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HSC 2015 MX2 Marathon (archive) (5 Viewers)

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InteGrand

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Re: HSC 2015 4U Marathon

im assuming this works with n amount of factors aswell?
The factors need to be prime though. So like I said in my earlier post, first we need to write the integer under consideration as a product of powers of its prime factors.
 

InteGrand

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Re: HSC 2015 4U Marathon


No, since the factors must be prime factors. So we'd need to rewrite the given integer in terms of prime factors first, and THEN we could apply that formula (or figure it out).
 

Drsoccerball

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Re: HSC 2015 4U Marathon

No, since the factors must be prime factors. So we'd need to rewrite the given integer in terms of prime factors first, and THEN we could apply that formula (or figure it out).
But theres no common prime factor (even though impossible) wouldnt that be the formula
 

InteGrand

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Re: HSC 2015 4U Marathon

But theres no common prime factor (even though impossible) wouldnt that be the formula

Not if they're not all primes.

E.g. in , 4 is not a PRIME factor, so we would need to rewrite 4 in terms of its prime factors, which is just 4= 22, then write (using index laws).

Now we have it in terms of PRIME factors, so we can apply the formula with the exponents in the primes in the last expression.
 

Sy123

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Re: HSC 2015 4U Marathon

Is that a typo?

The number would be of the form
Also try to do so without induction (though in an exam situation they probably wouldn't force you to think of a non-inductive solution)
 

Drsoccerball

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Re: HSC 2015 4U Marathon

Is that a typo?

The number would be of the form
You get 4n(n+1)


Since all even numbers are divisible by 2 if any of the numbers are even then proof is complete
Since adding one changes it to odd from even vice versa and there is an n it will always be divided by 2
soz for dodge proof
 

Ekman

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Re: HSC 2015 4U Marathon

You get 4n(n+1)


Since all even numbers are divisible by 2 if any of the numbers are even then proof is complete
Since adding one changes it to odd from even vice versa and there is an n it will always be divided by 2
soz for dodge proof
I would of mentioned that 2 is a prime number and that is able to divide all even numbers.
Hence: n(n+1) is always even for all real values of n. (As you get even number x odd number=even number)
As n(n+1) is always even, it is divisible by 2 as 2 is the base prime number for all even numbers
 

Sy123

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Re: HSC 2015 4U Marathon

You get 4n(n+1)


Since all even numbers are divisible by 2 if any of the numbers are even then proof is complete
Since adding one changes it to odd from even vice versa and there is an n it will always be divided by 2
soz for dodge proof
I would of mentioned that 2 is a prime number and that is able to divide all even numbers.
Hence: n(n+1) is always even for all real values of n. (As you get even number x odd number=even number)
As n(n+1) is always even, it is divisible by 2 as 2 is the base prime number for all even numbers
These are very similar proofs just worded differently. For future reference a more rigorous way of writing it would be (I'm not sure how nitpicky they would be in the HSC, its good to be safe though):









--------

Alternatively to avoid this lengthy wordy proof:

 
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Sy123

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Re: HSC 2015 4U Marathon



Try to prove it without induction (try to generalize the above method of using evens/odds, or go for a purely algebraic approach)
 

Speed6

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Re: HSC 2015 4U Marathon

How many marks will that question generally be?
 
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