Oxford MAT 2013/2014

I'm quite puzzled by the admissions statistics... Why is the competition for Mathematics & Statistics so much more intense than that for Maths/Maths & CompSci/Maths & Phil?

It's a shame we can't visit upstairs though



To be honest, 'Why Maths?' and 'Why Oxford?' doesn't sound like a great question to ask in an interview when they could be spending the limited time they have assessing your ability to do maths. Needless to say, do think about it but don't get too worried about it. Enjoy it!

How likely is it that we'll get asked questions on A2 level vectors/mechanics etc? I'm focusing my attention on sharpening up proof/calculus/graph sketching, should I revise the rest of the syllabus aswell?

Are all the colleges supposed to have replied by now since its a week before the interview period?

Yes.

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Holy **** Soutik - no wonder you are so good- you made it to India IMO shortlist according to stalking page- just reading through now and spotted it.

It's easier in India than in your country, trust me. India generally ranks 11-30 at the IMO, not in the top 10. Going to the selection camp here isn't half as big as going to the Trinity camp in the UK. Also, India shortlists 30+ people, much more than the UK's 20.

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The answer to the maximum limit is (N(N-1))/2 + floor (N/2). Have you managed to prove it yet? If so I'd be interested to see your method.

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You could have just said floor N^2/2.
Yes, I've proved it. I'll post the solution in LaTeX as soon as I get back home and have a proper internet connection. In very crude words, here's the main idea- when you expand V, you get each of 1 to N twice, preceded by either a + or a -. Number of + equals number of -=N. Best case is to ensure the highest N are + and the lowest N are -. Construction is easy when you get to the conditions.

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Yeah that sounds right. You can use the triangle inequality to deduce a similar argument but ultimately you construct one increasing sequence starting at the first term and one decreasing sequence starting with the second term I believe.

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I suppose that'll work. There are many valid constructions. Can you explain how you were using the triangle inequality for the first part? You said the sum is greater than or equal to twice a. What was your a, I never figured that out.

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Okay, so write out the arrangement as the sum of differences starting at mod(A-B) and ending at mod(Z-Y). By applying the triangle inequality on each pair of successive terms we get mod(A-C)...+mod(Z-X) + (mod(A-Z)) where we simply add on the last term. When we use the triangle inequality again we get the sum of the differences is greater than or equal to two differences between two random terms. At most a difference is N-1 so we get the max is 2a or 2(N-1), we can do this since one of them is between non-successive terms. A bit convoluted but it works nonetheless for even cases. For odd cases there are some minor alterations but the method is still valid.

What are we supposed to take with us btw?

Okay, that's what I thought, but I'm not sure if this is complete. How do you deal with the cases in which a=N-1 is not achieved between two consecutive terms? Even if you deal with that case, you need to show that V=2(N-1) can actually be achieved for some configuration. The construction is trivial, but necessary.

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Doubt they will ask questions on mechanics. I'm focusing my attention on graph sketching and number theory. What kind of proof are you doing?

Not really that difficult to fix tbh. I made a typo in my previous post which is probably why you picked up on this. It should be the sum of differences from mod(A-B) to mod(Y-Z) and we add on mod(A-Z) however applying the triangle inequality we get the differences are greater than 2*mod(A-Z)which means we only need 1 difference to be N-1 which can be easily constructed.

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