I think Peter Brown from UNSW wrote Q8b of the 2003 paper. Apparently no one got full marks for Q8 that year.
I can't see how tbh, when I did it whilst going through past papers I pretty much got all of it out (and this was quite a while back).
The pressure of doing the exam is a pretty big factor. I did around 10 ext2 HSC past papers and my lowest mark on those would still have scaled far higher than my actual exam mark. A friend of mine is a maths genius and for his study, he started doing past HSC papers from the early 90's and beat-or-equalled the first-in-state mark for every year except one, all the way to 2011. He panicked during the actual exam and lost a bunch of marks. Shit happens when you're under that kind of stress.
This. When I was doing past papers, I could hit 110+/120 pretty consistently. In the HSC, I think I got 82/100 raw. It wasn't even that I was stressed or anything, I felt fine but idk- it just happens.
I think he said that quite a few students got close but didn't quite manage it (were a few marks short). He marked Q8 too that year.
This was on the Wikipedia page for the Divisor function:
*steps back to make room for the pros*I just made this now:
Only knowledge from Polynomials really, can be done by anyone.
Nice work man. I will post another question soon (probably more about Palindromic polynomials), I first want to know what is fishy about the above question about the Harmonic numbers.IMO problems are floating around in this thread?
 Let $Q(x) = a_0 + a_1x + a_2x^2 + ... + a_{2n}x^{2n}$. \\ \\ $P(x) = (x+1)Q(x) = xQ(x) + Q(x) = a_0 + (a_0 + a_1)x + (a_1 + a_2)x^2 + ... + (a_{2n-1} + a_{2n})x^{2n} + a_{2n}x^{2n+1}$. \\ \\ Equating coefficients, $c_0 = a_0, c_1 = (a_0+a_1), ..., c_{2n} = (a_{2n-1} + a_{2n}), c_{2n+1} = a_{2n}$. \\ \\ Now $c_0=c_{2n+1} \quad \therefore a_0=a_{2n}$. \\ $c_1 = c_{2n} \quad \therefore a_0 + a_1 = a_{2n-1} + a_0 \quad \therefore a_1 = a_{2n-1} \\ \vdots$ \\ Proceeding as such, it can be shown that $a_i = a_{2n-i}$ for all integral $0 \leq i \leq 2n$ $\quad \therefore Q(x)$ is palindromic. $)
Yes, a solution to this question would be equivalent to proving Riemann.This was on the Wikipedia page for the Divisor function:
Note how it says: 'the Riemann Hypothesis is equivalent to the statement that'
Doesn't that mean that if we prove this that we prove the Riemann hypothesis, and as far as I know, the Riemann hypothesis is unsolved.....
What does the wikipedia page mean by what I bolded?
I'll let some one else try this, I got the answer but it's far too big to fit in the margin of this post.
I think he meant fishy in the sense that there is unlikely to be an mx2 level solution to this question. If there is, the person who finds it should probably spend his time sending his solution to the Clay Institute rather than typing it here.
I initially thought of taking the case of n prime and n not prime and somehow using the prime number theorem thing. But I can't get anywhere.
But we get 1000 000 for proving this then, I doubt the question is legit....
Freaking hell, why did I try and spend about an 1 hour trying to prove something that is out of the reach of even research mathematiciansObviously