HSC 2013 MX2 Marathon (archive) (1 Viewer)

[Sy123]

Re: HSC 2013 4U Marathon

Also for the 957 question, I am currently trying to find pythagorean triads that have multiples as 957 lol.

Like for the triple 3, 4, 5

957 is divisible by 3 into 319 times

Multiply 4 by 319 =/= square number

And it didnt work for the next couple of triples that would of worked, so back to square one I guess

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Lol, is the answer a = 479, b=478; c=31, d=2; a=45, b = 42; c=161, d=158

That is one answer, good work.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Also for the 957 question, I am currently trying to find pythagorean triads that have multiples as 957 lol.

Like for the triple 3, 4, 5

957 is divisible by 3 into 319 times

Multiply 4 by 319 =/= square number

And it didnt work for the next couple of triples that would of worked, so back to square one I guess

I would recommend swapping 957 for much smaller numbers, then trying to see a pattern (which is a really cool one at that imo).

 

[bobmcbob365]

Re: HSC 2013 4U Marathon

Actually I just remembered an awesome question (aye aye spiral )

If are all distinct positive integers, find what they are so that:

Expressing 957 in its prime factors: 3*11*29

Now, a^2 - b^2 = (a-b)(a+b)=957

So now, take turns making a+b and a-b, from the factors of 957.

1.
let a+b=957
a-b=1
2a=958
therefore, a = 479, b=478

2.
let a+b=3*11
a-b=29
2a=62
therefore, a=31, b=2

3.
let a+b=11*29
a-b=3
2a=322
therefore, a=161, b=158

 

[Sy123]

Re: HSC 2013 4U Marathon

Expressing 957 in its prime factors: 3*11*29

Now, a^2 - b^2 = (a-b)(a+b)=957

So now, take turns making a+b and a-b, from the factors of 957.

1.
let a+b=957
a-b=1
2a=958
therefore, a = 479, b=478

2.
let a+b=3*11
a-b=29
2a=62
therefore, a=31, b=2

3.
let a+b=11*29
a-b=3
2a=322
therefore, a=161, b=158

Very nice solution.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

The pattern I also wanted you to see was that any odd number can be expressed as a difference of two squares. To find these squares, you just add 1 and divide by 2, minus 1 and divide by 2. For example:

Say we want

 

[bobmcbob365]

Re: HSC 2013 4U Marathon

a) Find, in modulus-argument form, the roots of the equation z^(2n+1) =1.
b) Hence factorise z^2n + z^(2n-1) + ... + z^2 + z +1 into quadratic factors with real coefficients.
c) Deduce that 2^n * sin(pi/(2n+1)) * sin(2pi/(2n+1)) * sin(3pi/(2n+1)) ... sin(n*pi/(2n+1)) = sqrt(2n+1)

 

[bobmcbob365]

Re: HSC 2013 4U Marathon

The pattern I also wanted you to see was that any odd number can be expressed as a difference of two squares. To find these squares, you just add 1 and divide by 2, minus 1 and divide by 2. For example:

Say we want

aaah, I see =/

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

aaah, I see =/

Your way is good since it gets all the possible combinations for 'a' and 'b'.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

That is really cool, and the algebraic way to show it is:








The only way a and b can be integral is if k is odd.

The way I found it was I used the fact that squares go up by odd numbers. Like to get from to we add 3, then from to we add 5. Then we add 7, then 9, then 11, etc. So we are always adding if we let the number we are squaring be . Which is obvious by observing that:


Also this can be observed by knowing the series:

 

[Sy123]

Re: HSC 2013 4U Marathon

The way I found it was I used the fact that squares go up by odd numbers. Like to get from to we add 3, then from to we add 5. Then we add 7, then 9, then 11, etc. So we are always adding if we let the number we are squaring be . Which is obvious by observing that:


Also this can be observed by knowing the series:

Nice.

Also bob for part b do we need to factorise the whole thing into quadratic factors or only 1 quadratic factor?

 

[bobmcbob365]

Re: HSC 2013 4U Marathon

oh, factorise into a bunch of quadratic factors

 

[Sy123]

Re: HSC 2013 4U Marathon

Is it:



Making sure before I continue

 

[bobmcbob365]

Re: HSC 2013 4U Marathon

not quite; but i'm curious as to how you came to that

 

[Sy123]

Re: HSC 2013 4U Marathon

not quite; but i'm curious as to how you came to that

Assume that the factorisation is: (something that probably was wrong in the first place)


Equate co-efficient of z

on LHS we can get z an n number of times





I did something completely wrong somewhere

 

[bobmcbob365]

Re: HSC 2013 4U Marathon

Oh right, not sure if 1st line is right. But I think the question is leading on into expressing z^(2n+1) -1 in (z-z1)(z-z2)(z-z3)... form, where z1, z2, z3... are roots of z^(2n+1)=1

 

[SpiralFlex]

Re: HSC 2013 4U Marathon

I'm back from gaming. 14 hrs of L4D, let me think of something kwl.

 

[Sy123]

Re: HSC 2013 4U Marathon

Oh right, not sure if 1st line is right. But I think the question is leading on into expressing z^(2n+1) -1 in (z-z1)(z-z2)(z-z3)... form, where z1, z2, z3... are roots of z^(2n+1)=1

Oh wow, I cant believe I could not see that :/

 

[SpiralFlex]

Re: HSC 2013 4U Marathon

 

[bobmcbob365]

Re: HSC 2013 4U Marathon

haha yeah, dw its 12:30. Heads get all screwy.