HSC 2012 MX2 Marathon (archive) (3 Viewers)

rolpsy

Member
Joined
Apr 9, 2011
Messages
94
Gender
Male
HSC
2012
Re: 2012 HSC MX2 Marathon

centre: (1,√3)
radius: 2
√3.


 

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,256
Gender
Male
HSC
2012
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)
 
Last edited:

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
Re: 2012 HSC MX2 Marathon

Also smallest positive integral value.
 

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
Re: 2012 HSC MX2 Marathon

Level I


Level II


Level III

TBA tomorrow when I am bothered.
 
Last edited:

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
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

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
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

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
Re: 2012 HSC MX2 Marathon

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

IamBread

Member
Joined
Oct 24, 2011
Messages
757
Location
UNSW
Gender
Male
HSC
2011
Re: 2012 HSC MX2 Marathon

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



 

nightweaver066

Well-Known Member
Joined
Jul 7, 2010
Messages
1,585
Gender
Male
HSC
2012
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
 
Last edited:

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,256
Gender
Male
HSC
2012
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
 
Last edited:

nightweaver066

Well-Known Member
Joined
Jul 7, 2010
Messages
1,585
Gender
Male
HSC
2012
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.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 3)

Top