Mongoose528
Member
Re: Australian Maths Competition 2013
oh shiiit I thought it said 2018 lol
oh shiiit I thought it said 2018 lol
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Australian Maths Competition (1 Viewer)
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Mongoose528
Member
Re: Australian Maths Competition 2013
To whoever did the intermediate last year, which one did you think was harder, this years or last years?
To whoever did the intermediate last year, which one did you think was harder, this years or last years?
Mongoose528
Member
Re: Australian Maths Competition 2013
was the question with the chords in your exam? It was number 30 in the intermediate paper so I thought it might have been number 27/28/29 in the senior paper
Paradoxica
-insert title here-
Re: Australian Maths Competition 2013
We had a chord question yeah, not sure if it was the same one. There was 64 points on the circle and it was asking how many chords you could form without any crossing over. I couldn't work that out.
We had a chord question yeah, not sure if it was the same one. There was 64 points on the circle and it was asking how many chords you could form without any crossing over. I couldn't work that out.
Re: Australian Maths Competition 2013
If its asking for the maximum number of chords without intersection, not number of ways, then I think its 125, idk though, 75% sure its wrong.
Paradoxica
-insert title here-
Re: Australian Maths Competition 2013
No, I am sure it's 126. Unless I miscounted. Ceebs.
Re: Australian Maths Competition 2013
http://imgur.com/a/fsfVN
ye it is probably 126. Nevertheless this is what I did, probably I missed something haha
http://imgur.com/a/fsfVN
Paradoxica
-insert title here-
Re: Australian Maths Competition 2013
Also did you just assume the formula is true or carry a bit of inductive reasoning to demonstrate it?
Chords can intersect, but each one can only intersect with exactly one other chord. Pretty sure vertex contact does not count as intersection.
Also did you just assume the formula is true or carry a bit of inductive reasoning to demonstrate it?
Re: Australian Maths Competition 2013
yeah I don't think I fully got the question, and yeah I just assumed it haha, I didn't fully go into it, was just thinking of a quick answer by looking at simpler cases
Paradoxica
-insert title here-
Re: Australian Maths Competition 2013
Of course, I did not formally prove this statement, but meh.
I was generating a few cases and immediately deduced that you have to dissect the polygon into as many quadrilaterals as possible, then draw diagonals in each one, to maximise it.
Of course, I did not formally prove this statement, but meh.
SimpletonPrime
New Member
Re: Australian Maths Competition 2013
I think Q30 is an old question. I remembered this one from an AMO contest, may be. Although, I didn't get it during the test, I was out of time.
Do you guys have answer for question 27,28,29 in senior division?
I think Q30 is an old question. I remembered this one from an AMO contest, may be. Although, I didn't get it during the test, I was out of time.
Do you guys have answer for question 27,28,29 in senior division?
Paradoxica
-insert title here-
Re: Australian Maths Competition 2013
Knowing this, and the fact that f(61) = 63, we obtain the following series of function values:
f(63) = f(f(61)) = 122
f(122) = f(f(63)) = 126
f(126) = f(f(122)) = 244
f(244) = f(f(126)) = 252
f(252) = f(f(244)) = 488
f(488) = f(f(252)) = 504
f(504) = f(f(488)) = 976
f(976) = f(f(504)) = 1008
f(1008) = f(f(976)) = 1952
f(1952) = f(f(1008)) = 2016
f(2016) = f(f(1952)) = 3904
The prime factorisation of 2016 is 32*63I think Q30 is an old question. I remembered this one from an AMO contest, may be. Although, I didn't get it during the test, I was out of time.
Do you guys have answer for question 27,28,29 in senior division?
Knowing this, and the fact that f(61) = 63, we obtain the following series of function values:
f(63) = f(f(61)) = 122
f(122) = f(f(63)) = 126
f(126) = f(f(122)) = 244
f(244) = f(f(126)) = 252
f(252) = f(f(244)) = 488
f(488) = f(f(252)) = 504
f(504) = f(f(488)) = 976
f(976) = f(f(504)) = 1008
f(1008) = f(f(976)) = 1952
f(1952) = f(f(1008)) = 2016
f(2016) = f(f(1952)) = 3904
Paradoxica
-insert title here-
Re: Australian Maths Competition 2013
I don't remember any other questions. If you would kindly... provide them.I think Q30 is an old question. I remembered this one from an AMO contest, may be. Although, I didn't get it during the test, I was out of time.
Do you guys have answer for question 27,28,29 in senior division?
SimpletonPrime
New Member
Re: Australian Maths Competition 2013
After the test, I realise that we can figure out the pattern (or prove by induction) that)=2^k(4n+3))
and )=2^{k+1}(4n+1))
I was up to f(126)=252 and then time-out.
After the test, I realise that we can figure out the pattern (or prove by induction) that
SimpletonPrime
New Member
Re: Australian Maths Competition 2013
OK. I remembered two questions:
Given positive integer
and =0,f(n)=2016^2)
where =ax^2+bx+c)
. How many values of 
?
How many pairs (a,b) of positive integers such that
and 
.
OK. I remembered two questions:
Given positive integer
How many pairs (a,b) of positive integers such that
Paradoxica
-insert title here-
Re: Australian Maths Competition 2013
(a(n+m)+b)(n-m) = 2016² = 3⁴×7²×2¹⁰
n-m must be a perfect square so that means we should pull out all the possible square factors from the factorisation of 2016²
There's also the special case where the factors are equal.
If I'm reading that correctly, I'm seeing that m must also be less than all the integer coefficients of the quadratic, which is going to be a huge road block for people who like to bash problems.
The second one requires cases
a√b = ab/2
There are two ways this equation can be true.
1. The exponents are equal, so √b = b/2
=> b=4, a can be anything in the valid range.
2. a=1, therefore b can be anything in the valid range.
Observe the case a=1, b=4 is double counted, so subtract one from the final amount of solutions.
I think that's all, though I may be missing something glaringly obvious.
Well the first one is fairly simple in terms of the first step.OK. I remembered two questions:
Given positive integer
How many pairs (a,b) of positive integers such that
(a(n+m)+b)(n-m) = 2016² = 3⁴×7²×2¹⁰
n-m must be a perfect square so that means we should pull out all the possible square factors from the factorisation of 2016²
There's also the special case where the factors are equal.
If I'm reading that correctly, I'm seeing that m must also be less than all the integer coefficients of the quadratic, which is going to be a huge road block for people who like to bash problems.
The second one requires cases
a√b = ab/2
There are two ways this equation can be true.
1. The exponents are equal, so √b = b/2
=> b=4, a can be anything in the valid range.
2. a=1, therefore b can be anything in the valid range.
Observe the case a=1, b=4 is double counted, so subtract one from the final amount of solutions.
I think that's all, though I may be missing something glaringly obvious.
SimpletonPrime
New Member
Re: Australian Maths Competition 2013
Oh, in the test, m only less than n.
Why do you imply n-m must be a perfect square?
Oh, in the test, m only less than n.