Australian Maths Comp (1 Viewer)

[Constip8edSkunk]

How did you guys go? Anyone confident of getting a cheque or a medal?

 

[Affinity]

personally stuffed up big time

 

[BlackJack]

Medals are truly rare, aren't they? ;p The marking scheme is more complex so usually only a few people do get 1st places.

Cheques are much easier by comparison, I used to get them.

 

[McLake]

Originally posted by BlackJack
Cheques are much easier by comparison, I used to get them.

Oh yes, very easy to get BJ ...

 

[underthesun]

They'll probably give you $30 angus & robertson voucher, if you get the highest standardised score, or was it that its for high distinction? It's rather disappointing really..

 

[Harimau]

How many of the last 10 questions did people get? I still dont get how to do that n^12- n^8 - n^4 +1= 0 one... Did anyone do it?

 

[underthesun]

What? It was today?

 

[underthesun]

Damn I missed out.. give me some of the hard questions, since im curious

 

[Harimau]

Originally posted by underthesun
Damn I missed out.. give me some of the hard questions, since im curious

Umm alright.

Given: n^12- n^8 - n^4 +1= 0 for some positive odd integer n.

If k is a positive integer, what is the largest value of k such that 2^k divides into that equation?

 

[underthesun]

you mean for for 2^k to be a root of that equation?

 

[Harimau]

I think so... The largest possible k yeah.

 

[underthesun]

Then didn't you just check the answers one by one? As in which one fits etc..

 

[Harimau]

the answer were like 5,6,7,8,9 ... you try subbing 2^9 into that honking eqn when you're not allowed to use a calculator. I tried doing it using polynomials, but then it became too hard so i skipped that questions. That question was around Q 25-26

 

[underthesun]

wait a minute, how can those be the answer?

n¹² - n⁸ - n⁴ will always be even, and after you minus 1 it'll never be 0.

 

[McLake]

Originally posted by underthesun
wait a minute, how can those be the answer?

n¹² - n⁸ - n⁴ will always be even, and after you minus 1 it'll never be 0.

That's not true. Odd number to odd power is odd. Odd - odd is even, - odd is Odd + 1 is even (so possibly 0).

2^9 is easy, it's just 1024 (everyone should know 2^10) divided by 2.

 

[Affinity]

here's the proof for it

n^12 - n^8 -n^4 -1

= n^8(n^4 - 1) - (N^4 -1)

= (n^8 -1)(n^4 -1)

= (n^4 + 1)(n^2 + 1)(n+1)(n-1)(n^2+1)(n+1)(n-1)

you will have to realise that n^2 + 1, n^4 + 1 are not multiples of 4. (odd^2 is congruent to 1 modulo 4.)
or:

odd number^2 =(2n+1)^2 = 4n^2 + 4n +1 = 4(n^2 +n) + 1
= 4k + 1
so.. odd^2 +1 = 4k+2

so.. each of those only contribute 1 factor of 2.
now consider n+1 and n-1, one and only one is a multiple of 4.
the only merely multiple of 2.

so together there are 1+1+(1+2)+1+(1+2) = 9 factors of 2.

QED.

I placed 10 for my answer ... stupid me.

 

[Archman]

I just sub in n=3 and ..after a pretty long while.. found that 2^9 goes into it.

 

[underthesun]

it happens that, because there was a teacher strike yesterday, the maths comp was today. w00t

although i still didn't get that one right, anyways..

what was the answer to the 3x3 squares (similar to magic square thing)?

 

[Harimau]

Originally posted by underthesun
it happens that, because there was a teacher strike yesterday, the maths comp was today. w00t

although i still didn't get that one right, anyways..

what was the answer to the 3x3 squares (similar to magic square thing)?

I think i got that one... I put down 7. This is how i thought of it:

Consider the smallest and largest combination:

I.e 1+2= 3 and 9+8= 17.

Therefore there are 15 possible combinations (Or summations) that could be done.

Then consider all the ones that could be overcounted... Crap now i forgot the next step... my answer is 7 (Or was it 6?) though, and i think i got that right... Then again, maybe not.

 

[iambored]

wat did u people get 4 the car question>