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HSC 2013 MX2 Marathon (archive) (2 Viewers)

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Ikki

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Re: HSC 2013 4U Marathon

Replace the sqrt(2) with 2 and its correct :)

The proper way to do it is just consider:



We know that it is purely imaginary for



Therefore it is purely imaginary for
Cool, quick check...






Let Reals=0




Gotcha :)

Thanks sy :)
 
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Sy123

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Re: HSC 2013 4U Marathon

Mostly finished, I omitted most of seanieg's questions, I will compile them if I can solve them, if I cannot I will eventually make it a separate compilation for advanced problems.
I just need to sort it.


 

hit patel

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Re: HSC 2013 4U Marathon

damn sy, I lost my answer do you auto logout.
imma just type my second answer which looks pretty wrong since i did it on word
ii) l1+Ml=2l1-Ml
 

hit patel

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Re: HSC 2013 4U Marathon

Mostly finished, I omitted most of seanieg's questions, I will compile them if I can solve them, if I cannot I will eventually make it a separate compilation for advanced problems.
I just need to sort it.


That includes poly?
 

Sy123

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Re: HSC 2013 4U Marathon

I can help out with any you are stuck with that aren't too ridiculous.
Nice.


For the first rough version of the compilation, see the first post, I have edited it in there.


-------------


 

seanieg89

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Re: HSC 2014 4U Marathon

Cute little combinatorics problem:

A man is on the first floor of a building with n floors. He wants to use the elevator to stop at every floor exactly once. What are the minimum and maximum "lengths" of such a trip? How many trips of maximal length are there? You must justify all your answers carefully.

(Note: "Length"=number of floors traversed. Eg the trip 1->2->4->3 in a 4 storey building has length 1+2+1=4.)
 

Carrotsticks

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Just from a brief look, is the maximal length n(n-1)/2 and minimal length (n-1)?

Could be wrong, but that's what first came to mind.
 

seanieg89

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Re: HSC 2014 4U Marathon

Just from a brief look, is the maximal length n(n-1)/2 and minimal length (n-1)?

Could be wrong, but that's what first came to mind.
Yep, the main part of the problem is in determining the number of maximal trips (with proof) though.
 

Sy123

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Re: HSC 2014 4U Marathon

Cute little combinatorics problem:

A man is on the first floor of a building with n floors. He wants to use the elevator to stop at every floor exactly once. What are the minimum and maximum "lengths" of such a trip? How many trips of maximal length are there? You must justify all your answers carefully.

(Note: "Length"=number of floors traversed. Eg the trip 1->2->4->3 in a 4 storey building has length 1+2+1=4.)
Is it only 1?



Resulting in:

Now to prove that this is the only trip possible to yield maximal length:

Our initial moves are

This is the only way we can get a length of '(n-1)', if we make a different move first, say 1 -> 2, and then is NOT correct, since we already have been at 1.

Therefore the initial moves must be

The next move can be either to 2, 3, 4, .... , n-1

Likewise, we need to make a move of length (n-2) in order for the sum to work. The only way to do this is going to floor number 2, if we try to make the (n-1) length later on in the chain, we arrive at thet same scenario as the first. Therefore the next move must be:



Likewise, we proceed inductively, so that the only way to yield a move (n-3) in length, is to proceed from 2 -> n-1.

And so on, until we yeild a series of lengths:

We cannot permute the sum as proven above.

Now consider a general sum of moves to get to



There are n terms in the series because we need to make (n-1) moves to get to the other (n-1) floors.



The only series that can therefore add up to is



And since we have shown that we cannot permute this sum, therefore the only permutation allowed is:



Therefore we only have 1 trip of maximal length.

Likewise 1 trip of minimal length with a similar method, 1 + 1 + \dots + 1
 

seanieg89

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Re: HSC 2014 4U Marathon

Is it only 1?



Resulting in:

Now to prove that this is the only trip possible to yield maximal length:

Our initial moves are

This is the only way we can get a length of '(n-1)', if we make a different move first, say 1 -> 2, and then is NOT correct, since we already have been at 1.

Therefore the initial moves must be

The next move can be either to 2, 3, 4, .... , n-1

Likewise, we need to make a move of length (n-2) in order for the sum to work. The only way to do this is going to floor number 2, if we try to make the (n-1) length later on in the chain, we arrive at thet same scenario as the first. Therefore the next move must be:



Likewise, we proceed inductively, so that the only way to yield a move (n-3) in length, is to proceed from 2 -> n-1.

And so on, until we yeild a series of lengths:

We cannot permute the sum as proven above.

Now consider a general sum of moves to get to



There are n terms in the series because we need to make (n-1) moves to get to the other (n-1) floors.



The only series that can therefore add up to is



And since we have shown that we cannot permute this sum, therefore the only permutation allowed is:



Therefore we only have 1 trip of maximal length.

Likewise 1 trip of minimal length with a similar method, 1 + 1 + \dots + 1
Not quite, the logic is flawed at the start: we do not NEED a step of length n-1 in order to be maximal. Examples are not too tricky to find (the first example exists when n=6), but I think it might give a bit too much away if I provide one. The point is we don't need to form a sum of the form 1+2+...+(n-1) in order to attain the maximal length.
 
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RealiseNothing

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Re: HSC 2014 4U Marathon

Let for denote the set of prime numbers, such that:







etc

Show that has no solution for
 
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RealiseNothing

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Re: HSC 2014 4U Marathon

Not quite, the logic is flawed at the start: we do not NEED a step of length n-1 in order to be maximal. Examples are not too tricky to find (the first example exists when n=6), but I think it might give a bit too much away if I provide one. The point is we don't need to form a sum of the form 1+2+...+(n-1) in order to attain the maximal length.
For n=6, 1->5->2->6->3->4
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Show that the minimum value of



is



Where is a positive real number for and it is known that for integer
 
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Sy123

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Re: HSC 2014 4U Marathon

Let for denote the set of prime numbers, such that:







etc

Show that has no solution for
Essentially we must show that the sum cannot be prime.

We know firstly



Therefore,

Therefore,

Due to, (*), each is divisible by

Meaning, all is an integer, therefore:

for integer K

And since, , therefore does not satisfy the definition of a prime number, therefore there is no integral, k and n, such that
 

Sy123

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Re: HSC 2014 4U Marathon

Show that the minimum value of



is



Where is a positive real number for and it is known that for integer
Could you clarify what you mean by , there is no term in the sum beyond , so I was wondering if it were perhaps something like
 

seanieg89

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Re: HSC 2014 4U Marathon

Let for denote the set of prime numbers, such that:







etc

Show that has no solution for
Each term in the LHS sum is even, and the first term is 2. But 2 is the only even prime! So this sum cannot be prime.
 

Sy123

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Re: HSC 2014 4U Marathon

Show that the minimum value of



is



Where is a positive real number for and it is known that for integer
Letting, n=2, it should follow from this inequality:




EDIT: Actually it works my bad!
 
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