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HSC 2012 MX2 Marathon (archive) (1 Viewer)

bleakarcher

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Re: 2012 HSC MX2 Marathon

^I know right...why can't it be white?
 

IamBread

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Re: 2012 HSC MX2 Marathon

Guess I'll contribute a question as well.. :rolleyes:



 

nightweaver066

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Re: 2012 HSC MX2 Marathon

n=6 (assuming n=0 isn't allowed).

Given that w is the complex root of z^3=1 with smallest positive argument, evaluate:
(1-w)(1-w^2)(1-w^4)(1-w^8)
= (1 - w)(1 - w^2)(1 - w)(1 - w^2)

= (1 - 2w + w^2)(1 - 2w^2 + w^4)

= (-3w)(1 + w - 2w^2)

= (-3w)(-3w^2)

= 9w^3

= 9
 
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deswa1

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Re: 2012 HSC MX2 Marathon

Is this a past HSC question? I remember doing this exact question about two days ago. I'll let someone else post a solution though.

With your previous post, the answer is a nice integer. You made a mistake in the second line where you changed (1-w)(1-w)(1-w^2)^2 into (1-w^2)^3
 
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nightweaver066

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Re: 2012 HSC MX2 Marathon

Is this a past HSC question? I remember doing this exact question about two days ago. I'll let someone else post a solution though.

With your previous post, the answer is a nice integer. You made a mistake in the second line where you changed (1-w)(1-w)(1-w^2)^2 into (1-w^2)^3
Yeah it is a past HSC question.

Thanks. Got too excited looking for a difference of two squares.
 

deswa1

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Re: 2012 HSC MX2 Marathon

The answer to my question, a few posts back
 

deswa1

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Re: 2012 HSC MX2 Marathon

5 minutes to work out. 1 hour to type up. 30 minutes to post. I need to get better at latex.
1 hour's good by my standards for that length. It took me 25 minutes to type up a solution to the fourth roots of unity (about 4 lines of working). I've quit latex now...
 

IamBread

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Re: 2012 HSC MX2 Marathon

1 hour's good by my standards for that length. It took me 25 minutes to type up a solution to the fourth roots of unity (about 4 lines of working). I've quit latex now...
Haha yeah latex takes a bit to get used too...

 
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