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HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

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Chlee1998

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Re: HSC 2014 4U Marathon - Advanced Level

I'll prove the stronger result that



for positive integers a and b. Note that if this is true, then any common factor of your two quantities must also be a common factor of two distinct primes, and hence can only be 1.

The part was proven in the last page of this marathon. I'll show the other direction now.

We can write b=aq+r uniquely, with q,r non-negative integers and

If , then from the result.

So .

But since the LHS is odd, it is coprime to any power of 2. So we get



As the RHS is strictly less than the LHS, this is only possible if the RHS is 0. Ie r=0, which means that as claimed.
On the last page you never provrd that if a is relatively prime to b then 2^a -1 is relatively prime to 2^b -1. All u did was prove that if a divides b then 2^a-1 divides 2^b-1.
 

glittergal96

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Re: HSC 2014 4U Marathon - Advanced Level

On the last page you never provrd that if a is relatively prime to b then 2^a -1 is relatively prime to 2^b -1. All u did was prove that if a divides b then 2^a-1 divides 2^b-1.
I never said that I did?

In the last page I proved the half of the assertion I made in my first displayed equation in my most recent post.

In the rest of my most recent post, I proved the half, and explained why this result implies that a relatively prime to b implies 2^a-1 relatively prime to 2^b-1.
 

Chlee1998

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Re: HSC 2014 4U Marathon - Advanced Level

Sorry I mustve misunderstood then. Wait doesnt the strwjght vertical line refer to coprimality???
 
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Sy123

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Re: HSC 2014 4U Marathon - Advanced Level



 
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dan964

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Re: HSC 2014 4U Marathon - Advanced Level

I'd might raise both sides to the 'n' then probs induction for the resulting statement or original statement
n>0
 

Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

I'd might raise both sides to the 'n' then probs induction for the resulting statement or original statement
n>0
I don't think that would work

Besides, there's a much more elegant solution!
 

dan964

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Re: HSC 2014 4U Marathon - Advanced Level

I was thinking of comparing areas under the graph x^1/n (for x>0) using upper/lower rectangles and trapeziums after applied the AM-GM assumption to the LHS.
 

dan964

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Re: HSC 2014 4U Marathon - Advanced Level

Note that multiplying both and by some scales both sides of the equation by .

Hence we may assume wlog that for each i.
Are you multiplying each term, or just a single term?
If the former...
(same with b)
which scales the LHS by r??
 

dan964

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Re: HSC 2014 4U Marathon - Advanced Level

Are you multiplying each term, or just a single term?
If the former...
(same with b)
which scales the LHS by r??
never mind it is the latter.
what is "wlog" in the second line...
 

glittergal96

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Re: HSC 2014 4U Marathon - Advanced Level

when and where you use?
The meaning is literal.

We can assume (blah) and a proof of the claim under the assumption of (blah) implies that the claim is true even without the (blah) assumption.

In other words (blah) was just assumed to make the proof more concise.
 

Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

Note that multiplying both and by some scales both sides of the equation by .

Hence we may assume wlog that for each i.

Then by AM-GM

Nice proof!

Here is mine:





 

FrankXie

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Re: HSC 2014 4U Marathon - Advanced Level

Try this one:
Given that $a>0, b>0$ and $a^3+b^3=2$. Without using calculus prove that $a+b\leq2$.
 

Chlee1998

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Re: HSC 2014 4U Marathon - Advanced Level

Try this one:
Given that $a>0, b>0$ and $a^3+b^3=2$. Without using calculus prove that $a+b\leq2$.
These problems belong in another thread, which is just called the 4u marathon. anyways, a^3 + b^3 can be factoed to (a+b)(a^2-ab + b^2) =2

But from AM GM, a^2 + b^2 > 2ab > ab (for a,b>0). Hence the second factor is positive.
Thus (a+b) multiplied by a positive number equals to 2, meaning that a+b is less than 2.
 
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