HSC 2015 MX2 Marathon ADVANCED (archive) (13 Viewers)

[glittergal96]

Re: HSC 2015 4U Marathon - Advanced Level

Or is it the wording of the question that is throwing people off? Here is the exact wording of the question.

No, I understood it, am just not very good at / don't really like combinatorics so I thought I would let someone else do it . I am busy atm but can have a crack later tonight or tomorrow.

My guess is that it is a clever application of the pigeonhole principle.

 

[RealiseNothing]

Re: HSC 2015 4U Marathon - Advanced Level

No, I understood it, am just not very good at / don't really like combinatorics so I thought I would let someone else do it . I am busy atm but can have a crack later tonight or tomorrow.

My guess is that it is a clever application of the pigeonhole principle.

My exact thoughts.

Something to do with the fact that has 126 solutions for .

Compared to which is the amount of subsets size 5 of a 9 element set.

This was my first thought, not sure if it would arrive at a solution.

Edit: Oh and work in mod 5 too I should mention.

 

[RealiseNothing]

Re: HSC 2015 4U Marathon - Advanced Level

My exact thoughts.

Something to do with the fact that has 126 solutions for .

Compared to which is the amount of subsets size 5 of a 9 element set.

This was my first thought, not sure if it would arrive at a solution.

Edit: Oh and work in mod 5 too I should mention.

For example, if we choose an arbitrary subset of size 5 from 9 elements, then it has 126 possible combinations mod 5.

Since there are 126 possible subsets, then if each subset were unique we'd be done.

Though I'm fairly sure they aren't unique

And just note that out of the 126 possible combinations, exactly one gives you a multiple of 5.

 

[RealiseNothing]

Re: HSC 2015 4U Marathon - Advanced Level

For example, if we choose an arbitrary subset of size 5 from 9 elements, then it has 126 possible combinations mod 5.

Since there are 126 possible subsets, then if each subset were unique we'd be done.

Though I'm fairly sure they aren't unique

And just note that out of the 126 possible combinations, exactly one gives you a multiple of 5.

Wait this might actually work.

My brain is a bit shut off after exams though, gonna go through my solution and try clarify/debunk it.

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

While you guys work on sim's problem, I'll post another one: Find all integer solutions to x+x^2 +x^3= y+y^2

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

While you guys work on sim's problem, I'll post another one: Find all integer solutions to x+x^2 +x^3= y+y^2

only (0,0) (0,-1)
EDIT: woops

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

only (0,0) (0,-1)
EDIT: woops

even a year 10 student can merely state the answer without any kind of proof. So ill assume u couldnt solve this one!

 

[faisalabdul16]

Re: HSC 2015 4U Marathon - Advanced Level

even a year 10 student can merely state the answer without any kind of proof. So ill assume u couldnt solve this one!

what exactly are you looking for?

Hopefully nothing wrong with the way I've done it


<a href="http://www.codecogs.com/eqnedit.php?latex=f(x,y)&space;=&space;x&space;&plus;&space;x^{2}&space;&plus;&space;x^{3}&space;-&space;y&space;-&space;y^{2}&space;\\&space;f(x,y)&space;=&space;x(1&plus;x&plus;x^{2})&space;-y(1&plus;y)&space;\\&space;$Let&space;f(x,y)&space;=&space;0&space;\\&space;(1&plus;x&plus;x^{2})&space;\text{has&space;no&space;solution}\\&space;\\&space;\therefore&space;f(x,y)&space;=&space;0&space;\&space;\text{IFF&space;x&space;=&space;0&space;and&space;y&space;=0&space;or&space;-1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?f(x,y)&space;=&space;x&space;&plus;&space;x^{2}&space;&plus;&space;x^{3}&space;-&space;y&space;-&space;y^{2}&space;\\&space;f(x,y)&space;=&space;x(1&plus;x&plus;x^{2})&space;-y(1&plus;y)&space;\\&space;$Let&space;f(x,y)&space;=&space;0&space;\\&space;(1&plus;x&plus;x^{2})&space;\text{has&space;no&space;solution}\\&space;\\&space;\therefore&space;f(x,y)&space;=&space;0&space;\&space;\text{IFF&space;x&space;=&space;0&space;and&space;y&space;=0&space;or&space;-1}" title="f(x,y) = x + x^{2} + x^{3} - y - y^{2} \\ f(x,y) = x(1+x+x^{2}) -y(1+y) \\ $Let f(x,y) = 0 \\ (1+x+x^{2}) \text{has no solution}\\ \\ \therefore f(x,y) = 0 \ \text{IFF x = 0 and y =0 or -1}" /></a>

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

While you guys work on sim's problem, I'll post another one: Find all integer solutions to x+x^2 +x^3= y+y^2

[]

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

Just made this:

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

what exactly are you looking for?

Hopefully nothing wrong with the way I've done it


<a href="http://www.codecogs.com/eqnedit.php?latex=f(x,y)&space;=&space;x&space;+&space;x^{2}&space;+&space;x^{3}&space;-&space;y&space;-&space;y^{2}&space;\\&space;f(x,y)&space;=&space;x(1+x+x^{2})&space;-y(1+y)&space;\\&space;$Let&space;f(x,y)&space;=&space;0&space;\\&space;(1+x+x^{2})&space;\text{has&space;no&space;solution}\\&space;\\&space;\therefore&space;f(x,y)&space;=&space;0&space;\&space;\text{IFF&space;x&space;=&space;0&space;and&space;y&space;=0&space;or&space;-1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?f(x,y)&space;=&space;x&space;+&space;x^{2}&space;+&space;x^{3}&space;-&space;y&space;-&space;y^{2}&space;\\&space;f(x,y)&space;=&space;x(1+x+x^{2})&space;-y(1+y)&space;\\&space;$Let&space;f(x,y)&space;=&space;0&space;\\&space;(1+x+x^{2})&space;\text{has&space;no&space;solution}\\&space;\\&space;\therefore&space;f(x,y)&space;=&space;0&space;\&space;\text{IFF&space;x&space;=&space;0&space;and&space;y&space;=0&space;or&space;-1}" title="f(x,y) = x + x^{2} + x^{3} - y - y^{2} \\ f(x,y) = x(1+x+x^{2}) -y(1+y) \\ $Let f(x,y) = 0 \\ (1+x+x^{2}) \text{has no solution}\\ \\ \therefore f(x,y) = 0 \ \text{IFF x = 0 and y =0 or -1}" /></a>

why does f (x, y) =0 imply that both factors much equal to zero? This is clearly not necessarily true

 

[faisalabdul16]

Re: HSC 2015 4U Marathon - Advanced Level

why does f (x, y) =0 imply that both factors much equal to zero? This is clearly not necessarily true

Thats not saying that x and y must equal to 0.

f(x,y) = 0 is saying that the function in terms of x and y, IS equal to 0.

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

[]

likewise, nowhere in your working have u explained why those combinations are the ONLY possible solutions.

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

Thats not saying that x and y must equal to 0.

f(x,y) = 0 is saying that the function in terms of x and y, IS equal to 0.

Here it is by coincidence that x =0, y = 0 is a solution

where in your working have you shown that those are the ONLY possible solutions?

 

[faisalabdul16]

Re: HSC 2015 4U Marathon - Advanced Level

where in your working have you shown that those are the ONLY possible solutions?

Where i said that 1 + x + x^2 has no solution. Therefore x can only be equal to 0.

It is pretty obvious that they are the ONLY solutions... If you let y(1+y) = 0, is that going to have 134124323423421 solutions? Is it going to have 22432332 solutions? NO. It's going to have 2 solutions, and its the 2 that Drsoccerball and i have mentioned.

 

[InteGrand]

Re: HSC 2015 4U Marathon - Advanced Level

Where i said that 1 + x + x^2 has no solution. Therefore x can only be equal to 0.

It is pretty obvious that they are the ONLY solutions... If you let y(1+y) = 0, is that going to have 134124323423421 solutions? Is it going to have 22432332 solutions? NO. It's going to have 2 solutions, and its the 2 that Drsoccerball and i have mentioned.

Why MUST x equal 0 (that's the essence of the required proof basically)?

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

Where i said that 1 + x + x^2 has no solution. Therefore x can only be equal to 0.

It is pretty obvious that they are the ONLY solutions... If you let y(1+y) = 0, is that going to have 134124323423421 solutions? Is it going to have 22432332 solutions? NO. It's going to have 2 solutions, and its the 2 that Drsoccerball and i have mentioned.

so ur argument is, if g (x)=x (non zero factor) +h (y) =0, then these zeroes occur only when x=0 and h (y) =0? This is clearly not true, and both of yours + soccerball's solutions wouldn't get more than 1 mark in the actual exam!

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

so ur argument is, if g (x)=x (non zero factor) +h (y) =0, then these zeroes occur only when x=0 and h (y) =0? This is clearly not true, and both of yours + soccerball's solutions wouldn't get more than 1 mark in the actual exam!

Whats wrong with my solution...

 

[faisalabdul16]

Re: HSC 2015 4U Marathon - Advanced Level

Why MUST x equal 0 (that's the essence of the required proof basically)?

I see the flaw in my reasoning now

so ur argument is, if g (x)=x (non zero factor) +h (y) =0, then these zeroes occur only when x=0 and h (y) =0? This is clearly not true, and both of yours + soccerball's solutions wouldn't get more than 1 mark in the actual exam!

Yeap yeap now i get what you mean!

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

Why MUST x equal 0 (that's the essence of the required proof basically)?

For my proof im essentially proving when both sides are zero what values work. And since theres an extra x^3 it wont work any other way unless theyre both zero...
EDIT: Would that be "legit" enough as a proof