Re: HSC 2015 4U Marathon - Advanced Level
Just saw this.
Take any cube mod 10 and you get 0,1,8,7,6,5,4,3,2,9... as your cycle.
From here it is obvious that the equation can't have an integer solution.
Re: HSC 2015 3U Marathon
Take an arbitrary element and don't choose it. The you get nPk of choosing rest.
Now choose that arbitrary element, you get latter term.
Re: HSC 2015 4U Marathon - Advanced Level
Wait this might actually work.
My brain is a bit shut off after exams though, gonna go through my solution and try clarify/debunk it.
Re: HSC 2015 4U Marathon - Advanced Level
For example, if we choose an arbitrary subset of size 5 from 9 elements, then it has 126 possible combinations mod 5.
Since there are 126 possible subsets, then if each subset were unique we'd be done.
Though I'm fairly sure they aren't unique :(...
Re: HSC 2015 4U Marathon - Advanced Level
My exact thoughts.
Something to do with the fact that a_1+...+a_5=a has 126 solutions for a \in {0,1,2,3,4}.
Compared to \binom{9}{5}=126 which is the amount of subsets size 5 of a 9 element set.
This was my first thought, not sure if it would...
You'll never see the financial stuff in the 3U HSC.
Also I don't get what everyone finds so hard about perms and combs. There are much harder topics imo.