Search:

I just realised for question 6, I could have written down the general formula:

\binom{2k+1}{k+1}

which would have given me extra credit marks :(

1) Prove that 'n' has to be a multiple of 2 (fairly straight forward and obvious as you need an even amount of terms so that all 1's and -1's cancel out).

2)Prove that 'n' can be a multiple of 4....

My approach:

Prove that:

2^{3^n} + 1 = p.3^{n+1}

Where 'p' is not divisible by 3.

2^{3^2} + 1 = 2^9 + 1 = 513

\frac{513}{3^3} = \frac{513}{27} = 19

Me too.

Same here, I tried to prove it with my 4 page induction :p

Came so, so close as well, was disheartening having to give up and move on as time was running out.

I was the 1st ever.

I got k=n+1

I just said "all possible combinations of 4 people from 7, ie ABCD, ABCE, ABCF, ... , DEFG"

I'm pretty sure I got question 1, 2, 5, and 6 right, with partial solutions to 3 and 4 (came so close to...

Yep.

are we allowed to post solutions?

This is what I had for question 6, I hope it's actually right lol.

Basically we want any 4 to be able to unlock it, but any 3 can not. So if we give out 4 keys for each lock, then we will have 3...

I was the first person in my school to ever do it.

I think I got question 1, 2, 5, 6 right. Hopefully anyway lol. And hoping my partial solutions to questions 3 and 4 gets me a couple marks.

So who is pumped for the UNSW comp tomorrow?

The denominator of the first one should just be:

\binom{8}{1} + \binom{8}{2} + \binom{8}{3} + \binom{8}{4}

By having the rest you are double counting as choosing a group of 3 and a group of 5...

\frac{5!}{2!\times2!}

1am me wasn't thinking straight.

The area of the triangles will be \frac{1}{2} PQ.PR

So we need to find the average length of PQ and PR. Now if (n) is odd, then we can fix the point P to be the centre of the square. The average...

My solution is I can choose the point P in n^2 ways as it can take any point on the square. Then Q and R can both be chosen in (n-1) ways as they can take any point in the row or column that P was...

Yep I get the factored version first.

In that case I get n^4-2n^3+n^2

If this is right I'll post my solution.

edit: I just tried a different method and got \frac{1}{2} (n^4-2n^2+1)

Not sure if either of these are right or why I...

By different triangles, do you mean triangles with different co-ordinates as their vertices or triangles which are not congruent to each other, or they can be congruent if they are rotated?

Pairwise distinct integers just means that the integers are distinct with respect to one another I think. Simply put, you can not use an integer between 1 and 100 more than once.

~HSC science~