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HSC 2015 MX2 Marathon (archive) (1 Viewer)

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Drsoccerball

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Re: HSC 2015 4U Marathon

NEXT QUESTION

Each of a set of equal cubes has its six faces painted red, orange, yellow, green, blue and white respectively so that no two cubes are alike and every arrangement of colours is used. How many cubes are there in the set?
does the cube have to have all those colours at once ?
 

dan964

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Re: HSC 2015 4U Marathon

my guess
(1) One side will always be painted white, so it is irrelevant what side this is.
(2) We pick now the colour OPPOSITE the white side, as this will also be unique with rotation = 5
(3) And then we rearrange the remaining colours without repetition, which is 3! (we consider it as a circle, so not 4!)
5*3! gives 30.

method 2 is to pick the 4 colours excluding the ends which is 6C4
and then arrange the remaining two colours (which is 2!)
 
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InteGrand

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Re: HSC 2015 4U Marathon

Tried to make a question again:





Form a quartic equation with roots that are reciprocals of those for the given equation (when you do this, the coefficients get "reversed", you'll see).

Then use sum of roots for the given equation, divided by sum of roots for the new formed equation, it'll be 1, since when you reverse the coefficients of the given polynomial equation, the coefficients of and remain unchanged.
 

Drsoccerball

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Re: HSC 2015 4U Marathon

Form a quartic equation with roots that are reciprocals of those for the given equation (when you do this, the coefficients get "reversed", you'll see).

Then use sum of roots for the given equation, divided by sum of roots for the new formed equation, it'll be 1, since when you reverse the coefficients of the given polynomial equation, the coefficients of and remain unchanged.
DAMNNNNNN i didnt do it that way but thats genius
 

Ekman

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Re: HSC 2015 4U Marathon

You do realise that from product of roots
Anyway next question.






And for the next part just find the imaginary bit by realizing the denominator, which should be:

 
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braintic

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Re: HSC 2015 4U Marathon





And for the next part just find the imaginary bit by realizing the denominator, which should be:

The restriction for the limiting sum of a GP is r must lie between -1 and 1.

That is ... r must be REAL (To use that formula anyway)

This has no limiting sum.
If x is a rational multiple of pi, the partial sums follow a cyclically repeating pattern.
If x is a non-rational multiple of pi, there is no repeating pattern, but it certainly doesn't converge to a limit.

To have a limiting sum, the MODULUS of successive terms would at the very least have be reducing.

And sin x + sin 2x + sin3x + ... most definitely does not have a limiting sum. ( Except for the trivial case x = k pi )
 
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braintic

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Re: HSC 2015 4U Marathon

Your expressions include an 'n' representing the number of terms.
So they can't possibly be limiting (ie. infinite) sums.

You have to find limits as n->infinity, and these limits do not exist.

Try x=pi/2:

cisx=i
cisx+cis2x = i - 1
cisx+cis2x+cis3x = i - 1 - i = -1
cisx+cis2x+cis3x+cis4x = i - 1 - i + 1 = 0
And then the cycle repeats - there is NO limit.
 
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VBN2470

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Re: HSC 2015 4U Marathon

a little change would make a good question:



EDIT: Not sure if the RHS value is correct, or if I did something wrong and am missing something.
 
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