now that I'm aware of it it's obvious how useful such a skill would be. I'm already thinking about how it could work in computer programming.
@@puupipo same I’m in my first year in computer engineering and the hardest part for me in programming is to order your thoughts so the computer can understand it and when you get stuck on something how can you solve it.
@@Rogue_Route For the first problem you mentioned I've found it really useful to always write down the whole "skeleton" of your program in plain English before writing any code. It's a lot easier for your mind to handle when you're not trying to design the algorithm AND translate it into code simultaneously. For the second bit, taking a break to clear your mind can help. There's also something called "rubber duck debugging", give that a google if you're unfamiliar with the concept.
@@puupipo thanks for the advice never thought that writing the program in English first will help am going to try that before my exam in a few days in c++
My Oxford interview wasn’t scary at all, it was essentially a conversation about things related to the course I was applying to (in my case biology), I didn’t have a single question that wasn’t about biology. The academics are just looking for people who are passionate and interested in the course they’re applying to and can think creatively :)
Interesting. In the US you’re not applying to a course. You take a major, but you don’t have to immediately and half of your classes will be general education across a myriad of subjects both required and elective.
@@WaterYaDune Mate, the US System of being able to do whatever you want once you are in the University, from being able to change Major, to doing a Minor on something completely unrelated, or even double majoring is a much, much better system than the current European system.
I come from a humanities not a mathematics background (at Oxford), but I wanted to add: 1) the fact that they only focus on the subject - no questions about you or your aspirations - is very representative and imo a good thing 2) my interviews were really fun. The first two mins are always nervey, but I really got into them. They want someone who sits down and digs into the questions and gets really stuck in - where you really want to figure it out because you care about the problem, not just impressing the interviewers
Can I ask what humanities subject did you interview for? I would really love to apply to Oxford for Philosophy and Modern Languages and I am a little unsure about what questions they would ask..
I am INCREDIBLY curious to see what another discipline looks like in this process. Math is honestly something I've never understood, to the point that some of these questions were gibberish to begin with! I wonder what "extremely high level literature" (or philosophy, or art) looks like...
I do philosophy, politics, & economics at Oxford, & my philosophy interview was super precise, like if I said a single word which didn't quite fit with the rest of the sentence the interviewer would ask why I'd said it. At one point he asked me the same question 3x in a row (as in, once I'd answered he just asked the exact same question again) while just staring at me lol, was very stressful. The actual subject of the interview was just various ethical scenarios which he wanted me to reason through, so a pretty specific take on philosophy. Not something you could prepare for really
@@georgebarber3280 Omg PPE at Oxford. You're probs sat next to the next 3 PMs lol (Or I'll be able to say I replied to George Barber's comment on TH-cam once! lol)
So I had my english interview for Oxford a couple of months ago and they gave me a passage and 15 minutes to prep. They then just said talk about what you find interesting so I suppose the hardest bit was knowing what to talk about.
Well done! It was great to see someone else go through this… I applied for physics this year - I got through the entrance exam but was unsuccessful after interview!
I had to do 4 interviews for Maths and Compsci at Oxford. The pressure was INTENSE. My MAT score was very mediocre (despite a lot of preparation) so I can offer a perspective of someone who isn't as gifted as others. Many times I would just be staring at the question, not knowing where to start. Although I recognised the content and notation, it was put in way I had never seen before. I made quite a bit of progress on some questions, but I was guided by the interviewers a lot more than I liked. I can see that maximising your interview performance is a skill, but there is a point where, if you're just not smart enough (and I mean intelligence, not through lots of learning), you'll never be able to cope. Despite my passion for the subject, I didn't get an Offer unfortunately; competition is extremely fierce for the course. Don't feel bad if you don't get into Oxford or any other unis. It's probably not the uni for you, and you can always get the job of your dreams by working hard.
While there might be people that just don't have the intelligence due to some genetic or developmental issue, the vast, vast majority of people are able to achieve similar levels of inteligence through work. That said not all work is equal, if you only train yourself at solving a problem in a particular way then you're not working on the skill of abstracting the solutions.
I'd tend to somewhat second Spliter, above. I think that it's rarely a case of intelligence with these things, but more one of motivation. And I don't mean this in a 'oh, you were unmotivated, you should have just tried harder' kind of a way. I mean this as in: 'there are lots of different ways to handle maths, some of them suited to interviews, some not', and unless someone possesses god-tier motivational abilities, I think most people will tend towards working in the way which most suits them. So, in my case, I used to love teaching physics (slight detour from maths I know, but I've never been interviewed for maths, and figure the idea I'm trying to convey converts quite nicely), and so I associate vocalising my thought processes on the spot with being a rewarding experience, which meant that any prep I did would tend to use this skill, which meant I trained it more, which meant I would use it more. And this is even in sitting exams and in doing exam-prep - my tendency would be to try and handle a problem in a way which I could most easily explain/justify, which often wasted tonnes of time (in fact, this is a huge problem in the amount of time I spend on uni work to this day - multiple pages of explanation/thinking sometimes go into a 2 line algebraic solution). In other words, when it came to uni applications, I'd tend to do well in interviews, and far less well in my exams (managing to muck up my physics exam - by far my strongest subject - fairly badly). Point being, I _do_ think it's possible to train this, but I think it's a question of whether you find the most useful skills to solve the problems the most rewarding, and hence on how you tend to approach problems. I suspect, like me, you worked really hard, developed really useful skills, and whilst you will have developed your skills for interviews and the MAT, a tonne, you might have learnt _more_ in regions which were still practiced by MAT/interview prep, but which weren't as vital. And I think, most of all, my point with this is that you _still_ put in a tonne of work, you _still_ will have gotten a lot out of it. Whether or not it's the same stuff you need for the interview, honestly, who cares? I've got multiple friends at Oxbridge, and life there seems to value similar things as found in the entrance exams. If something else was the most rewarding bit of maths to you, then, you know what, the skills which you built up the most, but which didn't help the most, probably would've been neglected by the Oxford system, meaning you'd've had to put in even more work to make up the difference between the skills you found most rewarding, and the ones they wanted (this, again, from the experiences of a few friends). In short, huge congratulations for putting in all the prep - you probably developed lots of skills which you found more rewarding, and which Oxford wouldn't have nourished. So don't put yourself down, as 'not having been smart enough to learn what they wanted, despite all your work' - I think it's a huge fallacy to assume that repeating a particular action (even something such as diverse problem solving and abstract thinking) will produce a given particular result - it'll always vary person to person, and _can't_ be catered for by every university. Honestly, I remember finding past-papers and stuff to be a _huge_ drag, so congratulations on whatever prep you did - that in and of itself is worthy of commendation! You sound like you have a passion, so best of luck, I'm sure it will (have) take(n) you far! (Oh, and a slight addendum, RE the final part of Spliter's message - I often get the feeling from people, and often honestly perpetuate this myself, that learning by rote, or getting _really_ good at a particularly simple process, isn't very useful if you can't abstract it. That's not necessarily true: Often, clever algorithms and simplifications are only found by people who have just spent a lot of time working with a problem, and its processes.)
@@spliter88 I think it doesn't really matter how much you practice or study, if you can't take your knowledge of one thing and see a similarity in another to then apply it to that situation it's going to be tough. The number of smart people at my school that would get a question like "equally share 10 sweets between 2 people" and not be able to do it because the question wasn't "divide 10 by 2" was crazy (oversimplified example to get the point across). They were smart because they could just remember things and recite them, but couldn't actually apply that knowledge if it wasn't presented in the exact way they learned it. And throughout uni, doing computing, we would have 3 tutorials that taught us X-Y-Z, and then we'd be set to go do a task that required Z-X-X-Y, half the class would complain because "we were never taught how to do Z-X-X-Y"
@@scottg3192 My point was that knowing how to apply skills you've learned in one area and apply them to something else is itself a skill you can learn. You learn to abstract, it's not something that you're born with. Like, literally the tests they do on toddlers show that below a certain age they're just unable to abstract at all, and they learn that over their childhood. I don't think that it stops either, there's more and more complex abstractions we can learn as we age, a skill we can train.
@@scottg3192 It's nature vs nurture. Only one can get you so far. But I still think it is possible to learn it. Like with maths, i was never good at abstracting one problem's formula to another. But I have nowhere near that kind of problem with computing. But in theory it's the same, you learn one thing and apply it to the other. But some people struggle at computing but not maths problems.
I really love the attitude you have towards these "non-physical" type skills. I tutored math in high school and college and you would be the actual dream pupil. I had so many people that gave up before they even started or assumed that they were incapable of learning something and that was really difficult to work through. You have a great mix of respect for what you don't know (and for those who do know it!) balanced with the understanding that just like anything else you can improve with practice 💯
SOLUTIONS: If you are interested in the solutions of the 4 questions in this video here are my solutions but I cannot guarantee that everything is correct: Tin question: How would you choose the radius r in a tin (cylinder) with volume 1 to minimize the surface area of the tin? Solution: Since the tin has a volume of 1 and the volume is base area (circle) multiplied with the height h we can conclude that: V=πr^2*h=1 and thus h=1/(πr^2) The surface are consist of the top and bottom part each having the area of a circle and the cylindrical wall which is a rectangular with the circumference of the circle(2πr) and the height as length of the two sides. Thus we know: A=2*πr^2+2πr*1/(πr^2)=2πr^2+2/r To get the minimum we will differentiate: A'=4πr-2r^(-2) Minimum is where A'=0: 4πr-2r^(-2)=0 4πr=2r^(-2) r^3=2/(4π)=1/(2π)=(2π)^(-1) r=(2π)^(-1/3)=~0.54 Thus r=~0.54 and h=~1.1 f(x)+2f(1/1-x)=x: Insert 3 as x: f(3)+2f(-1/2)=3 Insert -1/2 as x: f(-1/2)+2f(2/3)=-1/2 Insert 2/3 as x: f(2/3)+2f(3)=2/3 Thus with substitution we know: f(3)=3-2f(-1/2)=3-2(-1/2-2f(2/3)=3+1+4f(2/3)=3+1+4(2/3-2f(3))=3+1+8/3-8f(3) 9f(3)=3+1+8/3=20/3 -> f(3)=20/27 Factorization of 840 with the largest sum (I assume only integers) so 840=a*b and a+b is maximized: Obviously to factorize a number in many parts will lead to a smaller sum therefore a factorization in two products will have the largest sum. To prove this we try to show with an inequality: a*b a*b-a a*(b-1) 422 210*4 -> 214 ... 42*20 -> 62 30*28 -> 58 Thus we can see that the factorization with the largest difference to each other but where still both products are an integer larger than 2 with the largest sum is 840=420*2 -> 422. Pizza slices: For k straight cuts through a pizza how many pieces n can you get? A single cut can always split at least the half of the pieces that are existing thus: n -> n+n/2 With a smart cut we can split even one piece more thus: n -> n+n/2+1 For cases where n is odd n/2 is being rounded down. We can write this up to k=4 as: k=0 -> n=0+0+1=1 k=1 -> n=1+0+1=2 k=2 -> n=2+1+1=4 k=3 -> n=4+2+1=7 k=4 -> n=7+3+1=11 We can write this always as arithmetic series added with 1, e.g.: k=4 -> n=(1+2+3+4)+1 n=(sum from 1 to k) +1 Thus we can write this arithmetic series as: n=1/2*k*(k+1)+1 With some simplifications: n=1/2(k^2+k+2) Hope this helped someone. Edit: Thanks for the comments guys I tried to implement them as far as I could.
From the PhD interview perspective- I had two for two slightly different projects. 1 went terrible- I had to do maths on a whiteboard in front of 6 academics and, surprise surprise, didn't get the place. The second was amazing- I clicked with the professor and we just chatted for an hour about their research area(s). Got that one :) I think it really is luck of the draw
This is giving me flashbacks to the oral exams I did a few times at uni - they were usually done as resits for failed exams, but they were 100 times more stressful than the exam itself even though the lecturer would sometimes help you through it. Ironically the only admissions interview I did wasn't stressful at all because I had no idea it was coming. I was there for the applicant day and literally got dragged into an office with two lecturers who grilled me for 30 minutes on the course, but because I didn't actually realise it was an admission interview until it was over I was totally calm through the whole thing, answered every question pretty casually, and got the acceptance letter the next week. I think there's something to be said for being relaxed because I've done far worse in easier interviews I've aggressively prepared for.
Just watching the interview and knowing that Mike was under a lot of pressure, stress, and anxiety, honestly has given me such ptsd of highschool and the overwhelmingness of having a 1 on 1 tutor and/or tutor
Just wanted to say, as a current Maths undergrad at Oxford, these videos are really cool and aswell as being entertaining are an amazing tool for people wanting to apply! And also pretty nuts to see Adele my pure maths tutor from first year! Friendliest tutor ever :)
Never expected there to be a part 2 of this! I absolutely love mathematics, but sadly found out too late. I'm following an Art history Research Master's instead.
@@lol360noscope6 Very true! That's why I focus on digital art history. There are some interdisciplinary opportunities and projects that combine computer science, mathematics and art history, so that's the direction I'm trying to take. I'll probably never be the main mathematician in the group, but I'll take what I can get, haha.
at every problem just pauzing the video and thinking for a little while, then watching the process of mike working thrue it in a different, or the same way, is a nice experience.
@@spiralingsphere3785 I think it depends on what you're studying, in medicine there's a lot of pressure from all directions. They don't tell people about the suicides that happen every year.
People prepare for weeks for this interview, and you did borderline in 2 days of practice. You're phenomenal, i believe if you prepare for a week or 2, you'll get into the university.
Loved this Mike! You're so right that the Oxford Admissions Process is a 'skill' you can improve at - I got rejected the first time, but honed the 'skill' over my gap year and got accepted second time round! Great video :)
I had an interview at Cambridge for natsci (Physics and and chemistry) and whilst I was unsuccessful and it was quite stressful at times, it was an incredible experience to have.
I think those question were easy. 1. Just fill in 3 in equation, after which you need to find f(-1/2), so you just fill in -1/2 after which you need to find f(2/3) after, so you just fill in 2/3 and you'll need to find f(3). Then you fill in everything backwards, and you get an equation for f(3). So f(3) = 20/27 (if I didn't miscalculate) 2. The second question you get maximum n EXTRA slices on the n-th cut. So you get n(n+1)/2+1 slices on the n-th cut. (the sum of the n first natural numbers is n(n+1)/2) 3. 420*2 sum 422 It's obvious. 840= 2*2*2*3*5*7 Just write out all possible factorizations and their sum. Or by proof: if one factor is a, the other is 840/a. Their sum is a+840/a. This function has a minimum at root(840) and rises after that. The highest possible integer value for a is thus 420.
So for the pizza problem my reflection is : we want to maximize the number of slices, so each cut should bisect the most amount of slices. Which means we want to have the maximum number of intersections within the circle. so the maximum number of intersections that is possible is equal to the number of lines, as you can only cross 1 or 0 times (parallel or crossing outside) now for the number of slices crossed, it is equal to 1+the number of intersections, which is the number of lines We split that in half, so we add that number to the previous number and we start with a single slice, so we end up with 1,2(1+1),4(2+2), 7(4+3), 11(7+4), 16 (11+5) etc... For a general formula for the number of slices after the nth cut We have the 1 at the start, so 1+??? And the rest is the addition equivalent of a factorial, i.e. triangle number, which is (n²+n)/2 So we end up with 1+(n²+n)/2 slices For the cylinder, we just take the area of both the tube and the ends, then take the formula for the area, and then move it around so that we have the formula for the two surfaces at a constant volume, and then we solve h and r for min(area) Another solution is to binary search it, which is probably faster, make two cylinders, one really tall and one really wide, look wich one has the smaller area, make one in the middle and look there, repeat a couple times and you probably will have an answer that's good enough. I think quick solution that may work in some cases is to look at the formulas and see if h or r is squared, if one is and the other is not that's a pretty good indication that you want to minimize the squared one (though maybe not as it's also affected by the area)
idk if this is mentioned further in the video, but the answer to the pizza problem(i think) is 1+(n²+n)/2, as with every crossing there is a new area created, and there are n-1 lines to cross if you don't go through existing intersections, and as exiting the pizza is also creating an extra area, this means the nth line adds n new areas at maximum, thus the amount of areas at any point can be written as 1+1+2+3+...+n-1+n, if you take that last part and add it to itself you can, rearange the terms to (1+n)+(2+n-1)+...+(n-1+2)+(n+1), as every term will then be n+1 and there are n terms the adding of this to itself (aka multiplying by 2) is equal to n(n+1), and thus the formula is 1+((n+1)n/2).
@@DavidHughesArchitect if wvery cut is concurrent yes, but non of the cuts are concurrent so if you have >2 cuts then there is more to consider than just the outer areas
I burst out laughing when he said that, always the optimist. I'll bet he would be fine if he made maths a hobby / passion like he's made learning random physical / mental tasks on the show all these years. Like playing chess a few hours a week, if he simply made maths his thing a few hours a week, he would get in next time.
not necessarily related, but I had my oxford law interview about a week ago - the pressure was crushing and although i'm typically a quite calm and confident person under pressure, it was a completely different story. It's hard to watch things like this and not compare myself to them, however for anyone in the same boat as me just remember that everything looks worse than it was in the post-morten; you look for what you did badly because you can more easily understand what the interviewers arent looking for, its much harder to appreciate what went well when you have little idea as to what they are looking for. To all the applicants who have made it similarly far, good luck!
@@joeyhardin5903 Yeah mine was online, I was sorta hyping myself up beforehand and mid-sentence the panel joined the meeting so that didnt help - in person would have been much more enjoyable!
I had my cambridge law interview mid december and i had done quite a few mocks but the real thing was so much scarier so you’re defo not alone :) best of luck for your application!
I'm currently an first year law student at Oxford and those interviews were hard. Caught myself having to think because I genuinely had no idea how to answer at first Like you said, you usually think you did worse than you actually did. Especially for law, it's not about knowing the law for the interviews but more your logic towards arriving to your conclusions
I think you can create n extra slices per cut. One extra slice after the first cut, two extra after the second cut, ... So if a recursive answer is allowed, the number of slices is S(n) = S(n-1) + n with S(0) = 1. (My strategy was just testing it and pattern recognision. No actual maths involved.)
@@jastejkohli5396 No, that is the correct answer. MrMatie725 is talking about it in the context of recursion formulas. MrMattie725 should have provided some examples to make his wording more clearer, but in recursion, he is referring to the previous answer, so it is not "3n-2" but the actual reiteration of the previous output in the function. For example: f(0) = 1 f(1) = 2 f(2)=4 f(3)=7 f(4)=11 f(5)=16 The pattern here: f(3) = f(2) + 3 = 4 + 3 = 7. Likewise, f(4) = f(3) + 4 = 7 + 4 = 11. Hopefully this helps to clarify what MrMattie725 meant to say. www.csun.edu/~ac53971/pump/20090922_pizza.pdf (See pages 1 to 2 here as well for more information on the pizza slice recursion formula question: "maximum number of slices with n straight cuts" kind of question.)
Mike - ya talk funny (kidding) and I am very much enjoying your channel. You cover so much that interests a wide variety of subscribers that there is no end to your content. Awesome job and thank you!
Despite of actually how much the episode is actually so nice, I loved how you used chill jazz for the most part. My suggestion actually keep trying more chill jazz for most parts and in other places of the video it’s gonna actually chill and spice the videos more. Such an amazing episode.
3:26 is pretty easy actually, All You need to do is add each line so that, no three lines go through the same "dot"/"place" and that no two lines are parallel, In this way with some general explanation you can proof ( by induction for those who want ) that when u add the "n" line, "n" new areas will be added. It is also quite clear this is the maximum amount of areas (I think at least )
for a formula lets say we mark S(n) the amout of areas/slices of pizza with n lines added as ive explained above , the formula will be S(n) = ( n^2 + n +2 )/2 , and this is what i said about proving with induction
3:26 is pretty easy actually, All You need to do is add each line so that, no three lines go through the same "dot"/"place" and that no two lines are parallel, In this way with some general explanation you can proof ( by induction for those who want ) that when u add the "n" line, "n" new areas will be added. It is also quite clear this is the maximum amount of areas (I think at least ) for a formula lets say we mark S(n) the amount of areas/slices of pizza with n lines added as ive explained above , the formula will be S(n) = ( n^2 + n +2 )/2 , and this is what i said about proving with induction
the answer of the problem 3:27 is this: I'm not originally from a English country so I can't explain well. think of it like this that the number of cuts are this; 2,4,7,11,... so the cuts are getting more by +2,+3,+4 , we don't need the additions right now but we need is that we can extend the num of the cuts, what i means is that 2=(1+1) , 4=(1+3) , 7=(1+6) and 11=(1+10) and so on. so the 1 is stable and ain't chanage at all, so what we need to do is, we need to find a formula for the 1,3,6,10 and when we find the formula then we plus it with a 1. we're gonna break or extended the 1,3,6,10 and gonna be like [1+(1)] + [1+(1+2)]+ [1+(1+2+3)] + [1+(1+2+3+4)]+.... and literary every nth term is 1 + sum of n terms, the formula to calculate sum of n terms is n(n+1)2. so as i said we gonna addition this formula with 1 so if we wanna find the num of the first cut till the n it's going to be like sth like this; The formula ➡➡➡ 1+[n(n+1)/2] 1+[1(1+1)/2]=1+1➡2 1+[2(2+1)/2]=1+3➡4 1+[3(3+1)/2]=1+6➡7 1+[4(4+1)/2]=1+10➡11 and so on .............. and the nth is ➡ANS: nth term = 1+n(n+1)2 hope that it was helpful and understandable
keep stepping up the length and keep doing more and more in depth things. shorter stuff shows how you learn, but longer content will teach us how to learn
For the cylinder question, you’d want a radius equal to the cube root of the volume divided by 2 pi and height being volume over pi * radius^2 since that’s when the derivative of the surface area is zero (showing a critical number). You could then search for the derivative on either side or find the second derivative to prove that the critical point is a relative minimum, showing that the equation is the optimal solution.
Solution to the f(x) + f(1/(1-x))=x question? I placed in the equation x = x, 1/(1-x), (x-1)/x and we get 3 equation with only "variables" f(x), f(1/(1-x)), f((x-1)/x) so we figure out f(x). The nice property that guaranteed us having 3 equations with at most 3 variables is that for g(y) = 1/(1-y), it holds that g(g(g(y))) = y
I got 20/27. I'll explain why. equation A: f(x) + 2f(1/(1-x)) = x substitute x = 3. You should arrive at f(3) + 2f(-1/2) = 3. This is equation B Next, consider the function g(x) = 1/(1-x). Find its inverse function, you should get g^-1(x) = (x-1)/x. substitute this into the very first equation, to arrive at equation C: f((x-1)/x) + 2f(x) = (x-1)/x. now, into equation C, substitute x = 3 to arrive at equation D: f(2/3) + 2f(3) = 2/3. Finally, substitute x = -1/2 into equation A to yield equation E: f(-1/2) + 2f(2/3) = -1/2 now write equations B, D, and E side by side: f(3) + 2f(-1/2) = 3 f(2/3) + 2f(3) = 2/3 f(-1/2) + 2f(2/3) = -1/2 now we can use either substitution or row elimination to solve for f(3): f(2/3) = 2/3 - 2f(3) f(-1/2) + 4/3 - 4f(3) = -1/2 f(-1/2) = 4f(3) - 11/6 f(3) + 8f(3) - 11/3 = 3 9f(3) = 20/3 therefore f(3) = 20/27
Maths is absolutely a skill that anyone can learn and be good at but like you say, it takes practice. If we don't use maths every day and therefore we don't practice maths then we forget stuff. For just 2 days practice this was good even if the interviewers were being generous. Imagine what you could have done with a week or two of practice.
I read Zoology 1989-1992. Got an upper class second degree and also picked up a boxing Blue. I did an entrance exam in Biology and Chemistry then got invited to interview at the Queen's college in December 1988. The interview was with my eventual tutor Dr Peter Miller and one of his postgraduate students. It was a mixture of gentle probing about interests then evolved quickly into analysis of specimens. Amino acid molecule models. Glycine and Alanine. Boxes of butterflies illustrating mimicry. Skulls of primates. If I knew I answered fully as I could. If I'd no idea I said so. Got home and on December 22'nd got my letter with the unconditional offer. 2 grade E's needed to get in. Amazing feeling. What an experience.
I solved the first problem by setting up a system of equations. Based on what Mike said, it sounds like he did it the same way. For question 2, I played with it and came up with a theorem: If a is at least 2 and b is at least 3, then ab>a+b. The implication of this is that we are looking for a sum of only two factors as the more we factor, the smaller the sum becomes. Lastly, a sum of two factors will look like a+840/a which is a decreasing function on the interval 0 to sqrt(840). Thus, the factor sum is maximal at a=2 so 2+420=422 is our answer.
I took maths at A-level and passed quite comfortably and I used to think I would do maths at university. But seeing this, I am confident I would’ve caved under that pressure and I’m glad I decided not to go to uni at all 😂
I may be the only one, but i always got an idea how to solve those problems almost immediately. Maybe my first semester Math at Uni helped out quite well, but if you actually try to formulate the questions into equations, they become certainly very easy.
To answer your question: • Theorem: The maximum possible number of pieces with n such cuts is exactly (n2 + n + 2)/2. We prove this theorem by induction on n. The base case says that we have (02 + 0 + 2)/2 = 1 piece with no cuts, which is true. • Suppose that we have (n2 + n + 2)/2 pieces with n cuts, and we want to make an n+1’st cut. By geometry, two straight lines meet in at most one point, so the new cut can cross each old cut at most once. This means that the new cut passes through at most n + 1 old pieces. Those pieces are divided into two and the others are not. So after n + 1 cuts we have at most (n2 + n + 2)/2 + (n + 1) = (n2 + 3n + 4)/2 = (n + 1)2 + (n + 1) + 2, at most the number we want. • We have to show that this number is always achievable. If the n + 1 cuts are in general position, meaning that every two cuts meet and that three or more cuts never meet in the same place, we achieve our bound as long as the pizza is big enough to include all the intersection points.
A great resource for the kind of problems Mike was doing is The Art of Problem Solving (AoPS) which funnily enough, was founded by a guy who graduated from the high school I’m in right now.
I'm not sure if I'm right but the pizza problem general formula is Pc is ((c)(c+1)+2)/2 (i used p cuz pizza and c cuz cuts hehe) so I'm not going to explain the whole thing but the point is to find a way to get the most intersection with the lines Edit: sorry i got overly excited, theres a part that showed you guys counting the intersections haha
Have you ever tried classic tetris?? A learn to play competitive tetris would be cool. Stopping at a maxout might be too hard so maybe 100k would be a good start and proof of concept? Super fun game and there's so much more to it than it might seem at first!
People that can do 1 on 1 tutors still amaze me to this day, the anxiety of not doing things right would be so overwhelming and just block learning for me
From what I gathered, Oxford interviews just basically down to comprehension and application of the desired subject or skill set. Not just, "do you know the answer?" Instead the real interview is, "Do you have the understanding and applicable skill set to solve this problem? Can you show that you can not only learn, but apply what you have learned, later in life?"
3:25 I can make theoretically infinite amount of slices by using 2 cuts. You just cut arbitrarily in random directions, curving the cut in all sorts of ways. Yes, you have 2 slices that are very weird shapes. But now you cut it again using the same method, and still using a random cut to go through the pizza in random directions. It’s still one cut. But now you have hundreds if not thousands of individual pieces. There is no formula to this as it’s based around chaos. 2 cuts can give you anything between 4-infinite slices
This is a university interview in the UK!? In America it's just "why do you want to come here?" and "who are your parents?" and all that stuff you said is like a job interview. It's a way for the school to get a personal profile of the applicant. And you'd better believe you have to dress and look the part. And if it's a state school, there's no interview at all!
Would love a followup video where you go through each math question in the video and explain how it's solved. Curious about the pizza slice question and the factorization of 840 :D
About the factorization of 840, the first question you may want to ask is *how many* factors you want to split the number into. For this, you can easily prove that one single factor a * b is at least as good as leaving the factors separate, which would only add a + b: a * b >= a + b | divide by b a >= a/b + 1 Because none of the factors can be 1, b is at least 2, and therefore, a >= a/2 + 1 >= a/b + 1 = a/2 + 1 | subtract a/2 a/2 >= 1 a >= 2 and this is of course also true because the factors still are at least 2. So now that we have proven that any two factors of a solution can be combined into a single factor to get a new solution that is at least as good, we know that at least one best solution will be a solution of the minimum number of factors that is allowed, which is in this case two factors. So we know that there are two factors x and y that multiply into 840, and their sum should be optimized. So you know that x * y = 840, and therefore, y = 840/x, which can be inserted into the to be optimized function f(x) = x + y = x + 840/x. So to optimize that function, you take a look at where the first derivative becomes 0, what the sign of the second derivative is at that position, and what the limit behaviour of the function is. f'(x) = 1 - 840*x^-2 f''(x) = 1680*x^-3 So because the second derivative is always positive for a positive value of x, you can already know that the value of x defined by f'(x)=0 will be a local minimum, which is not what we are looking for. What's left is the function's limit behaviour, which *would* be x=840, y=1 and x=1, y=840 if those were accepted values, but instead we'll have to look at the next smaller/larger acceptable x. Because the next x that is larger than 1 is x=2, you get the pair x=2, y=420 (and, due to the symmetric nature, x=420, y=2) which add up to 422, which is now proven to be the largest possible sum of a factorization. So the answer is that the factorization 2 * 420 = 840 gives us the largest sum, 422.
The pizza one is a recursion formula: f(n) = f(n-1) + n f(0) = 1 f(1) = 2 f(2) = 4 f(3) = 7 f(4) = 11 (if you try to draw this, one of the slices will look very, very tiny) f(5) = 16 (likewise, one of the slices will look extremely tiny) It would be nice to draw this on paper to show you, but thankfully this is also a question/answer you can look read about here: www.csun.edu/~ac53971/pump/20090922_pizza.pdf (see pages 1 and 2, recursion formula, pizza slice question). I also ended up with 2*420 = 422. My work for that one was really simple though. 840, I make "factorization trees" 840 ~ 2*420 ~ 2*210 ~2*105... and so forth (I used 2 here intentionally, but I simplfy and clarify below, e.g., just multply the 2's together respectively.) 2*420 = 422 4*210 = 214 8*105= 117 If I keep on going with this tree, the sum keeps on getting smaller, and it seems that 422 (from 2*420) is the largest one so far. Thus, 422 seems like the factorization of 840 that will provide the largest sum. You can also use KappaScopeZZ's (the TH-camr who also commented here) method of the derivatives to double-check some things, but that is not really necessary per se (you would usually use that in the context of a function, and/or a financial-related mathematics question from what I remember). The big picture is most important, as the interviewers mentioned.
These people are so nice. Imagine a grown man about half as smart as you trying really hard to do something you think is simple, say driving a car, failing, feeling really confident and asking "So, how did I do?". You'd like to think you'd stay cool.
:) the Oliver guy is a family member of mine, I didn't know he was in this video and just happened to be watching it and got the shock of my life when he appeared
Which factorization of 840 gives us the largest sum? I'd say the answer is 2*420 so 422 in a sum. assuming i did not misunderstand the question. for any 2 integers if one of them is at least 2 and the other at least 3 then the product of those will be greater than the sum. 2*2 = 2+2 but 2*3 > 2+3 and so on. since using 1 is not allowed and it has to be at least 2 pieces 242 is the obvious answer. for a number like 81 which is not divisible by 2 the largest sum would be 27*3 so 30 if you could include 2 there it would be 2*40,5 so 42,5 which is bigger than 30.
btw I was so interested in the pizza problem that I solved it on paint XD took me like 10 minutes hahaha (I haven't watched the entire video yet, idk if he solves it later, the problem caught my interest and I fired up paint and started drawing) the solution is P(C) = 1/2(C^2+C+2), where C is the number of cuts and P is the maximum number of pieces you get out of the pizza this is a Difference equation where every cut you add, adds how many cuts you have to the number of pieces u had in the previous cut, where the starting position is P(0) = 1 because if you do 0 cuts or never cut the pizza, the entire pizza can be seen as one large piece.
hey mike! love your vids, pretty consistent viewer for the past several years just a idea for a future challenge for yourself, how about learn how to competitively speedrun a video game you already like, there're a lot of really awesome heavily contested speedruns these days that don't get enough love in my opinion. just a idea, love the vids XD
The answer to the pizza problem is 1+(n(n+1))/2. Next, f(3) = 20/27. For the last problem, if we're talking about positive integers, then 2 and 420 for the factorization.
I think I have the answers to the second interview questions. The first question's answer is f(3) = - 2/27 and I got it from changing the equation to f(x) = x - 2f(1/(1-x)) and then tracing through the problem. I ended up with f(3) = 3 - 2[-0.5-2((2/3)-2f(3))] which simplifies to f(3) = 20/3 - 8f(3). Then I added 8f(3) to both sides and divided by 9 to get f(3) = 20/27. The second question's answer is the factors 2 and 420, because 2 is the smallest factor after 1, so then the other factor is the largest product of all the other factors. (Question 1 took me 17 minutes and question 2 took me 6 minutes. I did not use any outside help - no internet, no other people, just some scratch paper. I am currently in high school and am taking CalcBC, but I did not need any calculus to solve these problems.) Please reply if I made a mistake somewhere or if you have any other comments.
I was looking to see if anyone else posted the answer here. I also got 20/27 on the first problem, so that gives me some confidence :) I basically tried x = 3 abd x = 2/3 first. Then I went for x = -1/2 where I found f(2/3) again and then subsited it to have an equation with only one f(3).
Okay, I'd really like someone to check this for me. Would a generalizable formula for the pizza problem be this? p = (n^2+n)/2 + 1 P is the number of pieces. N is the number of slices. The (n^2+n)/2 is just a fancy way to calculate the additive factorial (that is, adding 1 + 2 + 3 ... etc., up to the number you're calculating; there's probably a fancy math name for this that I don't know). The +1 is needed because, from the outset, the pizza is one slice (at the 0th cut). Is this ... right? It works for one through four slices, I'm pretty sure, but I'm not 100% that it's universal.
The answer is 0.5(n^2+n+2) To maximise the number of slices you want to maximise the number of vertices created by the cuts, which means you really just have to avoid crossing several cuts in one spot. When you add a cut to the pizza you add a new slice for every slice you cut through. Assuming you don't cut through an already existent vertice you go through a different slice every time your cut passes another cut. And you go through as many cuts as possible, in order to maximize the number of slices. Well, the maximum number of cuts you can go through is the number of cuts you've already done, so the number of slices you add is n+1, with n being the number of cuts already on the pizza. So, the s number of slices for n+1 cuts is given by s[n+1]=s[n]+n+, with s[0]=1 (since a pizza with no cuts is a single very large slice). After this point the way I did it is probably not the best, but it still works: The number of slices added by each new cut is linear (n+1), which means the number of slices over the number of cuts is quadratic. From then you can just pick 3 points of the sequence and create a system of equations with s[n]=an^2+bn+c and solve it for a, b, and c.
you start with a function F where F(number of cuts)=number of slices F(0)=1 the whole pizza F(1)= 2 two slices of pizza F(2)= 4 4 slices of pizza F(3)= 7 7 slices of pizza From this we can see that our function is equal to the one previous plus our number! so 7 = 4(the previous answer) + 3(our current number) In maths we evaluate this as F(x)=x+f(x-1) expanding this F(x)=x+(x-1)+(x-2)+(x-3)...+1 -> this is where i got confused and had to look up some things. this is the arithmetic progression series which is basically just the sum of all numbers up to your number you could re-write the above as 1+2+3+....+x There is a formula for this which is f(x)=1+(x(x+1)/2) the arithmetic progression formula is fairly common and I remember learning it but I am not sure i could derive it on the spot. Im sure if i had just learned it or got some hints id be able to figure it out.
so for the pizza question I tried to make a general formula : 2n+ (n-1)/2 .[ applicable for odd no. of cuts , But for even no. of cut we have to substract 0.5 from our answer] for example if we want maximuim no. of pizza from 3 cut that is 7 , we have to put n=3 so the answer is 7 , if we want max no of piece from 4 cut that is 9 so we have to put n=4 thus answer will be 9.5 so substract 0.5 and so on. We can get max 14 piece from 6 cuts . ------Thank you,😊😊😊😊😊😊😁😁😁😁
I think the answer to the pizza question is (n(n+1)(n+2)/6)+1? Using the sum of the triangular numbers and adding one to make it work from the base case as it's 1 when n = 0. If I'm wrong please let me know because it seems like a super cool question!
I'm guessing it is more the sum of whole number + 1 because the way to maximise the nulber of slices is by cutting every line in each new cut, and when you cut a line a new pizza slice is "created". So when you have already n-1 cuts you create n-1 slices by cutting the cuts +1 because you started in an existing slice you cut. I hope it is clear I tried my best ; )
Coming soon: Mike graduates from Oxford… ?
Episode 18: Mike goes for OLIVER'S JOB?!
@@MikeBoyd XD
@🕊️ Peace 🕊️ hi
@@MikeBoyd 2032: Mike is an Oxford professor?
Yeah man Mike about to be like “Episode 43: Learning how to complete my Economics 203 homework in 28 minutes with help from twitch!”
Mike: "I think it went very well"
Tom: "It's very obvious that Mike hasn't done any maths for a while"
Harvard next
it was sarcasm
This came out worse than intended.
@@Xavier_Renegade_Angel he did
My brother got asked the distance to the moon in his Oxford University interview to study Physics. He said he didn't know, and they told him, "work it out".
From his time in air cadets, he knew that aeroplanes travel around 550 mph. The only other fact he knew was that the moon was roughly far enough away from Earth to fit 30 Earths inside the space between. So he applied the information he had. A flight to the opposite side of the World took roughly 24 hours. 24 hours x 550 mph = 13,200 miles. Double it for the circumference of the Earth = 26,400 miles. To turn a circumference into a diameter, divide by pi. I think he rounded down to 3.1 or something to make it easier, but then he times that diameter of Earth by 30 for an answer that was only a few thousand miles outside of the moon's maximum distance from Earth.
The interviewers loved him. He didn't just know the answer. He knew how to *apply* information that he already had to be able to *work out* the answer.
He got his place at Oxford.
That is actually such an amazing story, I absolutely adore that
I really like this question
there are so many ways to approximately solve it with close to no information
Definitely happened
I tried to do this myself using a different method, failed :), props to your family
There's a lot of different variations on questions like this. It's called a Fermi approximation, the idea is because your errors are randomly over/under you'll probably end up in the right ballpark
This series on Oxford is incredibly humbling.
It's incredible to think college grads are disciplined enough and have the drive to succeed in these admissions.
That sort of discipline only really kicked in for me during my masters (a bit too late lol).
Kudos to you guys
Yeah I had no idea what I was doing when I applied to University. The fact that people are this disciplined and intelligent at 18 is crazy to me.
Tbf, I was lucky that I was obsessed with what I wanted to do early on. Definitely had my hyperfixation play a part in that
This is whyi love Oxford 😌💯humble people everywhere 😌💁♀️💯
So interesting seeing you learn these type of “subjective/soft” skills where the success of learning it isn’t as clear - they’re actually a lot more relevant in day to day life
Yeah, for example it hadn't occurred to me before that getting "unstuck" while solving a problem is a skill that you could improve. And now that I'm aware of it it's obvious how useful such a skill would be. I'm already thinking about how it could work in computer programming.
@@puupipo Yes! I had the same thought and completely agree.
@@puupipo same I’m in my first year in computer engineering and the hardest part for me in programming is to order your thoughts so the computer can understand it and when you get stuck on something how can you solve it.
@@Rogue_Route For the first problem you mentioned I've found it really useful to always write down the whole "skeleton" of your program in plain English before writing any code. It's a lot easier for your mind to handle when you're not trying to design the algorithm AND translate it into code simultaneously. For the second bit, taking a break to clear your mind can help. There's also something called "rubber duck debugging", give that a google if you're unfamiliar with the concept.
@@puupipo thanks for the advice never thought that writing the program in English first will help am going to try that before my exam in a few days in c++
My Oxford interview wasn’t scary at all, it was essentially a conversation about things related to the course I was applying to (in my case biology), I didn’t have a single question that wasn’t about biology. The academics are just looking for people who are passionate and interested in the course they’re applying to and can think creatively :)
Interesting. In the US you’re not applying to a course. You take a major, but you don’t have to immediately and half of your classes will be general education across a myriad of subjects both required and elective.
@@SamAronow yeah it's usually 4 years in the US and 3 years in the UK. Maybe that's why.
In the US, they just want people who will give them money :)
@@WaterYaDune Mate, the US System of being able to do whatever you want once you are in the University, from being able to change Major, to doing a Minor on something completely unrelated, or even double majoring is a much, much better system than the current European system.
@@joaomiguelgoncalvesdematos6135 Agreed
I come from a humanities not a mathematics background (at Oxford), but I wanted to add:
1) the fact that they only focus on the subject - no questions about you or your aspirations - is very representative and imo a good thing
2) my interviews were really fun. The first two mins are always nervey, but I really got into them. They want someone who sits down and digs into the questions and gets really stuck in - where you really want to figure it out because you care about the problem, not just impressing the interviewers
Can I ask what humanities subject did you interview for? I would really love to apply to Oxford for Philosophy and Modern Languages and I am a little unsure about what questions they would ask..
If only real job interviews were like that
I am INCREDIBLY curious to see what another discipline looks like in this process. Math is honestly something I've never understood, to the point that some of these questions were gibberish to begin with! I wonder what "extremely high level literature" (or philosophy, or art) looks like...
THIS please make a video on it!
I do philosophy, politics, & economics at Oxford, & my philosophy interview was super precise, like if I said a single word which didn't quite fit with the rest of the sentence the interviewer would ask why I'd said it. At one point he asked me the same question 3x in a row (as in, once I'd answered he just asked the exact same question again) while just staring at me lol, was very stressful.
The actual subject of the interview was just various ethical scenarios which he wanted me to reason through, so a pretty specific take on philosophy. Not something you could prepare for really
@@georgebarber3280 Omg PPE at Oxford. You're probs sat next to the next 3 PMs lol (Or I'll be able to say I replied to George Barber's comment on TH-cam once! lol)
So I had my english interview for Oxford a couple of months ago and they gave me a passage and 15 minutes to prep. They then just said talk about what you find interesting so I suppose the hardest bit was knowing what to talk about.
@@georgebarber3280Is it true you do no work in Ppe?
Well done! It was great to see someone else go through this… I applied for physics this year - I got through the entrance exam but was unsuccessful after interview!
I’m in the same boat for the same subject. The interviews were interesting but very intense and focused
I had to do 4 interviews for Maths and Compsci at Oxford. The pressure was INTENSE. My MAT score was very mediocre (despite a lot of preparation) so I can offer a perspective of someone who isn't as gifted as others.
Many times I would just be staring at the question, not knowing where to start. Although I recognised the content and notation, it was put in way I had never seen before. I made quite a bit of progress on some questions, but I was guided by the interviewers a lot more than I liked.
I can see that maximising your interview performance is a skill, but there is a point where, if you're just not smart enough (and I mean intelligence, not through lots of learning), you'll never be able to cope. Despite my passion for the subject, I didn't get an Offer unfortunately; competition is extremely fierce for the course.
Don't feel bad if you don't get into Oxford or any other unis. It's probably not the uni for you, and you can always get the job of your dreams by working hard.
While there might be people that just don't have the intelligence due to some genetic or developmental issue, the vast, vast majority of people are able to achieve similar levels of inteligence through work.
That said not all work is equal, if you only train yourself at solving a problem in a particular way then you're not working on the skill of abstracting the solutions.
I'd tend to somewhat second Spliter, above. I think that it's rarely a case of intelligence with these things, but more one of motivation. And I don't mean this in a 'oh, you were unmotivated, you should have just tried harder' kind of a way. I mean this as in: 'there are lots of different ways to handle maths, some of them suited to interviews, some not', and unless someone possesses god-tier motivational abilities, I think most people will tend towards working in the way which most suits them.
So, in my case, I used to love teaching physics (slight detour from maths I know, but I've never been interviewed for maths, and figure the idea I'm trying to convey converts quite nicely), and so I associate vocalising my thought processes on the spot with being a rewarding experience, which meant that any prep I did would tend to use this skill, which meant I trained it more, which meant I would use it more. And this is even in sitting exams and in doing exam-prep - my tendency would be to try and handle a problem in a way which I could most easily explain/justify, which often wasted tonnes of time (in fact, this is a huge problem in the amount of time I spend on uni work to this day - multiple pages of explanation/thinking sometimes go into a 2 line algebraic solution).
In other words, when it came to uni applications, I'd tend to do well in interviews, and far less well in my exams (managing to muck up my physics exam - by far my strongest subject - fairly badly).
Point being, I _do_ think it's possible to train this, but I think it's a question of whether you find the most useful skills to solve the problems the most rewarding, and hence on how you tend to approach problems. I suspect, like me, you worked really hard, developed really useful skills, and whilst you will have developed your skills for interviews and the MAT, a tonne, you might have learnt _more_ in regions which were still practiced by MAT/interview prep, but which weren't as vital.
And I think, most of all, my point with this is that you _still_ put in a tonne of work, you _still_ will have gotten a lot out of it. Whether or not it's the same stuff you need for the interview, honestly, who cares? I've got multiple friends at Oxbridge, and life there seems to value similar things as found in the entrance exams. If something else was the most rewarding bit of maths to you, then, you know what, the skills which you built up the most, but which didn't help the most, probably would've been neglected by the Oxford system, meaning you'd've had to put in even more work to make up the difference between the skills you found most rewarding, and the ones they wanted (this, again, from the experiences of a few friends).
In short, huge congratulations for putting in all the prep - you probably developed lots of skills which you found more rewarding, and which Oxford wouldn't have nourished. So don't put yourself down, as 'not having been smart enough to learn what they wanted, despite all your work' - I think it's a huge fallacy to assume that repeating a particular action (even something such as diverse problem solving and abstract thinking) will produce a given particular result - it'll always vary person to person, and _can't_ be catered for by every university. Honestly, I remember finding past-papers and stuff to be a _huge_ drag, so congratulations on whatever prep you did - that in and of itself is worthy of commendation!
You sound like you have a passion, so best of luck, I'm sure it will (have) take(n) you far!
(Oh, and a slight addendum, RE the final part of Spliter's message - I often get the feeling from people, and often honestly perpetuate this myself, that learning by rote, or getting _really_ good at a particularly simple process, isn't very useful if you can't abstract it. That's not necessarily true: Often, clever algorithms and simplifications are only found by people who have just spent a lot of time working with a problem, and its processes.)
@@spliter88 I think it doesn't really matter how much you practice or study, if you can't take your knowledge of one thing and see a similarity in another to then apply it to that situation it's going to be tough.
The number of smart people at my school that would get a question like "equally share 10 sweets between 2 people" and not be able to do it because the question wasn't "divide 10 by 2" was crazy (oversimplified example to get the point across). They were smart because they could just remember things and recite them, but couldn't actually apply that knowledge if it wasn't presented in the exact way they learned it.
And throughout uni, doing computing, we would have 3 tutorials that taught us X-Y-Z, and then we'd be set to go do a task that required Z-X-X-Y, half the class would complain because "we were never taught how to do Z-X-X-Y"
@@scottg3192 My point was that knowing how to apply skills you've learned in one area and apply them to something else is itself a skill you can learn.
You learn to abstract, it's not something that you're born with.
Like, literally the tests they do on toddlers show that below a certain age they're just unable to abstract at all, and they learn that over their childhood.
I don't think that it stops either, there's more and more complex abstractions we can learn as we age, a skill we can train.
@@scottg3192 It's nature vs nurture. Only one can get you so far. But I still think it is possible to learn it. Like with maths, i was never good at abstracting one problem's formula to another. But I have nowhere near that kind of problem with computing. But in theory it's the same, you learn one thing and apply it to the other.
But some people struggle at computing but not maths problems.
I really love the attitude you have towards these "non-physical" type skills. I tutored math in high school and college and you would be the actual dream pupil. I had so many people that gave up before they even started or assumed that they were incapable of learning something and that was really difficult to work through. You have a great mix of respect for what you don't know (and for those who do know it!) balanced with the understanding that just like anything else you can improve with practice 💯
SOLUTIONS:
If you are interested in the solutions of the 4 questions in this video here are my solutions but I cannot guarantee that everything is correct:
Tin question: How would you choose the radius r in a tin (cylinder) with volume 1 to minimize the surface area of the tin?
Solution:
Since the tin has a volume of 1 and the volume is base area (circle) multiplied with the height h we can conclude that:
V=πr^2*h=1 and thus h=1/(πr^2)
The surface are consist of the top and bottom part each having the area of a circle and the cylindrical wall which is a rectangular with the circumference of the circle(2πr) and the height as length of the two sides. Thus we know:
A=2*πr^2+2πr*1/(πr^2)=2πr^2+2/r
To get the minimum we will differentiate:
A'=4πr-2r^(-2)
Minimum is where A'=0:
4πr-2r^(-2)=0
4πr=2r^(-2)
r^3=2/(4π)=1/(2π)=(2π)^(-1)
r=(2π)^(-1/3)=~0.54
Thus r=~0.54 and h=~1.1
f(x)+2f(1/1-x)=x:
Insert 3 as x:
f(3)+2f(-1/2)=3
Insert -1/2 as x:
f(-1/2)+2f(2/3)=-1/2
Insert 2/3 as x:
f(2/3)+2f(3)=2/3
Thus with substitution we know:
f(3)=3-2f(-1/2)=3-2(-1/2-2f(2/3)=3+1+4f(2/3)=3+1+4(2/3-2f(3))=3+1+8/3-8f(3)
9f(3)=3+1+8/3=20/3 -> f(3)=20/27
Factorization of 840 with the largest sum (I assume only integers) so 840=a*b and a+b is maximized:
Obviously to factorize a number in many parts will lead to a smaller sum therefore a factorization in two products will have the largest sum.
To prove this we try to show with an inequality:
a*b a*b-a a*(b-1) 422
210*4 -> 214
...
42*20 -> 62
30*28 -> 58
Thus we can see that the factorization with the largest difference to each other but where still both products are an integer larger than 2 with the largest sum is 840=420*2 -> 422.
Pizza slices: For k straight cuts through a pizza how many pieces n can you get?
A single cut can always split at least the half of the pieces that are existing thus:
n -> n+n/2
With a smart cut we can split even one piece more thus:
n -> n+n/2+1
For cases where n is odd n/2 is being rounded down.
We can write this up to k=4 as:
k=0 -> n=0+0+1=1
k=1 -> n=1+0+1=2
k=2 -> n=2+1+1=4
k=3 -> n=4+2+1=7
k=4 -> n=7+3+1=11
We can write this always as arithmetic series added with 1, e.g.:
k=4 -> n=(1+2+3+4)+1
n=(sum from 1 to k) +1
Thus we can write this arithmetic series as:
n=1/2*k*(k+1)+1
With some simplifications:
n=1/2(k^2+k+2)
Hope this helped someone.
Edit: Thanks for the comments guys I tried to implement them as far as I could.
For the third one I would just do a*b < a+b
Which implies a*(b-1)< b
a
this is exactly what I was looking for...thanks!
wait...there's an error here-
Insert 2/3 as x:
f(2/3)+2f(3)=3/2
shouldn't it be 2/3 rather than 3/2? (the =x bit)
F(3) is equal to 20/27
@@thecrafters6220 that is what I got
From the PhD interview perspective- I had two for two slightly different projects. 1 went terrible- I had to do maths on a whiteboard in front of 6 academics and, surprise surprise, didn't get the place. The second was amazing- I clicked with the professor and we just chatted for an hour about their research area(s). Got that one :) I think it really is luck of the draw
This is giving me flashbacks to the oral exams I did a few times at uni - they were usually done as resits for failed exams, but they were 100 times more stressful than the exam itself even though the lecturer would sometimes help you through it. Ironically the only admissions interview I did wasn't stressful at all because I had no idea it was coming. I was there for the applicant day and literally got dragged into an office with two lecturers who grilled me for 30 minutes on the course, but because I didn't actually realise it was an admission interview until it was over I was totally calm through the whole thing, answered every question pretty casually, and got the acceptance letter the next week. I think there's something to be said for being relaxed because I've done far worse in easier interviews I've aggressively prepared for.
Is tat what usually happens? They just randomly bring you in for an admissions interview? Or was it planned but you simply didn’t realise it?
Yes relaxation is key in not fumbling.
Just watching the interview and knowing that Mike was under a lot of pressure, stress, and anxiety, honestly has given me such ptsd of highschool and the overwhelmingness of having a 1 on 1 tutor and/or tutor
Just wanted to say, as a current Maths undergrad at Oxford, these videos are really cool and aswell as being entertaining are an amazing tool for people wanting to apply!
And also pretty nuts to see Adele my pure maths tutor from first year! Friendliest tutor ever :)
have you learnt differnetial geometry yet? if you havent, you should take a course on it! it's so beautiful!
Never expected there to be a part 2 of this! I absolutely love mathematics, but sadly found out too late. I'm following an Art history Research Master's instead.
It’s never too late
@@lol360noscope6 Very true! That's why I focus on digital art history. There are some interdisciplinary opportunities and projects that combine computer science, mathematics and art history, so that's the direction I'm trying to take. I'll probably never be the main mathematician in the group, but I'll take what I can get, haha.
Awesome video! I would've loved to see more of the process of "absorbing new knowledge and being able to use it quickly". Keep up the awesome work!
at every problem just pauzing the video and thinking for a little while, then watching the process of mike working thrue it in a different, or the same way, is a nice experience.
I got an offer from Cambridge this year and I can verify that the interview is the most stressful thing I have ever done
Went through the NatSciBio application process myself and I agree, it is stressful.
Luckily, it gets less stressful each time.
it's way less stressful than being a student there :/
@@hunterG60k Well yea, its because of the good old stress from exams and tests. For me in terms of stress level it has been: exams > interview > tests
@@spiralingsphere3785 I think it depends on what you're studying, in medicine there's a lot of pressure from all directions. They don't tell people about the suicides that happen every year.
@@hunterG60k are you a cam student?
People prepare for weeks for this interview, and you did borderline in 2 days of practice. You're phenomenal, i believe if you prepare for a week or 2, you'll get into the university.
Keep in mind he already has a degree in Engineering and work experience.
@@shrifrai1634 oh so a week of studying will give him a proper shot at the interview. And i believe people prepare for 2 or 3 months for this test
I really appreciate your positive attitude about learning mathematical skills.
Loved this Mike! You're so right that the Oxford Admissions Process is a 'skill' you can improve at - I got rejected the first time, but honed the 'skill' over my gap year and got accepted second time round! Great video :)
Can't believe you watch Mike too!!! Crazy crossover 😂
I hope you’re feeling alright at Oxford mate :)
@@karunk7050 Haha thanks 😅 guess u mean from the video
I had an interview at Cambridge for natsci (Physics and and chemistry) and whilst I was unsuccessful and it was quite stressful at times, it was an incredible experience to have.
I think those question were easy.
1. Just fill in 3 in equation, after which you need to find f(-1/2), so you just fill in -1/2 after which you need to find f(2/3) after, so you just fill in 2/3 and you'll need to find f(3). Then you fill in everything backwards, and you get an equation for f(3). So f(3) = 20/27 (if I didn't miscalculate)
2. The second question you get maximum n EXTRA slices on the n-th cut. So you get n(n+1)/2+1 slices on the n-th cut. (the sum of the n first natural numbers is n(n+1)/2)
3. 420*2 sum 422 It's obvious. 840= 2*2*2*3*5*7 Just write out all possible factorizations and their sum. Or by proof: if one factor is a, the other is 840/a. Their sum is a+840/a. This function has a minimum at root(840) and rises after that. The highest possible integer value for a is thus 420.
You should for sure do a similar format but for a software engineering interview (technical coding questions, systems design etc.)
I studied compsci here in the US and I would find that very interesting too.
So for the pizza problem my reflection is : we want to maximize the number of slices, so each cut should bisect the most amount of slices.
Which means we want to have the maximum number of intersections within the circle.
so the maximum number of intersections that is possible is equal to the number of lines, as you can only cross 1 or 0 times (parallel or crossing outside) now for the number of slices crossed, it is equal to 1+the number of intersections, which is the number of lines
We split that in half, so we add that number to the previous number and we start with a single slice, so we end up with 1,2(1+1),4(2+2), 7(4+3), 11(7+4), 16 (11+5) etc...
For a general formula for the number of slices after the nth cut
We have the 1 at the start, so 1+???
And the rest is the addition equivalent of a factorial, i.e. triangle number, which is (n²+n)/2
So we end up with 1+(n²+n)/2 slices
For the cylinder, we just take the area of both the tube and the ends, then take the formula for the area, and then move it around so that we have the formula for the two surfaces at a constant volume, and then we solve h and r for min(area)
Another solution is to binary search it, which is probably faster, make two cylinders, one really tall and one really wide, look wich one has the smaller area, make one in the middle and look there, repeat a couple times and you probably will have an answer that's good enough.
I think quick solution that may work in some cases is to look at the formulas and see if h or r is squared, if one is and the other is not that's a pretty good indication that you want to minimize the squared one (though maybe not as it's also affected by the area)
idk if this is mentioned further in the video, but the answer to the pizza problem(i think) is 1+(n²+n)/2, as with every crossing there is a new area created, and there are n-1 lines to cross if you don't go through existing intersections, and as exiting the pizza is also creating an extra area, this means the nth line adds n new areas at maximum, thus the amount of areas at any point can be written as 1+1+2+3+...+n-1+n, if you take that last part and add it to itself you can, rearange the terms to (1+n)+(2+n-1)+...+(n-1+2)+(n+1), as every term will then be n+1 and there are n terms the adding of this to itself (aka multiplying by 2) is equal to n(n+1), and thus the formula is 1+((n+1)n/2).
Is in not simply 1 cut 2 slices, 2 cuts 4 slices thus n cuts 2n slices?
@@DavidHughesArchitect if wvery cut is concurrent yes, but non of the cuts are concurrent so if you have >2 cuts then there is more to consider than just the outer areas
I love the mathematicians' calm intonation and way of speaking. Professor Riordan's voice reminds me of Grant Sanderson's videos
I love your learning videos. And that you teach us how to do it too. 👍
This was actually one of the best vids u have ever done.
Wow, was not expecting a return of the Oxford series!
14:51 Best line of the video:
“Borderline… so you’re saying there’s a chance.”
I burst out laughing when he said that, always the optimist. I'll bet he would be fine if he made maths a hobby / passion like he's made learning random physical / mental tasks on the show all these years. Like playing chess a few hours a week, if he simply made maths his thing a few hours a week, he would get in next time.
First heard this in 'Dumb and Dumber': th-cam.com/video/-9IgLueodZA/w-d-xo.html
That million dollar puzzle solve seems like something Tom Scott would make a video about.
I love how positive he was even at the end
not necessarily related, but I had my oxford law interview about a week ago - the pressure was crushing and although i'm typically a quite calm and confident person under pressure, it was a completely different story. It's hard to watch things like this and not compare myself to them, however for anyone in the same boat as me just remember that everything looks worse than it was in the post-morten; you look for what you did badly because you can more easily understand what the interviewers arent looking for, its much harder to appreciate what went well when you have little idea as to what they are looking for.
To all the applicants who have made it similarly far, good luck!
did you have to do yours online as well? it can make it so much more nervewracking when youre not even there in person!
@@joeyhardin5903 Yeah mine was online, I was sorta hyping myself up beforehand and mid-sentence the panel joined the meeting so that didnt help - in person would have been much more enjoyable!
I had my cambridge law interview mid december and i had done quite a few mocks but the real thing was so much scarier so you’re defo not alone :) best of luck for your application!
I'm currently an first year law student at Oxford and those interviews were hard. Caught myself having to think because I genuinely had no idea how to answer at first
Like you said, you usually think you did worse than you actually did. Especially for law, it's not about knowing the law for the interviews but more your logic towards arriving to your conclusions
@@brianj05 did you get an offer?
That "so you're saying there's a chance" got a much needed laugh out of me late at night rn, thank ya mike
ive cried on a math exam before. under the pressure of these interviews, i’m almost certain i would burst into tears after awhile 😂
I think you can create n extra slices per cut. One extra slice after the first cut, two extra after the second cut, ...
So if a recursive answer is allowed, the number of slices is S(n) = S(n-1) + n with S(0) = 1.
(My strategy was just testing it and pattern recognision. No actual maths involved.)
nah i dont know if that works because with 3 cuts you get 7 slices of pizza, i got slices (s) = 3n-2 but i dont know if thats correct
@@jastejkohli5396 No, that is the correct answer. MrMatie725 is talking about it in the context of recursion formulas. MrMattie725 should have provided some examples to make his wording more clearer, but in recursion, he is referring to the previous answer, so it is not "3n-2" but the actual reiteration of the previous output in the function.
For example:
f(0) = 1
f(1) = 2
f(2)=4
f(3)=7
f(4)=11
f(5)=16
The pattern here: f(3) = f(2) + 3 = 4 + 3 = 7.
Likewise, f(4) = f(3) + 4 = 7 + 4 = 11.
Hopefully this helps to clarify what MrMattie725 meant to say.
www.csun.edu/~ac53971/pump/20090922_pizza.pdf
(See pages 1 to 2 here as well for more information on the pizza slice recursion formula question: "maximum number of slices with n straight cuts" kind of question.)
2^n-1 + n
Mike - ya talk funny (kidding) and I am very much enjoying your channel. You cover so much that interests a wide variety of subscribers that there is no end to your content. Awesome job and thank you!
Despite of actually how much the episode is actually so nice, I loved how you used chill jazz for the most part.
My suggestion actually keep trying more chill jazz for most parts and in other places of the video it’s gonna actually chill and spice the videos more.
Such an amazing episode.
I love your vids and I respect your work you never fail to impress me 👍❤️
th-cam.com/video/0_S0TN5-15w/w-d-xo.html
Finally it's here
3:26 is pretty easy actually, All You need to do is add each line so that, no three lines go through the same "dot"/"place" and that no two lines are parallel, In this way with some general explanation you can proof ( by induction for those who want ) that when u add the "n" line, "n" new areas will be added. It is also quite clear this is the maximum amount of areas (I think at least )
for a formula lets say we mark S(n) the amout of areas/slices of pizza with n lines added as ive explained above , the formula will be S(n) = ( n^2 + n +2 )/2 , and this is what i said about proving with induction
3:26
is pretty easy actually, All You need to do is add each line so that, no three lines go through the same "dot"/"place" and that no two lines are parallel, In this way with some general
explanation you can proof ( by induction for those who want ) that when u add the "n" line, "n" new areas will be added. It is also quite clear this is the maximum amount of areas (I think at least )
for a formula lets say we mark S(n) the amount of areas/slices of pizza with n lines added as ive explained above , the formula will be S(n) = ( n^2 + n +2 )/2 , and this is what i said about proving with induction
the answer of the problem 3:27 is this: I'm not originally from a English country so I can't explain well. think of it like this that the number of cuts are this; 2,4,7,11,... so the cuts are getting more by +2,+3,+4 , we don't need the additions right now but we need is that we can extend the num of the cuts, what i means is that 2=(1+1) , 4=(1+3) , 7=(1+6) and 11=(1+10) and so on.
so the 1 is stable and ain't chanage at all, so what we need to do is, we need to find a formula for the 1,3,6,10 and when we find the formula then we plus it with a 1. we're gonna break or extended the 1,3,6,10 and gonna be like [1+(1)] + [1+(1+2)]+ [1+(1+2+3)] + [1+(1+2+3+4)]+.... and literary every nth term is 1 + sum of n terms, the formula to calculate sum of n terms is n(n+1)2. so as i said we gonna addition this formula with 1 so if we wanna find the num of the first cut till the n it's going to be like sth like this;
The formula ➡➡➡ 1+[n(n+1)/2]
1+[1(1+1)/2]=1+1➡2
1+[2(2+1)/2]=1+3➡4
1+[3(3+1)/2]=1+6➡7
1+[4(4+1)/2]=1+10➡11
and so on ..............
and the nth is ➡ANS: nth term = 1+n(n+1)2
hope that it was helpful and understandable
Love your dedication
keep stepping up the length and keep doing more and more in depth things. shorter stuff shows how you learn, but longer content will teach us how to learn
Great episode! Also, I am very much looking forward to that kendama video ;D
Thanks for the sponsor segment, actually got me interested and subscribed
Imagine this is all an elaborate plan for him to get into Oxford.
Lol
I actually love this type of content! So interesting, so insightful, please keep it coming!!!
For the cylinder question, you’d want a radius equal to the cube root of the volume divided by 2 pi and height being volume over pi * radius^2 since that’s when the derivative of the surface area is zero (showing a critical number). You could then search for the derivative on either side or find the second derivative to prove that the critical point is a relative minimum, showing that the equation is the optimal solution.
Watched episode 1 like 3 times because it was really insightful, been waiting for this episode
Great work 🥳🥳🥳 Thank you 💜💜💜
Solution to the f(x) + f(1/(1-x))=x question?
I placed in the equation x = x, 1/(1-x), (x-1)/x and we get 3 equation with only "variables" f(x), f(1/(1-x)), f((x-1)/x) so we figure out f(x).
The nice property that guaranteed us having 3 equations with at most 3 variables is that for g(y) = 1/(1-y), it holds that g(g(g(y))) = y
I got 20/27. I'll explain why. equation A: f(x) + 2f(1/(1-x)) = x
substitute x = 3. You should arrive at
f(3) + 2f(-1/2) = 3. This is equation B
Next, consider the function g(x) = 1/(1-x). Find its inverse function, you should get g^-1(x) = (x-1)/x. substitute this into the very first equation, to arrive at equation C:
f((x-1)/x) + 2f(x) = (x-1)/x.
now, into equation C, substitute x = 3 to arrive at equation D:
f(2/3) + 2f(3) = 2/3.
Finally, substitute x = -1/2 into equation A to yield equation E:
f(-1/2) + 2f(2/3) = -1/2
now write equations B, D, and E side by side:
f(3) + 2f(-1/2) = 3
f(2/3) + 2f(3) = 2/3
f(-1/2) + 2f(2/3) = -1/2
now we can use either substitution or row elimination to solve for f(3):
f(2/3) = 2/3 - 2f(3)
f(-1/2) + 4/3 - 4f(3) = -1/2
f(-1/2) = 4f(3) - 11/6
f(3) + 8f(3) - 11/3 = 3
9f(3) = 20/3
therefore f(3) = 20/27
Maths is absolutely a skill that anyone can learn and be good at but like you say, it takes practice. If we don't use maths every day and therefore we don't practice maths then we forget stuff. For just 2 days practice this was good even if the interviewers were being generous. Imagine what you could have done with a week or two of practice.
Great work 🥳🥳🥳 Thank youuu 💜💜💜
OMG keep this content is so good!!
I read Zoology 1989-1992. Got an upper class second degree and also picked up a boxing Blue. I did an entrance exam in Biology and Chemistry then got invited to interview at the Queen's college in December 1988. The interview was with my eventual tutor Dr Peter Miller and one of his postgraduate students. It was a mixture of gentle probing about interests then evolved quickly into analysis of specimens. Amino acid molecule models. Glycine and Alanine. Boxes of butterflies illustrating mimicry. Skulls of primates. If I knew I answered fully as I could. If I'd no idea I said so. Got home and on December 22'nd got my letter with the unconditional offer. 2 grade E's needed to get in. Amazing feeling. What an experience.
Hey I personally believe that you got in I just had that constant feeling for no reason but I was kinda doubting it but you got in!
Tom rocks maths plays football for the college too, I'm gutted to have missed you Mike even tho I go to Teddy
Love all your vids, keep it up!
I solved the first problem by setting up a system of equations. Based on what Mike said, it sounds like he did it the same way. For question 2, I played with it and came up with a theorem: If a is at least 2 and b is at least 3, then ab>a+b. The implication of this is that we are looking for a sum of only two factors as the more we factor, the smaller the sum becomes. Lastly, a sum of two factors will look like a+840/a which is a decreasing function on the interval 0 to sqrt(840). Thus, the factor sum is maximal at a=2 so 2+420=422 is our answer.
It’s 841
@@jimmysyar889 you can’t use 1 and 840…
@@GreenMeansGOF Oh I wasn't really paying attention to the specifics. In that case, obviously you're right
15:21 I see you have a tetris game there, have you heard of NES tetris and if so have you heard about rolling
I took maths at A-level and passed quite comfortably and I used to think I would do maths at university. But seeing this, I am confident I would’ve caved under that pressure and I’m glad I decided not to go to uni at all 😂
I may be the only one, but i always got an idea how to solve those problems almost immediately. Maybe my first semester Math at Uni helped out quite well, but if you actually try to formulate the questions into equations, they become certainly very easy.
Your expression when you said, "so you're saying there's a chance", i felt that emotion
A great high quality video!
To answer your question:
• Theorem: The maximum possible number of pieces with n such cuts is
exactly (n2 + n + 2)/2. We prove this theorem by induction on n. The base
case says that we have (02 + 0 + 2)/2 = 1 piece with no cuts, which is true.
• Suppose that we have (n2 + n + 2)/2 pieces with n cuts, and we want to make
an n+1’st cut. By geometry, two straight lines meet in at most one point, so
the new cut can cross each old cut at most once. This means that the new
cut passes through at most n + 1 old pieces. Those pieces are divided into
two and the others are not. So after n + 1 cuts we have at most (n2 + n + 2)/2
+ (n + 1) = (n2 + 3n + 4)/2 = (n + 1)2 + (n + 1) + 2, at most the number we want.
• We have to show that this number is always achievable. If the n + 1 cuts are
in general position, meaning that every two cuts meet and that three or more
cuts never meet in the same place, we achieve our bound as long as the
pizza is big enough to include all the intersection points.
A great resource for the kind of problems Mike was doing is The Art of Problem Solving (AoPS) which funnily enough, was founded by a guy who graduated from the high school I’m in right now.
I'm not sure if I'm right but the pizza problem general formula is Pc is
((c)(c+1)+2)/2 (i used p cuz pizza and c cuz cuts hehe) so I'm not going to explain the whole thing but the point is to find a way to get the most intersection with the lines
Edit: sorry i got overly excited, theres a part that showed you guys counting the intersections haha
ayy borderline isnt that bad, i had borderline syndrom for 4 years now and if i watch your videos i feel top notch :)
Love this informative video 💛💚💙
Have you ever tried classic tetris?? A learn to play competitive tetris would be cool. Stopping at a maxout might be too hard so maybe 100k would be a good start and proof of concept? Super fun game and there's so much more to it than it might seem at first!
I feel like Oxbridge just gives me war flashbacks. But a super interesting video at the BTS of Oxford
An honour seeing your comment here! I also get flashbacks about my idiotic mistakes in an interview that cost me my place. How are you doing?
Never thought a maths teacher could be this cool 😎
Then you have been robbed of the potential of education
@@theregalproletariat 🤣
People that can do 1 on 1 tutors still amaze me to this day, the anxiety of not doing things right would be so overwhelming and just block learning for me
From what I gathered, Oxford interviews just basically down to comprehension and application of the desired subject or skill set. Not just, "do you know the answer?" Instead the real interview is, "Do you have the understanding and applicable skill set to solve this problem? Can you show that you can not only learn, but apply what you have learned, later in life?"
Yeah. A lot of it is they'd rather have someone who doesn't know the answer but can work towards it instead of someone who knows it off by heart
3:25
I can make theoretically infinite amount of slices by using 2 cuts. You just cut arbitrarily in random directions, curving the cut in all sorts of ways. Yes, you have 2 slices that are very weird shapes. But now you cut it again using the same method, and still using a random cut to go through the pizza in random directions. It’s still one cut.
But now you have hundreds if not thousands of individual pieces.
There is no formula to this as it’s based around chaos. 2 cuts can give you anything between 4-infinite slices
This is a university interview in the UK!? In America it's just "why do you want to come here?" and "who are your parents?" and all that stuff you said is like a job interview. It's a way for the school to get a personal profile of the applicant. And you'd better believe you have to dress and look the part. And if it's a state school, there's no interview at all!
Dr Tom looks like he fresh off a Devon beach after catching some waves
Would love a followup video where you go through each math question in the video and explain how it's solved. Curious about the pizza slice question and the factorization of 840 :D
About the factorization of 840, the first question you may want to ask is *how many* factors you want to split the number into. For this, you can easily prove that one single factor a * b is at least as good as leaving the factors separate, which would only add a + b:
a * b >= a + b | divide by b
a >= a/b + 1
Because none of the factors can be 1, b is at least 2, and therefore,
a >= a/2 + 1 >= a/b + 1
= a/2 + 1 | subtract a/2
a/2 >= 1
a >= 2
and this is of course also true because the factors still are at least 2. So now that we have proven that any two factors of a solution can be combined into a single factor to get a new solution that is at least as good, we know that at least one best solution will be a solution of the minimum number of factors that is allowed, which is in this case two factors.
So we know that there are two factors x and y that multiply into 840, and their sum should be optimized. So you know that x * y = 840, and therefore, y = 840/x, which can be inserted into the to be optimized function f(x) = x + y = x + 840/x. So to optimize that function, you take a look at where the first derivative becomes 0, what the sign of the second derivative is at that position, and what the limit behaviour of the function is.
f'(x) = 1 - 840*x^-2
f''(x) = 1680*x^-3
So because the second derivative is always positive for a positive value of x, you can already know that the value of x defined by f'(x)=0 will be a local minimum, which is not what we are looking for. What's left is the function's limit behaviour, which *would* be x=840, y=1 and x=1, y=840 if those were accepted values, but instead we'll have to look at the next smaller/larger acceptable x. Because the next x that is larger than 1 is x=2, you get the pair x=2, y=420 (and, due to the symmetric nature, x=420, y=2) which add up to 422, which is now proven to be the largest possible sum of a factorization.
So the answer is that the factorization 2 * 420 = 840 gives us the largest sum, 422.
The pizza one is a recursion formula:
f(n) = f(n-1) + n
f(0) = 1
f(1) = 2
f(2) = 4
f(3) = 7
f(4) = 11 (if you try to draw this, one of the slices will look very, very tiny)
f(5) = 16 (likewise, one of the slices will look extremely tiny)
It would be nice to draw this on paper to show you, but thankfully this is also a question/answer you can look read about here: www.csun.edu/~ac53971/pump/20090922_pizza.pdf (see pages 1 and 2, recursion formula, pizza slice question).
I also ended up with 2*420 = 422. My work for that one was really simple though.
840, I make "factorization trees"
840 ~ 2*420
~ 2*210
~2*105... and so forth
(I used 2 here intentionally, but I simplfy and clarify below, e.g., just multply the 2's together respectively.)
2*420 = 422
4*210 = 214
8*105= 117
If I keep on going with this tree, the sum keeps on getting smaller, and it seems that 422 (from 2*420) is the largest one so far.
Thus, 422 seems like the factorization of 840 that will provide the largest sum. You can also use KappaScopeZZ's (the TH-camr who also commented here) method of the derivatives to double-check some things, but that is not really necessary per se (you would usually use that in the context of a function, and/or a financial-related mathematics question from what I remember). The big picture is most important, as the interviewers mentioned.
These people are so nice. Imagine a grown man about half as smart as you trying really hard to do something you think is simple, say driving a car, failing, feeling really confident and asking "So, how did I do?". You'd like to think you'd stay cool.
:) the Oliver guy is a family member of mine, I didn't know he was in this video and just happened to be watching it and got the shock of my life when he appeared
Which factorization of 840 gives us the largest sum? I'd say the answer is 2*420 so 422 in a sum. assuming i did not misunderstand the question. for any 2 integers if one of them is at least 2 and the other at least 3 then the product of those will be greater than the sum. 2*2 = 2+2 but 2*3 > 2+3 and so on. since using 1 is not allowed and it has to be at least 2 pieces 242 is the obvious answer.
for a number like 81 which is not divisible by 2 the largest sum would be 27*3 so 30 if you could include 2 there it would be 2*40,5 so 42,5 which is bigger than 30.
That formula will be :-
T(n) = n + 2^(n-1)
Where :-
n is the number of cuts.
Edited :-
for 1
Very nice Kendama you have there Mike!
btw I was so interested in the pizza problem that I solved it on paint XD took me like 10 minutes hahaha
(I haven't watched the entire video yet, idk if he solves it later, the problem caught my interest and I fired up paint and started drawing)
the solution is P(C) = 1/2(C^2+C+2), where C is the number of cuts and P is the maximum number of pieces you get out of the pizza
this is a Difference equation where every cut you add, adds how many cuts you have to the number of pieces u had in the previous cut, where the starting position is P(0) = 1 because if you do 0 cuts or never cut the pizza, the entire pizza can be seen as one large piece.
hey mike! love your vids, pretty consistent viewer for the past several years just a idea for a future challenge for yourself, how about learn how to competitively speedrun a video game you already like, there're a lot of really awesome heavily contested speedruns these days that don't get enough love in my opinion. just a idea, love the vids XD
The answer to the pizza problem is 1+(n(n+1))/2. Next, f(3) = 20/27. For the last problem, if we're talking about positive integers, then 2 and 420 for the factorization.
3:38 it's an arthemetic progression, so An = A + (n-1)d ?
I think I have the answers to the second interview questions.
The first question's answer is f(3) = - 2/27 and I got it from changing the equation to f(x) = x - 2f(1/(1-x)) and then tracing through the problem. I ended up with f(3) = 3 - 2[-0.5-2((2/3)-2f(3))] which simplifies to f(3) = 20/3 - 8f(3). Then I added 8f(3) to both sides and divided by 9 to get f(3) = 20/27.
The second question's answer is the factors 2 and 420, because 2 is the smallest factor after 1, so then the other factor is the largest product of all the other factors.
(Question 1 took me 17 minutes and question 2 took me 6 minutes. I did not use any outside help - no internet, no other people, just some scratch paper. I am currently in high school and am taking CalcBC, but I did not need any calculus to solve these problems.)
Please reply if I made a mistake somewhere or if you have any other comments.
I was looking to see if anyone else posted the answer here. I also got 20/27 on the first problem, so that gives me some confidence :) I basically tried x = 3 abd x = 2/3 first. Then I went for x = -1/2 where I found f(2/3) again and then subsited it to have an equation with only one f(3).
I also got 20/27. Had to substitute 1/(1-x) = u and then also x=-z. Then z=x=y=3 for all 3 equations. Honestly pretty easy.
Concerning the pizza problem, Tom actually made a video on that.
Can confirm
Okay, I'd really like someone to check this for me. Would a generalizable formula for the pizza problem be this?
p = (n^2+n)/2 + 1
P is the number of pieces. N is the number of slices. The (n^2+n)/2 is just a fancy way to calculate the additive factorial (that is, adding 1 + 2 + 3 ... etc., up to the number you're calculating; there's probably a fancy math name for this that I don't know). The +1 is needed because, from the outset, the pizza is one slice (at the 0th cut). Is this ... right? It works for one through four slices, I'm pretty sure, but I'm not 100% that it's universal.
@@timshort9787 Looks like I got there a bit differently but wound up with the right equation! Yay maths. :) Thanks for pointing me to that video.
Fantastic episode 👍
What was the answer to the pizza slice question? Would be interesting to see what you should have done to get the answer
The answer is 0.5(n^2+n+2)
To maximise the number of slices you want to maximise the number of vertices created by the cuts, which means you really just have to avoid crossing several cuts in one spot.
When you add a cut to the pizza you add a new slice for every slice you cut through.
Assuming you don't cut through an already existent vertice you go through a different slice every time your cut passes another cut.
And you go through as many cuts as possible, in order to maximize the number of slices.
Well, the maximum number of cuts you can go through is the number of cuts you've already done, so the number of slices you add is n+1, with n being the number of cuts already on the pizza.
So, the s number of slices for n+1 cuts is given by s[n+1]=s[n]+n+, with s[0]=1 (since a pizza with no cuts is a single very large slice).
After this point the way I did it is probably not the best, but it still works:
The number of slices added by each new cut is linear (n+1), which means the number of slices over the number of cuts is quadratic.
From then you can just pick 3 points of the sequence and create a system of equations with s[n]=an^2+bn+c and solve it for a, b, and c.
you start with
a function F where F(number of cuts)=number of slices
F(0)=1 the whole pizza
F(1)= 2 two slices of pizza
F(2)= 4 4 slices of pizza
F(3)= 7 7 slices of pizza
From this we can see that our function is equal to the one previous plus our number!
so 7 = 4(the previous answer) + 3(our current number)
In maths we evaluate this as
F(x)=x+f(x-1)
expanding this
F(x)=x+(x-1)+(x-2)+(x-3)...+1 -> this is where i got confused and had to look up some things.
this is the arithmetic progression series which is basically just the sum of all numbers up to your number
you could re-write the above as 1+2+3+....+x
There is a formula for this which is
f(x)=1+(x(x+1)/2)
the arithmetic progression formula is fairly common and I remember learning it but I am not sure i could derive it on the spot. Im sure if i had just learned it or got some hints id be able to figure it out.
this is called the Lazy Caterer's problem.
@@apologeticanarchist326 wow that was very thorough. Thanks for the answer!
I wonder when this was filmed, I'd love to see Mike Boyd and Prof. Crawford around Oxford!
so for the pizza question I tried to make a general formula : 2n+ (n-1)/2 .[ applicable for odd no. of cuts , But for even no. of cut we have to substract 0.5 from our answer]
for example if we want maximuim no. of pizza from 3 cut that is 7 , we have to put n=3 so the answer is 7 , if we want max no of piece from 4 cut that is 9 so we have to put n=4 thus answer will be 9.5 so substract 0.5 and so on. We can get max 14 piece from 6 cuts .
------Thank you,😊😊😊😊😊😊😁😁😁😁
I think the answer to the pizza question is (n(n+1)(n+2)/6)+1? Using the sum of the triangular numbers and adding one to make it work from the base case as it's 1 when n = 0. If I'm wrong please let me know because it seems like a super cool question!
I'm guessing it is more the sum of whole number + 1 because the way to maximise the nulber of slices is by cutting every line in each new cut, and when you cut a line a new pizza slice is "created".
So when you have already n-1 cuts you create n-1 slices by cutting the cuts +1 because you started in an existing slice you cut.
I hope it is clear
I tried my best ; )
I want you to learn how to do those cool racing drone tricks!
Observational learner here, To become more efficient at math, stay focused and don't waste time doing the basics like combing your hair.
And check the tattoos on your forearm for formulas