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HSC 2015 MX2 Marathon (archive) (2 Viewers)

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Drsoccerball

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Re: HSC 2015 4U Marathon

Er, are you really sure you have a nice expression for this?

I am a bit sceptical. You should probably phrase it as a "prove that ..." question so people can easily check that the question works before spending time on it.
If we give the answer the solution is easy
 

seanieg89

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Re: HSC 2015 4U Marathon

I only ask how sure you are because of how often I see people (especially students) make up their own questions that don't work.

Hopefully (mostly for the sake of students time) this is not such a question, but I am still a little doubtful that you will be able to find an objectively nicer expression than a (k+1)-fold product for this quantity.

Happy to be proven wrong though!
 

Ekman

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Re: HSC 2015 4U Marathon

Check your solution by substituting some outlandish values of n (such as n=4, 7, 11 etc.)
An issue that sean raised up from the question is that k isn't a random value, but instead k=n-1. And n must always remain even.
 

FrankXie

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Re: HSC 2015 4U Marathon

An issue that sean raised up from the question is that k isn't a random value, but instead k=n-1. And n must always remain even.
yes only good for even number n and k is half of n.
the idea is absolutely neat. well done.
 

Sy123

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Re: HSC 2015 4U Marathon

Solution:










Good question and nice solution!

I don't think you have a full solution though, as your "k" is only "n/2" (what if n is odd?), instead of it being (n-1)
So I would square that result you have, because

And then you would have to account for double counting and make separate cases for when n is even or odd

Also, these questions are better suited for the Advanced Level thread, if your question is too difficult for Q16 of the HSC, then it is better to put it in the Advanced Thread
 

Drsoccerball

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Re: HSC 2015 4U Marathon

Good question and nice solution!

I don't think you have a full solution though, as your "k" is only "n/2" (what if n is odd?), instead of it being (n-1)
So I would square that result you have, because

And then you would have to account for double counting and make separate cases for when n is even or odd

Also, these questions are better suited for the Advanced Level thread, if your question is too difficult for Q16 of the HSC, then it is better to put it in the Advanced Thread
When making it, it didnt seem that hard ahaha sorry bout that
 

seanieg89

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Re: HSC 2015 4U Marathon

When making it, it didnt seem that hard ahaha sorry bout that
The issue was not so much difficulty as you not saying that:

a) k=n-1
b) n is even.


b) is not a big deal, but a) certainly is. I still don't think that the question (as originally posed) is doable.
 

Drsoccerball

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Re: HSC 2015 4U Marathon

The issue was not so much difficulty as you not saying that:

a) k=n-1
b) n is even.


b) is not a big deal, but a) certainly is. I still don't think that the question (as originally posed) is doable.
a)Franky edited that from the beginning. It was the first question we made so go easy on us
 

seanieg89

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Re: HSC 2015 4U Marathon

a)Franky edited that from the beginning. It was the first question we made so go easy on us
Neither of you confirmed his guess or corrected my post "well, the first (k+1) of them anyway".

I am not trying to be mean or anything, I'm just trying to impress on you the importance of certainty when it comes to making your own questions. People won't want to invest time trying to solve questions you post in the future if there is a history of the questions themselves not being correct. (Especially if the nature of the question is such that they cannot "test" it).

It's great to see students being creative with high school tools :).
 

Drsoccerball

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Re: HSC 2015 4U Marathon

Made this question by myself:
 
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VBN2470

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Re: HSC 2015 4U Marathon

NEW QUESTION

 
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