Australian Maths Competition (1 Viewer)

[Vadevalor]

Re: Australian Maths Competition 2013

Anyone got back their paper? Can anyone tell me the real answers?

 

[hayabusaboston]

Re: Australian Maths Competition 2013

I always do it. Just as I'm about to fall asleep it hits me lol. Must be a subconscious thing, or maybe our minds just become very clear at that time when we are relaxed?

I have photographic memory in that phase right before sleep. Also during lucid dreaming, of which I do a lot. I can visualise an event crystal clear, from months or yeeaaars ago, and just play it back in my mind, lurid as ever.

 

[tywebb]

Re: Australian Maths Competition 2013

There is a proof of the AM-GM inequality published in the 1920s by George Pólya. There are many proofs of this, but Pólya's method is called Pólya's Dream.

I'm delighted that it's in the 4 unit Cambridge text book, but disappointed that they did not appropriately accredit it to Pólya.

It's called Pólya's Dream because he actually dreamt the proof!

 

[iBibah]

Re: Australian Maths Competition 2013

There is a proof of the AM-GM inequality published in the 1920s by George Pólya. There are many proofs of this, but Pólya's method is called Pólya's Dream.

I'm delighted that it's in the 4 unit Cambridge text book, but disappointed that they did not appropriately accredit it to Pólya.

It's called Pólya's Dream because he actually dreamt the proof!

Wow, that would make studying so much easier if we used sleep time lol

EDIT: Just googled it, and I think I just found a video you uploaded.


Is that you? the teacher?

 

[Vadevalor]

Re: Australian Maths Competition 2013

I thought the answers were provided by AMT?
Anw do you have the alphabet answers can you post them here?

 

[iBibah]

Re: Australian Maths Competition 2013

I thought the answers were provided by AMT?
Anw do you have the alphabet answers can you post them here?

Nah mate the answers are provided later when awards are distributed.

 

[RealiseNothing]

Re: Australian Maths Competition 2013

I thought the answers were provided by AMT?
Anw do you have the alphabet answers can you post them here?

I know the answers for all the multiple choice if you want them.

 

[Vadevalor]

Re: Australian Maths Competition 2013

I know the answers for all the multiple choice if you want them.

Yes, please. Thanks!

 

[hayabusaboston]

Re: Australian Maths Competition 2013

I know the answers for all the multiple choice if you want them.

PM me so I can check my own plz?

 

[Gnowyhtac]

Re: Australian Maths Competition 2013

Yes please

 

[MathsGuru]

Re: Australian Maths Competition 2013

Q28: A positive integer N is made up of only 0's and 1's. When divided by 37, the remainder is 18. Find the smallest amount of 1's N can contain.

Five 1s are enough:

1101101 = 29759 x 37 + 18

 

[Gnowyhtac]

Re: Australian Maths Competition 2013

I know the answers for all the multiple choice if you want them.

PM me please thanks

 

[RealiseNothing]

Re: Australian Maths Competition 2013

Five 1s are enough:

1101101 = 29759 x 37 + 18

How exactly would you work it out though?

 

[iBibah]

Re: Australian Maths Competition 2013

How exactly would you work it out though?

Working out is too mainstream.

 

[Web Addict]

Re: Australian Maths Competition 2013

How exactly would you work it out though?

Aren't most of the questions involving integers just trial and error? The only way that I attempt them is to substitute in numbers until I get the answer, although this takes a very long time.

 

[MathsGuru]

Re: Australian Maths Competition 2013

Aren't most of the questions involving integers just trial and error? The only way that I attempt them is to substitute in numbers until I get the answer, although this takes a very long time.

None of the questions are ever intended to be solved by trial and error, though a bit of experimenting with numbers can often set you on the right track

For the 37 & 18 question, note first that 37 x 3 = 111 which seems to be heading vaguely in the right direction.
Then 999 is also a multiple of 37, so as well as 100 being 11 less than a multiple of 37, 1000 is 1 more than a multiple of 37.

Then looking just at the remainders after division by 37:
1 leaves 1 ; 10 leaves 10 ; 100 leaves 26 (effectively -11) ; 1000 leaves 1 again, so the pattern repeats:
10,000 leaves 10 ; 100,000 leaves 26 or -11 ; 1,000,000 leaves 1 remainder

Then you just need to combine as few of these remainders as possible to get an overall remainder of 18

 

[enigma_1]

Re: Australian Maths Competition 2013

Whoaaa how do you guys actually do the questions? They freak me out, legit they're like essays towards the end. Please share your techniques, or how you even approach the questions. Fannkss!!

 

[Makematics]

Re: Australian Maths Competition 2013

Let the heights of the trapezium be 'a' and 'b'. You can then find 'a' by considering one of the larger trapeziums, and then add up all the individual areas of the trapeziums to find an expression for 'b' in terms of 'a' and substitute your value of 'a' to find 'b'. Then just use the formula for a trapezium.

i just split it up into a rectangle and a triangle and did tan (angle) =8/(5+3+8)=1/2 and then the angles are corresponding angles and then messed around to get 55/4. sounds easier than what you did.

 

[Makematics]

Re: Australian Maths Competition 2013

I get 714 m^2 as my answer for Q27. has anyone else got an answer?

 

[tywebb]

Re: Australian Maths Competition 2013

I got 728.

Solution attached.