Official TSR Mathematical Society

Here is my attempt at question 4:

Based on fiddling about using N=8 and N=10 and arranging them in a circle different ways, I am convinced the minimum $V=2(N-1)$ and the maximum is $V=\sum_{1}^{N-1}(N)+\frac{N}{2}$

I have no rigorous proof.

My thinking is that if you place the numbers side by side in the order 1, 2, 3, 4, ..., N then there is a difference of 1 between each pair, and there are N-1 of these pairs. There is also the pair at the 'end' of the circle, which consists of (N, 1) and this pair has a difference of N-1 too. So $V=1(N-1)+(N-1)=2(N-1)$




If I am wrong (which I expect I am given its a round 2 BMO question) I would love to be shown how to answer these sorts of questions correctly!

That is what I had in mind for the maximum too, I just didnt know how to express it as a proof. While toying round, I arranged my numbers like this for my 'maximums':
..........2
...8............7
1..................3
...5............6
..........4

......2.....9
..10..........3
1.................8
...6...........4
......5.....7

Which is basically this idea of 'peaks' and 'valleys'.

$\displaystyle\frac{1}{r}\frac{d}{dr}(r\frac{dv}{dr}) = -k$ where k is a constant. Find the general solution of v.

-kr^2/4+A?

Working out?

in my head



Too easy bruv

I have a simple puzzle which I doubt anyone here will be able to solve.

7 men, A B C D E F G are staying in a hotel for 15 days. Every day the men gather to dine round a circular table. t No man is never sat next to the same 2 neighbours on more than 1 day. Can you come up with a seating arrangement for the 7 men on the 15 days. There is a solution, but none found it.

Gambling puzzle


There's a gamble that costs £4 to enter. In the gamble you must flip a coin n-times. The number x is chosen at your will. You start with £2 and every time the coin lands heads you quadruple the money you currently have. If the coin lands tails you get nothing at all. So What number x should you pick?

After the first integral you'll get an additive constant, so dv/dr = -kr/2 + c/r. So there's an extra c log r term in the answer.

this might not be an interesting maths problem, but I find it hard and would appreciate a step by step guide as to how to approach and solve this problem

so it's simultaneous equations, -

x^2 + 4y^2 = 10
2y+3x =10