HSC 2012 MX2 Marathon (archive) (4 Viewers)

[rolpsy]

Re: 2012 HSC MX2 Marathon

centre: (1,√3)
radius: 2√3.

 

[deswa1]

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)

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

Also smallest positive integral value.

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

Level I


Level II


Level III

TBA tomorrow when I am bothered.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Level I


Level II


Level III

TBA tomorrow when I am bothered.

This is a good idea. Having difficulty 'levels' when posting up questions is a good way to cater for different skill levels on the forums.

However, I suggest you co-ordinate the colours with the level of difficulty.

ie: Green = Easy, Blue = Medium, Red = Hard.

Black = Destroy.

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

This is a good idea. Having difficulty 'levels' when posting up questions is a good way to cater for different skill levels on the forums.

However, I suggest you co-ordinate the colours with the level of difficulty.

ie: Green = Easy, Blue = Medium, Red = Hard.

Black = Destroy.

It's always black when it comes to negative. =.=

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

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

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

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

Because nobody can read white.

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

This is a bit of a tricky question.

However looking at the quality of the questions and solutions so far, I'm quite convinced that one of ther 2012'ers will get it:

Preuni question?

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Preuni question?

Do you go to Pre Uni?

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

Do you go to Pre Uni?

Nope.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

Guess I'll contribute a question as well..

 

[cutemouse]

Re: 2012 HSC MX2 Marathon

The 'little circle' is the universal symbol for 'function composition'.

Hahahaha. That made LOL. Owned him

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

Guess I'll contribute a question as well..

Question

 

[nightweaver066]

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

 

[deswa1]

Re: 2012 HSC MX2 Marathon

Question

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

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

yeh its 9.

 

[Riproot]

yeh its 9.

No, it's 8.
No, it's 9.
Yes, that's right, I was testing you. It's 9.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

No, it's 8.
No, it's 9.
Yes, that's right, I was testing you. It's 9.

And that's a magic number.....

 

[nightweaver066]

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.