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Re: MX2 Integration Marathon Actually dw, was about to suggest the possibility of a typo...but IBP does work.

Re: MX2 Integration Marathon What is the source of the question as posted here?

The exact formulae for this class of problems can be found here by the way: http://en.wikipedia.org/wiki/Necklace_(combinatorics). I have not tried to prove these myself, but the idea would be to separately count how many strings are in each equivalence class using elementary number theory.

An amusing corollary of FLT is that the n-th root of 2 is irrational for n > 2. For if (a/b)^n = 2 for some integers a,b, then a^n = b^n + b^n provides a solution to the Fermat equation ;). Sadly, Fermat's last theorem is not strong enough to prove that the SQUARE root of 2 is irrational.

It's a shame. I think elementary number theory would be an excellent thing to teach high school students.

Discovery Freelancer 4.86, my housemate started up a private server. An excellent mod of a very underrated game.

Don't study 24/7, you will hate life. Just study smart and be efficient when you actually are supposed to be working, for the HSC that is enough to achieve all but the very highest of goals.

Of course there are several unanswered questions like how things change when we consider bracelets instead of tables, and how things change when beads are of > 2 different colours that might be fun to think about.

I will try to explain what's going on here. (Apologies if it is not entirely coherent, I am not particularly good at combinatorics.) One way of thinking about the division when counting things like circular arrangements is that we are counting groups of objects that are in some sense "the...

http://www.xamuel.com/inverse-graphing-calculator.php?phrase=79.9

Of course, I meant the question to read as if you (the reader) are the contestant, and hence know everything that has been written.

Many of you will have heard of the Monty Hall problem, presented below: 1. There are three briefcases, two are empty and one contains one million dollars. Monty knows which one contains the prize. 2. The contestant chooses a case but does not open it. 3. Monty opens an empty case out of the two...

Re: HSC 2013 4U Marathon It doesn't even have to have those roots. Eg P(x)=x+1.

Re: HSC 2013 4U Marathon And the fact that you are taking the sine/cosine of a geometric sequence of angles.

Yeah I think there are, but a lot require manual intervention etc. And as the modern world will be pretty different starting from 1066 (or earlier with old gods) I think I may as well just play each game on its own...they all offer something different.

Re: HSC 2013 4U Marathon Haha it is quite nice.

Re: HSC 2013 4U Marathon By the product condition, the inequality can be rewritten as: (abc-1) + (a+b+c) - (ab+bc+ac) > 0 <=> (a-1)(b-1)(c-1) > 0. Hence either one or all three of the LHS factors must be positive. They cannot all be positive because of the product condition, and so we can...

Yep. Although I suspect engine differences may make EU4->Vic2->HOI3 more problematic. In any case, these games all seem excellent as standalones. Am ready to embrace my inner aspie and sink many hours into them.

Have caught the Paradox Interactive bug. Currently playing CK2 and am very interested in playing around with the others, possibly doing some kind of mega campaign.

memorising and repeating procedures without really understanding.