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Re: 2012 HSC MX2 Marathon Ah right I see, okay I guess it is a fair stretch then, maybe a bit of wishful thinking can lead somewhere ;)

Re: 2012 HSC MX2 Marathon Thanks, but it is related to the HSC albeit a fair extension. You guys do combinatorics in the HSC right? The above question can be solved purely combinatorially.

Re: 2012 HSC MX2 Marathon Why don't we add some pure maths questions in here, here's a fave one of mine, combinatorics A permutation \{x_1, \ldots, x_{2n}\} of the set \{1,2, \ldots, 2n\} where \mbox{n} is a positive integer, is said to have property T if |x_i - x_{i + 1}| = n for at least...

Re: 2012 HSC MX2 Marathon There is a stronger result, that every natural number greater than 1 has a prime divisor. Anyhow, several methods, first as per the question states: Assume a set Q = \{2, 3, \cdots, k-1\} where \forall \ n \in Q, then n > 1 and n has a prime divisor. Now to show that...

Reduction formulas are based on recurrence relationships. What you need to realise is that we are essentially computing some integral \int f(x, n) dx by continually reducing it down into an integral, \int f(x, q) where q<n. To find the integral, we first set n to its initial value while then...