How hard was my Cambridge interview? (ft.
ฝัง
- เผยแพร่เมื่อ 1 มิ.ย. 2024
- To prove once and for all that Cambridge is better than O*ford 😤 (just kidding)
Tom's channel: / tomrocksmaths
100k Q and A Google form: forms.gle/WtpM3HqhkdPuwsbP9
You might recognize Tom from his Navier-Stokes equations on the Numberphile channel, and he is a tutor at Oxford teaching students at St. Edmund Hall college in Oxford. I just thought it was funny to have a Cambridge student to interview a teacher at Oxford.
*I HAVE TO EMPHASISE THAT THIS IS ON A MUCH MORE DIFFICULT LEVEL THAN THE OTHER QUESTIONS I HAVE HEARD FROM MY FRIENDS. THIS IS AGREED BY MY FRIENDS, SO IT'S NOT JUST ME SAYING MINE ARE HARDER.*
I am a fourth-year Cambridge math student, and so I thought I would share with you the interview questions that I have gone through! I have signed a sort of NDA for not disclosing the exact interview questions, but it has now been 4 years since my last interview, so it is fine!
Actually, my interviewer knows about this channel, but I am not sure whether he watches it or not. I hope I don't get into trouble, because in the spirit of the confidentiality agreement I signed, I should be able to disclose the details of the interview questions now. It has been 4 years since my interview, so unless they are so uncreative to come up with new interview questions, I think this should be fine.
A bit of notes:
(1) Sorry for a lot of the technical glitches - we were very unlucky, because Tom has also used the same setup in his exam series without any problems.
(2) There is an organ playing in the background - they are practising for the evening service at the college! If you don’t like it, I’m sorry; if you like it, consider it background music.
(3) For Cambridge applicants: the interview format differs from college to college. If you are interviewing for St John’s, there are separate pure and applied interviews; for Trinity (which I strongly discourage any potential applicants to apply to, simply due to the competitive nature), your interview will be based on some questions you’ve attempted in a test prior to the interview. However, because of COVID, I am not sure if all these have changed.
(4) For the people who are here for the maths: for the second question, what I meant to say was that because there are 999 consecutive integers, when you shift by one place, then you are either (a) adding an even number and deleting an odd number, or (b) adding an odd number and deleting an even number.
In the case of (a), adding an even number does not change the number of primes, but deleting an odd number might or might not decrease the number of primes by 1, so the number of primes in the interval either changes by 0 or -1.
In the case of (b), deleting an even number again does not change the number of primes, but adding an odd number might or might not increase the number of primes by 1, so the number of primes in the interval either changes by 0 or +1.
Actually, even if it is 1000 integers, a similar argument applies, though I would say this parity argument is slightly more complicated. Anyway, the idea is that initially you have more than 10 primes, and then you have also constructed an interval with exactly 0 prime, so somewhere in between, there must be an interval with exactly 10 primes, because every time you are only either changing the number of primes by +1,0, or -1.
Video chapters:
00:00 Introduction
01:24 First question: graph sketching
08:54 Second question (Part I): 0 prime
19:59 Second question (Part II): 10 primes
29:47 Third question: physics
50:04 Final point: SAQ (now called My Cambridge)
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I’m really out of my comfort zone with this video: revealing my face, and being unscripted, and I was really nervous (actually I am still second-guessing a lot of things I said in the video!). Hopefully this is still something that you will enjoy! I actually reached out to Tom first, just because it seems such a funny idea for a Cambridge student to interview an Oxford teacher, and thanks Tom for agreeing to help me out in his very busy schedule. There are also some notes about this video in the description, and please check it out!
Notes:
(1) Sorry for a lot of the technical glitches - we were very unlucky, because Tom has also used the same setup in his exam series without any problems.
(2) There is an organ playing in the background - they are practising for the evening service at the college! If you don’t like it, I’m sorry; if you like it, consider it background music.
(3) For Cambridge applicants: the interview format differs from college to college. If you are interviewing for St John’s, there are separate pure and applied interviews; for Trinity (which I strongly discourage any potential applicants to apply to, simply due to the competitive nature), your interview will be based on some questions you’ve attempted in a test prior to the interview. However, because of COVID, I am not sure if all these have changed.
(4) For the people who are here for the maths: for the second question, what I meant to say was that because there are 999 consecutive integers, when you shift by one place, then you are either (a) adding an even number and deleting an odd number, or (b) adding an odd number and deleting an even number.
In the case of (a), adding an even number does not change the number of primes, but deleting an odd number might or might not decrease the number of primes by 1, so the number of primes in the interval either changes by 0 or -1.
In the case of (b), deleting an even number again does not change the number of primes, but adding an odd number might or might not increase the number of primes by 1, so the number of primes in the interval either changes by 0 or +1.
Actually, even if it is 1000 integers, a similar argument applies, though I would say this parity argument is slightly more complicated. Anyway, the idea is that initially you have more than 10 primes, and then you have also constructed an interval with exactly 0 prime, so somewhere in between, there must be an interval with exactly 10 primes, because every time you are only either changing the number of primes by +1,0, or -1.
Worry not, your videos have always been of the highest quality.
Well done for doing the face reveal despite it putting you out of your comfort zone 👍
Only Respect for you man 🙌🏽🤙🏽
It's a rather hard thing to do, being fair, reasonable, kind, correct, strict and quick-thinking enough live on camera. Well done!
I remember I discovered you writing answers on Quora. Looking at how far you've come evokes a different feeling in me...
Thanks for having me - this was a lot of fun!
Hello! I subscribed!
@Sanjoy Kundu well he cares, he gained 1 sub
@Sanjoy Kundu Why must you chose to spread hate, it's so unnecessary...
Hi Tom,
I really enjoyed the video this week and it was good to see how you adapted to something where you were under pressure and not in control :)
I wanted to jot down a couple of comments I had whilst I watched ...
Q1 - Fair question and one entrants should be able to solve even with a little guidance in my view
Q2 - One thing i was thinking when you were tacking this ... given the problem didn't specify a "positive" interval, you could have specified the interval (-1000,0) as this contains 1000 integers where non of them are prime (as they have a factor of -1) ???
Also, for the 2nd part, where you you need to specify an interval with exactly 10, then again, using the interval from part (a), you could then list the first 10 +ve primes (as you did) which are (2,3,5,7,11,13,17,19,23,29) and then use this interval to give the interval of (-971,29) ???
A bit pedantic and perhaps a cheeky response and definitely not what the interviewer was looking for of course, but it's all in the way the question is phrased. Perhaps shows a little lateral thinking too which must be a good thing in an interview.
Q3 - Sounded straight-forward but was actually tricky. I got there but it was a tricky one. Lots going on in this question.
You were in a tough spot, Tom. Nice video
I have a lot of respect for Tom. It takes lots of courage to do these questions as a professor and risk your reputation.
when you’re just enthusiastic about math and not much else, reputation means nothing really
Lol the guy has a phd and studied at both Oxford and Cambridge he’s intelligent and it’s a fact he’s proven himself by peer review and academic record just because he can’t answer every single question doesn’t make him even the slightest dumb even at the top level of maths people may forget very simple and basic mathematics not to say this was simple at all but definitely questions that aren’t standard and require thinking regardless of your mathematical abilities
Don’t you think if he failed miserably he would not have the video shown?
@@karagi101 I think he failed miserably at the second problem.
He did fail miserably, but these interviews are designed to watch you think through a problem while you are lost.
Nice problems! Kinda jealous as an American because these math interviews seem significantly more fun than any part of the admissions processes we have here :D and in my opinion, watching how a student talks through a problem that's difficult but actually geared to what they've learned so far is a much better way of assessing whether they're suited for a top math program, especially compared to a multiple-choice standardized test where "you either know it or you don't"
Well, there are definitely Americans applying to Cambridge - I have a friend (now graduated) from California also studying maths at Cambridge. **Of course, it depends on your financial situation and the availability of scholarships/funding.**
swings and roundabouts I think, your standardised tests have some pretty huge flaws in terms of assessing knowledge, but your university system seems more freeform, and better designed.
@@paultapping9510 Their admissions process is an embarrassment though. In the UK people often complain about people from less privileged background not being able to perform as well as people from more privileged background, but America has gone and taken that to a whole new level.
They prefer extracurriculars (not supercirriculars) over academics meaning that poor students who have to work and can't afford any extra classes (which you have to pay for in lots of schools) won't be as desirable as middle-class students due to their now lowered GPA. UK universities prefer academics and subject-related extra-circulars over the other things which makes them infinitely better. Harvard takes this to another level with their legacy system which is there to encourage donations from alumni.
@@shaaravguha3760 Late to the convo, but if you think prioritising extracurriculars is a bad thing I sincerely doubt your judgement. Anybody can game a high school syllabus: including people without a shred of intelligence and/ or social skill. America's admissions process is far superior to just looking at GPA.
The legacy system (which is really bad) is the only problem I see. Should stop demonising privilege and becoming irrational.
@@geizuko9927 'anyone can game a high school syllabus'. and anyone can game extracurriculars with enough money. you can't however game an interview like this without geuine knowledge, curiosity and smarts.
Thanks for going outside your comfort zone, Trevor! It can be intimidating to watch your highly polished and brilliant lessons -- but I have to say your insight is also very well expressed intuitively so I think they are also very accessible.
Anyway it is awesome to watch you two struggling a bit with questions out of the blue -- this being the way that maths is always learned in experience!
Also great to put your face to your voice!
Its so nice to finally have a face to the wonderful maths education I've been getting from this channel over the last gosh year or so. I started watching when you talked about the dream scandal. Here I come to find its with one of my favorite numberphile and computerphile youtubers. This is truly a great day, thank you for the high quality content you always produce!
It's fascinating that both Tom and Trevor considered the third question to be the hardest, when that was the only question that I felt was fairly straight forward. I doubt I would ever have been able to complete the prime number question, no matter how long I had to think about it!
Completely agree. Even after the massive "hint" I had no idea how to use factorials to get anywhere. Tom still needed nudging, which made me feel better!
I agree. The physics question was pretty simple in my opinion compared to the others. I sort of got the idea for the first part of the prime question but inspired by Tom. I had no idea how to start for the second part. I solved the physics question out of interest in like 3-5 minutes but my main interest is physics so that helped a lot.
As someone currently undergrad student at Cambridge, I can tell u it is extremely nerve wracking sitting in that waiting room for so long, however as soon as the interview starts I promise u they will settle and you will feel like you are in flow state. Also, it is perfectly fine to not know an answer. In fact, most people who come out thinking they flunked the interview and aren’t going to get offers are actually the people who get them more frequently or not
@Tomrocksmaths I love that you take on these challenges. It shows that the great math minds don't always just wave their wand and "poof" out a proof. You actually do have to take the time like anyone would with a puzzle to figure out the solution. The secret to being good at math is to break it down and take the time for details. I was a horrible math student in primary school but then something clicked and I had the drive and patience to work through it. I went on to successfully complete about every Math class I could take with a B.S. in Electrical Engineering and M.S. in Engineering. Thank you for not making it look like a magic trick!
When I went for my Cambridge interview back in 1981, I wasn't asked any questions in the Math part of the interview. My interviewer told me that he knew enough from my entrance exam results. Despite getting in, I was disappointed not to have the opportunity to discuss Mathematics at the interview. So now I watch these videos to see how well I would do.
The person who interviewed me was to later become my PhD advisor. He layer told me his typical interview question, which was to plot sin^2(x). He said invariably people plot it like |sin(x)| with sharp corners on the bottom. He points this out, and if they understand their mistake, he accepts them. If they see from their figure that it is (1-cos(2x))/2, he gives them a scholarship. I remember having thought about this long before the interview, and I would have answered it so quickly that I think he would have thought I was trained for this question.
Cool collab! Bring more of such videos. They really help.
Thanks for video. This brought back memories of my Oxford interview in I think 1982. At that interview I was asked to show that integration was the inverse of differentiation which was a nice question that I hadn’t considered but that was within the scope of A-level knowledge.
The prime number one was really nice. I couldn't think how to approach it at all, but as soon as you said factorials, it all became clear. Just shows how one key insight or thought can be critical to solving a problem.
Love Q2! [spoilers]
So brave of you guys to do this unprepared, especially Tom. Even after the explanation, I'm still not convinced Tom got it. I was also thrown by part 2 being a yes/no existence question, not a constructive one (is there a nice construction for the interval?) But once that is clear, it is relatively easy. But the explanation in the vid is not super clear.
Part I. This is kinda like the flip side of Euclid's proof that there are infinitely many primes: n! + 1 has a prime factor greater than n, therefore there is a prime greater than n for all n ( ~ Euclid), .
For this question, n! + 2 has factor 2, n! + 3 has factor 3 ... n! + n has factor n => there are no primes among the n - 1 consecutive integers from n! + 2 thru n! + n. Set n = 1000 and you are done.
Part II. The explanation was basically right but not clear. Consider the 999 consecutive integers in the interval [2, 1000]. The interval contains more than 10 primes. We know from Part I that the interval [1000! + 2, 1000! + 1000] contains 999 consecutive integers but no primes. Now, start sliding the lower interval upwards by integer increments: [3, 1001], [4, 1002] ... With each shift, the number of primes in the sliding interval will do one of the following:
1/ stay the same (prime comes in at the top and prime drops out the bottom, or composite comes in at the top and composite goes out the bottom), [because 1001 is not prime, the former in fact never happens for windows of odd size, like 999]
2/ reduce by one (prime drops out the bottom, and composite comes in at the top), or
3/ increase by one (prime comes in at the top, composite drops out at the bottom).
Thus, the count of primes in this shifting window of 999 consecutive integers begins above 10, and (from Part I) eventually drops to 0. Since it changes only by increments or decrements of 1, it must at some point be exactly equal to 10. (NOTE: there is no need to assume that it is a decreasing function, even though the overall trend is downward. Also, 999 being odd makes no difference to the proof.)
As stated in the vid, this assumes the Discrete Intermediate Value Theorem (intuitive, and "common sense", but not totally trivial). Proof here: anuragbishnoi.wordpress.com/2015/06/25/discrete-version-of-the-intermediate-value-theorem/
ALSO IM KINDA CONFUSED, WE'RE ASSUMING THEY CAN BE INTEGERS SO WHY NOT. -1 TOWARDS -999. I MEAN THEY ARE "INTEGERS"
@@egoxagony4623 you can, but preferably we're going to stick to a simpler solution
@Adam Very nicely explained.. Thanks!
the second part isn't really about primes the argument essentially works for any sufficiently sparse sequence, the first part is just there to show that the prime numbers are indeed sufficiently sparse.
@@existenceispain2074 Correct. Any monotonic increasing integer sequence that goes from dense to sparse (not necessarily monotonically) on the consecutive integers will attain all intermediate densities in the way described.
So close to interview season, it’s refreshing to see this!
That's exactly why I am doing this now!
Thanks for this, it's good to see how a mathematician working out the solutions to these problems
For Q1, you can note that after completing the square on both sides and simplifying, we obtain (y-0.5)^2=(x-0.5)^2, which is a translation (of 0.5 units) along both x and y axes of y^2 = x^2. And y^2 = x^2 trivially has solns y = +/- x.
hi, We can also use the slope-intercept formula to draw the graph. Here we use the general format y=mx + b (sometimes written y=mx + c). This equation can be written as y = - 1/1x +1
Greate video ! And sir, your interviewing skills are good 😊 fun to watch. Thanks
I was quite challenged by the prime number questions but understood them in terms of elimination of terms in factorials… the one involving the coreolis effect was much more straight forward. I’m a 71 year old mechanical engineer.
"elimination of terms in factorials" - similar thought here!
You actually don’t even need to consider the Coriolis effect- you can do it in the lab frame and it’ll be easier
As a 3rd year pure maths focused Cambridge student, I haven’t done any physics since A level physics really (I didn’t really manage to follow the dynamics and relativity course in first year). But the physics question still felt doable without many formulae (only knowing what things are):
Don’t consider the rotating frame at all, just consider the “x and y” components at the moment of shooting the gun that is its going 5m/s forwards and since the gun is moving at 5m/s sideways, it also has a perpendicular component of 5m/s (which corresponds precisely with the r and θ components, but thinking from a more A level physics perspective since rotating frames aren’t really covered there). This gets the 45° angle and the speed of 5√2 m/s. This avoids all the x’ = r’e_r + rθ’e_θ (I can’t write dot notation in TH-cam comments).
Nice to see another mathmo here! I think what Tom was struggling is the relationship between the angular velocity and the actual tangential velocity, so if you don't know the "formula" v = r*omega, then you could not really deduce that the gun is moving at 5 m/s sideways; and you would somehow have to rely on that r’e_r + rθ’e_θ stuff (which is almost equivalent anyway).
I think I’m helped by the intuition that ω is in rad/s and we have the radius, so you kinda just input what the radius is. (Which works for instantaneous tangential velocity, but probably not for other things 😂).
@@mathemaniac Didn't Tom say he taught dynamics? How do you teach dynamics without knowing v = r*omega? Genuine question, does dynamics mean something else in the UK because no shot you could teach dynamics without that knowledge in the US.
@@mathemaniac Thanks for the video. I didn’t know about dots and managed question 3 ok…
A very common method used in maths A-level is to resolve a physical scenario into orthogonal components. If we follow the diagrams in the video: let the origin be the point of release of the bullet, let x be the direction tangential to the circle at the point of releasing the bullet and y be the direction towards the centre of the circle at the point of releasing the bullet then: We can represent velocities as (x,y) where x is the speed in the x direction and y is the speed in the y direction.
The speed of the shooter at the point of release is r ( which is 5) because that’s what 1 radian per second means. The direction is horizontal because at the point of release that is the direction of the shooter. So the velocity = (5,0).
The gun provides the bullet with an additional relative speed 5 and the direction is up because it is towards the centre so relative velocity is (0,5). The overall velocity of the bullet is the sum of the two velocities = (0,5) + (5,0) = (5,5) which is as described in the video at 45 degrees (i.e. x=y). So the bullet will next hit the circle at position (5,5). To find the time when this will happen we can keep the maths simple by just considering one of the components. Let’s just consider the y component. The bullet is travelling up at 5 m/s so to travel a vertical distance of 5m will take 1 second. Now for the last stage let’s talk in terms of fractions of a full rotation. The bullet hits the circle after 1/4 of a full rotation (Call that B). Because the rotation is 1 radian per second, the people have turned r metres in 1 second which is r/(2 PI r) = 1/(2 PI) of a full rotation (Call that T). We can approximate PI = 22/7 so T = 1/2 x 7/22 = 7/44. The bullet hits the person at a fraction of the full circle = B-T = 1/4 = 7/44 = ((11-7)/44 = 4/44 = 1/11. Because there are 22 seats this is the seat two to the right of the shooter.
@@mathemaniac I've actually never heard of v=r*omega at all. But you have a circle of circumference 10*pi meters and if you are rotating at 1 rad/s then you rotate round the entire circumference in 2*pi seconds, dividing one by the other you have a velocity of 5m/s. The bullet will have the muzzle velocity of 5m/s in the direction of the centre of the circle and your tangential velocity of 5/ms as per your rotation speed, so obviously travelling at 45 degrees and hitting a quarter of the way round the circle from you in 1 second.
Amazing collab!
I absolutely adore this guy, he is such a cool guy! I love tom
It is amazing how Pythagoras crops up so often. As the student pointed out the key to the first question is to recognize the difference of two squares.
The point about it being much easier sitting comfortably at home is interesting. It seems our surroundings can affect what we see. I have been caught out in an interview with a simple question. It was a piece of code and I was supposed to notice that variables had been used without being initialized. Before the relevant code there was some other strange code which caught my attention. I don't remember what it was but it was not important to the interviewer but it was important to me. Another case happened to-day and is even more trivial. Our local Aldi has been rearranged and I am still re-learning to find things. I could not find what I was looking for and asked one of the staff. She took me to the relevant section pointed it out and I still did not see it. I had looked at the item at least three times and still had not seen it. It was the wrong colour because it was part of their gourmet brand. I was expecting it to look similar to other items in the same display.
I think if this was a genuine interview, the interviewer would be most interested in observing the thought processes demonstrated by the candidate. Not finding the complete answer may not necessarily mean a failure.
Yeah that's what a good interview's all about. It's not about the what but about the how and why; "How did you come to this conclusion and why did you take the assumptions that you did?"
Its great to see you in front of the cam and to see people like you struggling in action! Give it a Reload sometimes, Matt Parker is surely also open for that...
I've never heard of 'r theta dot' but just converted radians to segments of the circle and then did normal high school vector addition. I got the same number and whilst my method was a little yucky and had to use some ugly numbers, I think the question is fairly good for not requiring too much prior knowledge.
In the second question I think this should be the way to make proof more neat: by shifting interval by 1, number of primes can either +1 -1 or don't change. (0, 1000) has more than 10 , (1000!, 1000! + 1000) has 0. So there is an interval where we would get exactly 10.
Yes - I was too nervous on the shoot, and I have pointed this out in the pinned comments and the description.
@@mathemaniac Sorry😅, didn't pay attention to the description
wow respect, my head would catch fire from overload if i had to sit there :'D just best wishes for both of you.
Great stuff guys.👍🏿👍🏿👍🏿👍🏿 I’m actually teaching functions in P3, I’ll give this question to my students next week 😁
Love your videos
Keep up the good work
for question one, you can factor the euqation and you'll get (y-(x+1))(y-x)=0 and vice versa, then you'll get the two lines symmetric
Hey! Great video! This will come in handy for interview prep (if i even get one lol)! Questions seem really hard, but it is Cambridge at the end of the day. I was just curious, if you don't mind me asking, what College did you apply to and what college do you study at? Thanks!
Great video! Thank you
Great and interesting video. Thanks!
The problem sketching of graph can be solved by completing the square on both sides by adding 1/4, which results in two equations y=x, x+y=1
For question2b, my initial thought was to list all of the primes and look at the 1st prime number, 10th and 11th on the list. Then always consider the kth, k+9th and k+10th. When the difference between k+9th and kth number is less than 999 while the difference between k+10th and kth is greater than 999, then you can start with that kth prime number which is an integer and then write 999 consecutive integers and we’re done! As the question ask for whether or not we can find, then the answer is yes by this approach.
By the way I’m a 23 fall maths applicants so the video is very great and useful!
I don't think this argument is complete - how would you guarantee that there is a (k+9)th prime is less than 999 from the kth prime, and the (k+10)th is more than 999 from the kth prime? You could have the prime numbers being more than 1000 apart forever after some point! (Of course that does not happen, but you would have to additionally show that this does not happen.)
@@mathemaniac Thank you so much for your response! And please let me explain my thought in detail: The question asks whether we can find an interval with 999 consecutive integers with exactly 10 primes. So my approach is list all the prime numbers (for example use p(k) to denote the kth prime number). Then if there is a k such that p(k+9)-p(k)999, then from p(k) to p(k)+998 (a total of 999 consecutive integers) it is guaranteed to have exactly 10 primes, which are p(k),p(k+1)…p(k+9), and both p(k-1) and p(k+10) will not be included in this interval and so on. Hopefully we can use computer program to find that k after putting all the prime numbers from small to infinity.
So this is just an method to find the required interval, I haven’t proved whether there exists such interval (probably computer program can find one), but my intuition told me when k is very large, the prime numbers will be more spread out so there exist k and can be found by my method.
Does my explanation convince you now? I am really happy and excited to talk about maths with you!!
It was excruciating watching Tom struggling.
The fact he could look at a piece of paper with just numbers and symbols and almost instantly picture what shape the graph will resemble before solving the equation is crazy to me. I'll never understand math like that.
Great video. Thank you
Awesome video. I am very interested in math and sciences
Tom looked so done half way through the fourth question lolllll
Thanks so much sir!!! @Mathemaniac Rocks Math
Feeling so good rn bc I had the right idead for problem 2 part B 🥰🥰🥰
Thanks for having me - this was a lot of fun!. Thanks for having me - this was a lot of fun!.
In the physics question, looking at the problem in the inertial frame of the lab turns it into a simple geometry problem
For Q1 I just thought if you set them equal to some third variable z, then x and y are on equal terms, you have to treat them equally. The line y = x is clearly part of the solution. Next I would wonder if there was some additional trickery with sign, and after plugging in some numbers I'd probably get the other line. Then I would think if I was done or not, but on the spot I might have over-thought it.
I did my Cambridge maths interview 5 years ago (sadly I didn’t pass). Since the interview is facing students from different courses, the questions were really concise and not difficult at all, but they were not very straight forward and require a bit of thinking. Moreover the interviewers will keep throwing questions at you within the time limit, so you must be extremely efficient in solving problems from all disciplines of mathematics - they will give you hints, but you have to be able to immediately understand and apply the hint. The methods you apply to solve a problem also matter since it shows to what extent you can make some key observations.
Another thing to mention is that you may get “nature of science” sorts of questions. So it’s really up to you to show your understanding of the fundamental stuff underneath the subject itself.
Edit: you may skip a problem if it involves something you’ve never seen before, but from what I have experienced, you cannot skip a number theory problem, probably because they think you should be able to do it as long as you know what a prime number is 😂 and leaving everything else to algebraic manipulations and factorisation techniques etc.
_"and not difficult at all, but they were not very straight forward and require a bit of thinking. Moreover the interviewers will keep throwing questions at you within the time limit, so you must be extremely efficient in solving problems from all disciplines of mathematics - they will give you hints, but you have to be able to immediately understand and apply the hint. "_
Well yes, if you only have, say, 30 seconds, or even 60 seconds to solve one of it, you will easily flunk. But the man in the video needed somthing around 10 minutes. How he ever got to teach a course in anything is totally amazing!?
The second question is something that I enjoy showing to people with a moderate about of math. Once you explain it, most understand it.
It is great the way dumb down these maths questions so we can follow along. I would like to point out an error in the last physics problem although given the short ranges involved it would not effect the solution in the "real world". The distance traveled is in a curve therefore it is not travelling 10 meters. I can see that someone a lot cleverer than I could actually come up with a formula. Ir would be really interesting if you factor in velocity loss as well.
The prime questions are really nice! I got part a) and b) real quick, so happy for that, and discovered the last one, c), only when it was explained. c) is really nice though! (using both a) and b))
I really enjoyed this video.
I have a question.
For the third question i think it is easier if you think of the perspective of someone outside the wheel. That person would see the bullet in a straight line and it would take it 2s to reach the other side (because v=5m/s and 2r=10). In that time the wheel would rotate 2 rad (since the angular velocista is 1 rad/s). And since each person is pi/11 rad apart and the person opposite to you is the 11th (opposite is pi rad appart and pi/11*11=pi) then the person that would get hit is in the place 11-n-1 where n is such that: n*pi/11=2
And that is,aprox, 7. Therefor the 3rd person is the unlucky one. No need to consider the relative velocity (sqrt(50)).
But im not sure if this is correct or was just a lucky mistake cause Tom and other people in the comments use the inside perspective. So my question is: is my method wrong? And if so why?
P.D love this video seeing the reaction, thinking with Tom and lisent to your interviews questions was great.
_"So my question is: is my method wrong?"_
Yes it is. And even your calculation is wrong, even if your first assumption was right.
Of course we may argue that the premises of the question are not stated in a complete or concise way. But if he says "you have a gun firing paintballs", we may well assume that he is holding this gun in his hands and fires it in the usual way. We should _not_ assume that this gun is fixed apart from this carousel and fired remotely controlled. So the gun - and the paintball - has a _tangential velocity of 5m/s additional to its radial velocity of 5m/s. So the ball does not hit the point 180° from the starting point but 90° apart.
Well done Tom, I paused the video at 3:25 and had a go. I spotted the difference of two squares (in the first question y^2 - x^2 = y - x which normally is (y-x)(y+x) ) which then reveals the line y = 1 - x . My nephew is studying for his Leaving cert Maths maths exam so these "tricks" are in my mind as I help him. Nothing like being able to sort it out ab initio as you did.
Am I the only one who saw that y=x is an obvious solution to this?
How do you get the second line y=x using difference of two squares
@@aniruddhnls No I also saw it right up front
@@aniruddhnls Lol no that was very obvious from both sides of the equation being exactly the same, was amusing that he completely missed that even to the end.
The way I worked it was a) calculate horizontal speed (2*pi*r)/ 1 radians per second = 5m/s. Therefore resultant velocity = sqrt(50) at 45 degrees. Then using the angle across = half from the centre means the impact is 90 degrees ....ie at 90 degrees while the shot was fired at 180. Calculate time to travel which is 1 second ...by which time you've travelled 1 radian (360/2*pi) = 57.3 degrees so 22.7 degrees away from shooter. Each seat is 16 (ish) degrees apart...so if the shooter is first you hit the 3rd person. No messy equations really....calculating tangential velocity was key though which was d/t is rotational form. Nice question.
Interesting extension-type question.
close it should actually be 4th person to your right.
Gracias por el video.😊👍🏽
I can't think of anything more frustrating than having someone in my ear telling me I'm on the right track when in comes to a math problem.
FR but its also reassuring
I’d say no matter the parity of the interval length, when you shift by 1 you include a number and exclude another one, so the count can only change by -1, 0, 1.
i don’t get that though because it can have 0 for ages then get 2 and it’ll stay at 2 2 2 2 until 100 shifts to get rid of the 2 primes from the count
@@edzty318 how would it gain 2 in a shift of 1 integer each time? You can either A) Gain a prime and not lose a prime, +1 B) Gain a prime and lose a prime, +0, or C) No new primes and lose a prime, -1.
For the point at 19:35 about this being the smallest starting number using this method, couldn't you argue that using [(999! + 2), (999! + 1000)] works too, you just have to carefully consider the 999! + 1000 edge case, which is pretty easy because there are many factors of 1000 that can be spotted easily that are smaller than 999 and hence the whole expression is easily factorable.
There are very probably much smaller numbers satisfying the requirements.
I got the centrifuge one immediately after your hint that we don't even need F = ma. I'd been racking my brain trying to remember what I know about non-inertial reference frames and centrifugal force, and that made me realize there are no forces on the paintball when it's in the interior of the centrifuge (well, assuming a vacuum so that there are no fluid dynamics considerations) so the Coriolis effect image we had of the paintball curving away doesn't apply here: the paintball travels in a straight line!
The whole time I was like "the paintball doesn't care what the surroundings are, as long as nothing's pulling it, it will continue to travel in a straight line" and the curvy line is just the "footsteps in the sand" of the paintball that happen as the disk of the arena rotates. It only depends on what time the projectile reaches the other side, and what the speed of rotation is.
Example: if speed = 10m/s and radius = 5m and speed of rotation is 0m/s where will the ball land? It will land exactly 180° away from your current position. What about if the rotation speed is 180°/s? Well, the ball will move to the place 180° away in 1 sec, but in that time the arena will also have rotated 180°, so +180°-180° = 0° so it will land right back at you!
What about half the speed? That means it will only have rotated by 90° so it will be 180°-90° = 90°. So it will land on the person 3 hours east of you.
And in general,
Θ = 180° - 2rS₁/S₂
where:
Θ: the degrees you have to rotate from the initial position to get to the position of the projectile's landing spot.
r: the radius of the disk on whom you're rotating.
S₁: the speed of the rotation of the disk.
S₂: the speed of the projectile.
Or if you prefer,
θ = π - RS₁/S₂
where:
θ: the angle you have to rotate from the initial position to get to the position of the projectile's landing spot.
R: the diameter of the disk on whom you're rotating.
S₁: the speed of the rotation of the disk.
S₂: the speed of the projectile.
What about suvat?
Isn't problem 2b actually easier? I mean from 2a you have an interval of the right size with no primes. You can obviously slide it down, by 1 at a time, until you cover the right number of primes.
You are certain to manage because to he count can change only by at most 1 at a time and there are more than 10 primes in the first 1000 numbers.
I don't think it matters whether the interval is of even or odd length.
A nice problem for sure. Thanks!
you know that moment when a math video is almost an hour long
It's worth noting that you don't need physics equations on Question 3. You can reason about it like so:
I am rotating 1 rad/s
A full rotation is 2pi rads, so I am traveling 1 / 2pi ROTATIONS per second.
1 rotation in this example is 2 * pi * radius = 2 * pi * 5 meters
and because you are traversing the circumference of the circle,
the tangential speed = proportional rotations per second * the distance in a full rotation = (1 / 2pi) * (2 * pi * 5) = 5 m/s (at a 45 degree angle because the tangential and inward velocities are perpendicular and equal in magnitude)
NOTE: The above is still the formula for tangential velocity, but it can be reached with reasoning instead of memorization. "If i go around 10% of a circle in 1 second, and the circle is 100 meters around, then I go 10% of 100 meters in a second, or 10 meters/second"
then reason that the inward velocity of 5 m/s + the horizontal velocity (rightward as drawn in this video) of 5 m/s will cause the paintball to move 1 radius vertically and 1 radius horizontally in 1 second, placing the impact point 1 quarter turn from the start point after a travel time of 1 second.
So you have an impact point pi/2 rads in front of you, 1 second after you fire.
BUT, after 1 second, you have moved around 1 radian, and the impact point is pi/2 - 1radians in front of you. There are 22 seats per 2 pi radians = 22 / (2 (22/7)) so 7/2 seats per radian.
So the paintball hits ((pi/2) - 1) * 7/2 = ~(22/7/2 - 1) * 7/2 = (11/7-1) * 7/2 = 11/2 -7/2 = 4/2 = 2 seats in front of you
And this way, you don't need physics equations, square roots, calculators. The only foundational things to know are:
circumference = 2pi*r
pi ~= 22/7
how to simplify fractional math.
After writing this wall of text, I hope this clarifies the reasoning for, well, at least someone!
Answer is actually 4 seats to your right.
@@macmos1 stop posting this everywhere, it's not 4
For part one of question 2, why couldn’t negative numbers be used? Were nonnegative integers implied?
Q3 you can do without the ‘theta dot’ stuff. In one second you rotate 1 radian, so in 2pi seconds you do a full rotation, which is distance 10pi, so your tangential velocity is 5m/s. Which gets you to the 45deg which Tom got to.
Yep, way easier like that. Then you get that the difference between the person and the landing spot is (pi/2 - 1) which lets you easily calculate the seat relative to the person.
You can derive this without the formula. We are travelling at an angular velocity of 1 rad/s, so to travel entirely around the circle it takes 2*pi seconds(time period). The circumference of the circle is 10*pi (distance), so we could just use speed =distance/time to get that the tangential velocity is 5m/s, before then using pythagoras to determine the speed as rt50
at 49:58
@Dnomyar Akunawik how?
The third question inspired me to work out the problem for the general case:
n = [2φ - 2ωr sin(φ)/√(v²+(ωr)²)] • N/(2π) + 1
with φ = arctan(v/(ωr))
Chill
Love Q1. [spoilers]
The smoothest way to get this one would probably be to see that 1/ equation is symmetric in x and y, and all y=x are in the solution set, so graph includes line y=x, and any other components must be symmetric about y=x, 2/ it is the equation of a conic section (no powers higher than two), 3/ it includes the four "zero points" mentioned in the video. => (degenerate conic section consisting of) two straight lines through the two pairs of points.
But I took a more roundabout route on first attempt. Like Tom, I found the four zeroes, and like Tom wondered whether there was "a circle" involved (coz it must be a conic section of some kind - I guess that is why he said that) But has to include the entire line y=x since symmetric in x and y. Implicit differentiation was a wild stab in the dark: dy/dx = (2x - 1)/(2y - 1). Slope = 1 when on line y = x (obviously). The slope at the other two zeroes is -1, so def not a circle. Not differentiable at (1/2, 1/2) even though this point is in the solution set and is on y=x. dy/dx = 0/0. So maybe something weird at that point like self-intersection. => degenerate conic section of two lines intersecting at (1/2, 1/2), a.k.a. limiting case of hyperbola.
But y(y-1) = x(x-1) is just a quadratic equation y^2-y-x^2+x=0 which can be solved by y = 1/2 ± sqrt{1\4+x^2-x}=1/2 ± (x-1/2). The rest follows easily.
@@peterpankert3810 Definitely a nice and disciplined approach.
I don't think it would have occurred to me (it didn't), because 1/ "symmetry in x and y", 2/ "obvious zeros", and 3/ "conic section" seemed more immediately motivating.
When Tom says he gets it I still don’t lol😂😂
These were my approaches to quentions 1, 2
Q1: (y - 0.5)^2 = (x - 0.5)^2
that's the same as |y - 0.5| =|x - 0.5| , which is a cross with a vertical shift up to 0,5 and a horizontal shift to 0,5 to the right
Q2.1: negative numbers are integers and aren't primes. -999 to 0
Q2.2: The tenth prime is 29, therefore: -970 to 29
very smart approach to Q2!
I remember struggling with some symbolic algebra software on a Tektronix 4010 terminal in a computer lab in DAMTP and looking through the open door to see Stephen Hawking passing by. That would have been in 1980 or so.
Great video! Super cool
It does not actually matter that the interval for the 10 primes question is odd. By shifting 1, you can only change the number of primes by one, no matter of whether the size of the interval is even or odd. You can "win" a prime at the head and/or "lose" a prime at the tail and that's it.
Yes, I realise that, and that's why I pointed this out in the pinned comment.
Nice vid!
I don’t think that our interval consists of an odd number of consecutive integers is crucial. I think the following argument works (correct me if you find a flaw). We know that there is at least one interval of 999 consecutive integers below ours that contains at least 10 primes (just consider the interval starting at 2). We also know that we can construct an interval at least as large as this that contains no primes. Each shift down by one of such an interval that initially contains no primes has a net result of either no increases in the number of primes, one less prime, or or one more prime. Therefore just keep shifting down until you have 10 primes. (BTW: Number theory makes me slightly nauseous! 😢) ;)
Surprised the physics problem gave so much trouble
Q3 [spoilers]
Ok, so yes this question is annoying coz I forgot to add the velocity of the gun from which the paintball is fired.... But otherwise, fairly straightforward.
Assume centrifuge is in cartesian plane with its center at (0,0), bullet is fired from point (0, -5).
Let's call the seat from which bullet is fired "seat 1". So, "seat 12" would be directly opposite.
pi = 22/7
22 seats around the circumference of 2 x pi radians, so 22 seats = 2 x 22 / 7 radians.
1 seat = 2/7 radians.
1 radian = 7/2 seats.
Velocity of bullet is sum of two components: 5m/s toward center of circle (+ve y direction), plus 5m/s tangential to circle (+ve x direction). NOTE: knowing the instantaneous velocity of circular motion is a prerequisite for this question. These two components of the velocity are equal and perpendicular to one another. So, path of bullet is at 45 degrees (pi/2 radians) from axes, and crosses circumference of centrifuge again at (5, 0). It takes 1 second to reach (5, 0) from (0, -5), since bullet is traveling 5m/s in both x and y directions.
So, bullet hits the seat that is at pi/2 radians (90 degrees) around the circumference from the starting position, but in that time the seats have rotated 1 radian. So bullet hits the seat at pi/2 - 1 radians from the seat from which it was fired. pi/2 -1 = 11/7 - 7/7 = 4/7 = two seats around from the guy who fired the bullet.
Answer is actually 4 seats to your right.
@@macmos1 I think you'll need to explain how you got that.
@@TheNothingNihilates Sure. Say at theta = 0rad and t = 0sec you fire the bullet. The bullet travels at 5m/s. The radius of the centrifuge is 5m, so diameter is 10 m. In the inertial frame of reference the bullet will just travel straight a total of 10m from theta = 0 before it hits the other side of the centrifuge at theta = pi rad, meaning it will reach the other side in time = d/v = 10m/5m/s = 2s. Because the centrifuge rotates at 1rad/s, your seat after 2 seconds since you fired the bullet will be at theta = 1 rad/s * 2 s = 2 rads. We know that by dividing up the number of seats over the entire circumference that 1 seat = 2/7 radians. So at t = 2 seconds when the bullet reaches theta = pi, that seat will be 2 (your seat) + 2/7 + 2/7 + 2/7 + 2/7 rad = 2 + 4(2/7) rad which is approx. equal to pi. Therefore, seat 4 from your right is the seat that will get hit.
@@macmos1 It travels 5√2 m at a speed of 5√2 m/s. So one second of travel time. See the reasoning given above. So the answer is two seats to the right, same as given in the video. The bullet does not travel along the diameter, but along a 45° chord, because it already has velocity of 5m/s perpendicular to diameter before it is fired, . That is the "annoying" bit. I missed it first time too!
@@TheNothingNihilates On second thought you might be right. There is a tangential component to the initial velocity of the bullet I'm not accounting for.
n!+1 must be a prime number too. If we think about a prime number as numbers such that if represented in a base from 2 to (n-1) will have the least significative digit as 1 for each base then n! will have surely all 0 in the least significative place so increasing n! by 1 will make the least significative digits to 1
My geometric approach on 1 question:
x(x - 1) = y(y - 1)
It is basically the same equation on both sides. One can reformulate the question like that:
On which points does the function f(t) = t * (t - 1) get the same value?
If we plot the graph we see that it is parabola, so the function is symmetric relative to the middle point of the roots (0 and 1) of the equation. So f(x) = f(y) when x and y are symmetric relative to point 1/2 => x - 1/ 2 = - (y - 1/2) => x = 1 - y
One can show that there is no other 2 points on which f gets the same value, using monotonicity of f(t) on (-infinity, 1/2] and [1 /2, + infinity).
And there is one more case when x and y are the same point, i.e x = y.
Well that's a good approach, but again, I doubt it would be easy to come up with it when you are on the spot when you are nervous during the interview.
@@mathemaniac Totally agree, I just thought it was interesting approach so decided to share. Love your videos btw.
Are you still in St Edmund Hall, I would love to meet up
Shouldn’t be telling him he’s right on the steps along the way and giving hints - unless that occurs in the actual interviews.
I was feeling so stressed while Tom was doing the second problem, I can't imagine how stressed he might have been...
How to find the center of a cercle using just a compus? 🤔
Hello, I am a Chinese students that are applying to study math at cambridge, I really want help with my interview. Can I sincerely ask for chances to do mock interview with you? (I am willing to pay for this)☺
Regarding Q1: y(y-1) = x(x-1) is just a quadratic equation y^2-y-x^2+x=0 which can be solved by y = 1/2 ± sqrt{1\4+x^2-x}=1/2 ± (x-1/2). The rest follows easily.
Q2 was the most interesting IMHO. The only integer possibly prime in the interval [1000!, 1000!+1000] is 1000!+1. Every other integer 1000!+k is divisible by k. In Q2b (the trick with interval shifting) I thought of a function P(n):= (# of primes in [n, n+999]) for which P(n+1) = P(n) + x with x ∈ {-1,0,1}. So every integer between P(1) and P(1000!+1) must be hit at some n. 10 of course also.
I solved Q1 in less than 2 minutes, Q 2 in about 5 minutes, Q3 also. Q3 was not so much about physics rather than geometry.
Step by learning math from visualisation is it possible for every branch of mathematics?
@Dnomyar Akunawik 😂😂
The way I would write a proof for the "can you find 10 primes?" question: If you write the primes in order {p_i} and then take 1000!+2 as the upper bound to the list. We know that there are no primes in the interval (1000!+2,1000!+1000) - and there are more than 10 in the interval (1,1000), so shifting the window down from (1000!+2,1000!+1000) and considering the function p(n) to be the number of primes in the interval, it always increases or decreases by 1 for each number we slide down (either a prime slides in to the bottom or slides out of the top), and the interval (2,1000) contains more than 10, there will be some interval in between with exactly 10 primes.
Nice. Thanks for writing out, clearer than explanation in vid. It is a bit like intermediate value theorem reasoning.
What you suggest some one who want to become master in maths from zero to hero?
For the second question, can't you just use negative numbers for the intervals, seeing as no negative number can be prime?
So in the case of part I, we can have an infinite number of intervals with only consecutive negative numbers that fit the requirements (can also throw in 0 and 1 into the intervals for fun, since they're not prime).
And in the case of part II, just start from 29 (which is the tenth prime number, starting from 0) and go backwards, i.e. [-969; 29] and [-968; 30] being correct solutions.
Trivial solutions should be discounted/if you were to say this they'll just say what about in the positive domain
It is quite likely that in the real question (the one from the Cambridge test) they used the notation "positive integer", and this may well have been left out accidentally by the videomaker.
This question gets to be a kind of a trick question if you admit negative integers, as many people would instinctively only consider positive integers. But trick questions are not usually part of the test, as rumor goes.
l love the fact that pi becomes 22/7 as soon as physics becomes the topic
The physics question could be solved with a quite simple method that only uses techniques up to what is learned in Algebra II/Geometry, as long as you have a basic intuition on physics.
First note that dealing with a circle and the number 22 should hint at pi being approximated as 22/7.
Since the radius is 5m, the diameter is 10m. The bullet travels at 5m/s, therefore it takes 2 seconds to reach the opposite side. We know that the angular velocity is 1 rad/s, so after 2 seconds the bullet had an angular displacement of 2 rad. To travel a full circle, the angular displacement would be 2pi radians, meaning that the bullet hit the other side after 1/pi rotations of the centrifuge. If we now approximate pi as 22/7, our 1/pi expression turns into 7/22, meaning that the bullet hit the other side after the seats had rotated seven seats in the direction of rotation. Therefore, the seventh seat from seat 12 (the seat opposite the first seat) in the direction opposite rotation is the seat that the bullet hits. Since 12-7=5, the fifth seat would be the one that is hit.
Hope this gives a somewhat simple explanation to what may seem like a daunting problem!
Bro at timestamp 47:38 he says "4/7 rads thus it is person 3". Can you help me explain to me why that is 3 not 5. Cause according to my understanding, the 4/7 suggests it is 4 gaps between me and the person I hit. Since I am person 1, 4+1=5 is the person I hit. So why is it person 3?
@@stinkypinky6010 I'm not sure why he said that because a few minutes later he went on to change his answer to person 7. Honestly, I just did the question and skipped to the end to see if I got the same answer so I didn't originally see this part. However, 4/7 radians would be person 3 (although this is not the correct answer, I believe you are trying to understand why this is the case), and we know that there are 2pi radians in a full rotation/circle. Earlier, we approximated pi to be 22/7, meaning that there are about 2*22/7 or 44/7 radians in one circle. Since we have 22 seats, we would have to split this circle up evenly into 22 parts, so we could divide 44/7 by 22 and get 2/7. 2/7 would signify the number of radians between seats. Since we are assuming the person shooting the bullet is person 1, if we were to move 4/7 radians (the same as 2*(2/7) radians), we would know we moved over two seats. 1+2=3, hence why he said it would be person 3. Remember that person 3 is the wrong answer, but I hope this helped you understand why person 3 would be 4/7 radians!
@@dannychristensen295 thanks bro I really needed that
@@stinkypinky6010 No problem!
That physics problem is actually very very easy, since you are sitting opposite person 11 and the bullet has traveled 7 seats "left" (22/7 is pi in this example and we have a time of 1s for a lenght of 5m with a constant/average velocity of 5m/s) than you would be hitting person 4 because that's 7 seats to the left of the person you were sitting opposite to at the time of firing. Could anyone explain why I'm maybe wrong/misunderstanding the question?
Since on question 2 we were talking about integers and not naturals, aren't negative numbers all composite?
As such part a could be with any interval containing purely negative numbers and part b could be answered with (10th prime number - 1000, 10th prime number)
Or did I miss something?
I'm assuming you mean that because a negative number is -1 * number, then it is not prime. However, if we were to allow this definition then for some prime such as 5 you could say it is equal to (-1 * -5), and as both -1 and -5 are composite according to your definition, the product is also composite hence 5 is not prime which is a rather silly argument.
It all depends on whether your definition of a prime includes negative numbers or not. Furthermore, even if you were to pull this trick out in an interview, I'd wager they'd just tell you to consider positive integers or natural numbers.
@@mouldyvinegar1854 True, I'm not super well versed in number theory. I just said what came to mind.
Still, under the assumption of including negative numbers in our definition of primes, only -1 could be considered prime as -1=1 x -1 but not -5 as -1x5=-5 therefore being composite. So 5=-1x-5 is not a valid prime factorization.
However I agree that negative primes cause some issues: -1 x -1 x 5 is also 5. Maybe it would make sense to make 1 composite in the integer prime set
But thinking again, 1 has infinite 'prime' factorizations as (-1)^2n so maybe integer primes are doomed frome the start
Either way, the fact they mentioned integers and not naturals in the question was strange to me.
@@mightyhelper8336 The most likely explanation is that saying integers instead of positive integers or natural numbers was a slip of the tongue. If it was intended, I feel they would have made a remark about the negative integers.
For 2° problem: What about this: take the last 10 prime numbers from 1 to 999 and exclude them from the calculation of 999! (So, you multiply 989 numbere instead of 999. Now, this number+999 will have exactly 10 primes
Got the same physics in my Cambridge interview even though I didn’t take physics nor advanced mechanics so rip 💔
But very interesting questions!
I hope you don't mean that it is the exact same physics question... just that it means you have physics in your interview.
Hmmm that really should not happen that Cambridge actually tests you something you don't know. Let this be known for any future applicants: you can say that you haven't covered this in your courses before in the interview, if this happens.
The easiest way to think about the interval of 999 non-primes problem is: If we find a highly composite number n,
n is divisible by 2, so n + 2 is divisible by 2
n is divisible by 3, so n + 3 is divisible by 3
n is divisible by 4, so n + 4 is divisible by 4
n is divisible by 5, so n + 5 is divisible by 5
...
congratulations. jolly good fellow.
I’m confused on the physics section.
Why couldn’t we just say that the bullet will travel the distance in 6m/5m/s -> 1.2 seconds
Then say that the centrifuge will rotate 1.2 radians in that time. Realize that each person is 1/7 radians so it’s
1.2/(1/7) meaning that it would have rotated 8.57 chairs hence 8 chairs because half of a chair is still on 8. Then know that the person directly across from you is just half the circle, which is 11 chairs. Since we’re moving counter clockwise, it will hit person 3.
This is unnecessarily hard, I have never had to use anything this hard in the real world.
in the last question shouldn't 45° be π/4 radians?
bro was sweatingggg