Australian Maths Competition (2 Viewers)

RealiseNothing

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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.
 

MathsGuru

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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
 

Web Addict

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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

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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

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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

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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.
 
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Makematics

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Re: Australian Maths Competition 2013

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

James lee

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Re: Australian Maths Competition 2013

Does anyone have the intermediate answers for this years one? I'd like to double check my answers, thanks!
 

Makematics

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Re: Australian Maths Competition 2013

I got 728.

Solution attached.
OMG arghhh i made a silly mistake, accidentally did (1/2)x51x28=714... thanks anyway. so they expect you to know all those pythagorean triads in the AMC or what?
 

jihad1234

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Re: Australian Maths Competition 2013

Hi, does anyone know the answer for Questions 18, 26, 27 and 30 in the Junior Paper? It would be really appreciated if you could provide those. Thanks a lot!
 

iBibah

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Re: Australian Maths Competition 2013

18. D
26. 972
27. 728
not sure about 30, would be interested to see a solution.
He said junior paper? Arent some of the last questions different?
 

MathsGuru

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Re: Australian Maths Competition 2013

On the junior paper, the easiest way to see that 18 is D is to observe that triangle DNC is equilateral, TM is parallel to DC, and AM is parallel to DN
 

MathsGuru

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Re: Australian Maths Competition 2013

Hi, does anyone know the answer for Questions 18, 26, 27 and 30 in the Junior Paper? It would be really appreciated if you could provide those. Thanks a lot!
For 26, 27 and 30 I get 512, 162 and 765 respectively
 

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