Finding yourself at the culmination of two years’ anticipation and attempted preparation is oddly giddying. I’d visualised the beginning of my interviews so many times and in so many ways that standing in a dingy staircase in Blue Boar quad counting down the clock was absurdly low-key. For something that I myself had long-since defined as an important crossroads in my life, there should have been more drama. Maybe a murder. Something like that.

There I was though, sitting in one of those mass-produced plastic chairs that are ubiquitous through school, making forced conversation with some posh twat interviewing for History of Art or some bullshit, whose Mum turned out to be from the same town as me. Looking back, it was sort of disheartening how utterly flabbergasted he was to have come across another person who came from the town.

The interviewee before me scuttled out of the room and gave me no more than an inscrutable glance before going down the stairs. My competition; my rival. It’s strange to think that now, as we became, and remain to this day, great friends and shared most of the ups and down of uni. But in that moment, statistically, there was half a place that the two of us were going head-to-head for. Grr.

None of this was helpful to be thinking about, of course. A few minutes later I was sitting across the desk from the two interviewers, with a few sheets of blank A4 and a couple of biros between us. One was a tutorial fellow of the college who I’d stalked at length, the other a younger post-doc type. Someone had drawn a Gaussian and the number “1089” on the whiteboard. The manner of my showing in struck me at the time as bizarrely muted – was I in the bad books already? It didn’t occur to me until years later that the majority of professional physicists are simply awkward AF.

With very little in the way of preamble, the first question was forthcoming.

# Question 1

Imagine that you have a length of fence, which you can bend at arbitrary points to make “corners”. You also have a wall, whose length is very much greater than the length of the fence. You’re going to use the fence to make three sides of a rectangle, and the fourth side will be some of the wall. What’s the maximum area you can enclose, and how do you arrange the fence to do that?

That was such an odd thing to hear said out loud that I thought I’d better draw a diagram to confirm whether I’d even understood the question…

(I know. Should have been an artist.)

The free variable was to be the length of wall incorporated, labelled x. Given that our enclosed area is rectangular, then the opposite side of the rectangle must also be of length x. If we denote the length of our fence by a, then we must have (a – x) length of fence left over for the “sides”. This must be shared equally, so each “side” must have a length of $(a-x)/2$.

Bringing out some pure Year 7 Maths gold, I proceeded to declare that the area must then be

$A=\frac{x(a-x)}{2}$

All going well. There were a few weary, affirmative nods. The lad is shit hot on rectangles. No worries there.

So what I wanted to find out was what value of x would result in the maximum A, which sounded comfortingly like a “differentiate, set to zero to find stationary point” thing. I announced that this was what I was going to do and sincerely hoped that they wouldn’t want me to go to first principles and show why that gave rise to a max/min. Thankfully they were happy with me just going ahead and doing it, so I wrote down the agreeable-looking expression

$\frac{d}{dx}\frac{x(a-x)}{2}=0$

Firmly in my comfort zone by this stage, I ploughed on…

$a-2x=0$

$x=a/2$

I underlined this with a flourish. There were frowns and tilting of heads as the tutors attempted to read upside down what I’d written. “So… what’s the area?” one said. Good point. I hadn’t actually answered the question. My value for x represented the length of wall which gave you the maximum area, rather than the area itself.

$A_{max}=\frac{(a/2)(a-(a/2))}{2}$

$A_{max}=\frac{a(a/2)}{4}$

$A_{max}=\frac{a^2}{8}$

Yes? Yes. Great. That went well.

Can you draw a graph of the area enclosed as a function of x?

Erm… Well… My eyes strayed towards the Gaussian on the whiteboard. But that couldn’t be right. It must be a parabola. And it must have zero value when $x=0$ and $x=a$. AND I’VE MOTHERFUCKIN JUST FOUND the max point! With those three points I sketched something. With $a=4$ it looks like this:

The interviewers nodded to each other. From that, I took that my answer sufficed. HA! Gaussian, pfft.

The younger of the two interviewers piped up to move us onto the next question.

Right, let’s go on to something that’s not easy.

A layer or two beneath the nerves I was feeling, I felt a ripple of anger at this. “Don’t feel pleased to have done that because it was easy. This won’t be, though.” Since that night I’ve been on the other side of the interview table many, many times, and I can’t imagine being motivated to undermine or discourage someone like that. Anyway, there was no time for that, because…

# Question 2

Choose any 3 digit number. The only constraint is that the first digit within it has to be bigger than the third. (Eg, 321). Reverse the order of its digits, and subtract this from the original number. (Eg, 321 – 123 = 198). Reverse the digits of this new number, and add the result of that, to the new number. (Eg, 198 + 891 = 1089). You’ll notice that the answer has been written on the whiteboard behind you for the duration of the interview. Why is it always 1089?

Now, I won’t lie, my initial reaction was simply to think “oh shit”, because I had absolutely no idea and didn’t really know where to start. The proof is as follows, and eventually I got to it, but it was very much with the prompting and cues or the interviewers. It would not be fair in any way to say I proved it myself, even laboriously.

The reason it happens is because the process of reversing and subtracting twice effectively recovers and cancels out the original digits you chose, but with a few artefacts left over from all the dicking about reversing them (flipping digits between the hundreds and units columns), which sum to 1089.

Denote your initial 3 digits (comprising your number) as A, B and C. (Recalling that it is stipulated that $A>C$.) The number is then given by

$100A+10B+C$

Following the sequence of steps set out in the question, then,

$(100A+10B+C)-(100C+10B+a)$

$=100(A-C)+C-A$

Now we are instructed to reverse this number, which will be a 3 digit number because of the constraint $A>C$. We need to therefore rearrange it slightly to represent it in units of hundreds, tens and ones, and therefore be able to algebraically reverse it…

$=100(A-C-1)+10(10)+(C-A)$

But again, $A>C$, so we have to borrow a 10:

$=100(A-C-1)+10(9)+(C-A+10)$

Reversing again,

$100(C-A+10) +10(9) +(A-C-1)$

And adding this to the last variation, we get

$((100(A-C-1)+10(9)+(C-A+10))+(100(C-A+10)+10(9)+(A-C-1))$

$=100(A-C-1+C-A+10)+10(9+9)+1(C-A+10+A-C-1)$

$=100(9)+10(18)+1(9)$

$=900+180+9$

$=1089$

With the tutors prompting me through, I was as surprised as anyone when this number appeared on my page, but made some agreeable “ahhh” sounds of understanding. You can see just from the above that all it boils down to is addition and subtraction algebra, but of the kind that’s awkward to keep track of when you’re nervous and under pressure.

There was an awkward pause where nobody made eye contact. Obviously I felt pretty foolish. Apparently not being able to bring themselves to speak to me about maths and physics any further, The Man Who Found Things Easy piped up with “I’ll show you out, then” and proceeded to walk carefully to the door, open it, and watch me through it, all in utter silence. It was weird.

And with that, I was one interview down. I had literally no idea how I’d done. It was about half past five in Oxford in the pit of mid-winter. I headed back to the room I’d been loaned to de-suit and whiled away the 21 hours until my next interview. Read all about it in part 2…

## One thought on “Oxford Physics Admission Interview 1: Questions and Answers”

1. When I actually used this question in interviews, no-one actually got as far as an actual ‘X degrees C’ answer in the ten minutes or so we allowed for it, nor did we expect them to. We use this sort of question to try to find how applicants think about problems, and how they might operate within a tutorial.