Oxford MAT 1992-1995 solutions thread.

Anyone have any idea where I can find the answers to the 1999 and 2002 papers? the papers are here http://www.mathshelper.co.uk/oxb.htm

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1994 paper I B5

1995 Paper I Question B5

I though this one was a bit different to B2,B3 (which are bookwork tbh). Not totally confident of my waffly solution. So would appreciate anyone checking it, and if correct please add it to the first post Farhan.

Looks good to me- I jsut worked through it without looking at your solutions so it's interesting to see where we were different: I did part b) using (alpha,beta)=(1,-1). I think part e) is much shorter than you made it: d is defined as the min such that d>0, and r=<d-1. Hence r=<0. But we know r>=0 from c). Hence r=0.

On part c) I said Choose $\lambda \in \mathbb{N}$ such that $\lambda d \leq p<(\lambda+1)d$ and it quickly follows

Finally, on part d) I said $d=\frac{p-r}{\lambda}$ and equated that with the first equation they gave

1995 Paper I Question B5

I though this one was a bit different to B2,B3 (which are bookwork tbh). Not totally confident of my waffly solution. So would appreciate anyone checking it, and if correct please add it to the first post Farhan.

Thanks for checking..

(Maybe I'm failing to see something obvious, but...) For part e): "d is defined as the min such that d>0, and r=<d-1. ... ... Hence r=<0", how you deduce that "r=<0" for the first part? Surely d does not always have to be 1 for all p and q?

You need to be careful here - it could happen that this gives d < 0.

If r < d, and r > 0, it would not be true that "d is defined as the minimum ... with d > 0".

You have 1/Det A, not Det A (you should not have taken the reciprocal).

Yeah I know. Considering that p and q are symmetric in this problem I assumed WLOG that $p \geq q$ and proved the stronger result that $d \leq p$ or $q$

You could still have d = 0 though. (I haven't seen what you actually wrote, so you may have ruled that out). It's definitely simplest to just go with $\alpha = 1, \beta = 0$.

Comments inline, in red.

I did them, so could compare answers.