Australian Maths Competition (4 Viewers)
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Makematics
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Re: Australian Maths Competition 2013
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?
iBibah
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Re: Australian Maths Competition 2013
He said junior paper? Arent some of the last questions different?
Makematics
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Re: Australian Maths Competition 2013
Just solve these 4 equations in 4 unknowns:
p+q=r+s
p<sup>2</sup>+r<sup>2</sup>=53<sup>2</sup>
p<sup>2</sup>+s<sup>2</sup>=51<sup>2</sup>
q<sup>2</sup>+s<sup>2</sup>=25<sup>2</sup>
and given that p,q,r,s are positive integers, you get p=45, q=7, r=28, s=24.
It's quite simple really.
No.
Just solve these 4 equations in 4 unknowns:
p+q=r+s
p<sup>2</sup>+r<sup>2</sup>=53<sup>2</sup>
p<sup>2</sup>+s<sup>2</sup>=51<sup>2</sup>
q<sup>2</sup>+s<sup>2</sup>=25<sup>2</sup>
and given that p,q,r,s are positive integers, you get p=45, q=7, r=28, s=24.
It's quite simple really.
Re: Australian Maths Competition 2013
For 26, 27 and 30 I get 512, 162 and 765 respectively
Re: Australian Maths Competition 2013
For Q28, a sum of 9 consecutive positive integers is always a multiple of 9 (it's 9 times the middle number).
Similarly a sum of 11 consecutive positive integers is always a multiple of 11.
For even numbers (when there's no middle number), it works slightly differently.
Try some and you'll see that the sum of 10 consecutive positive integers is always 5 more than a multiple of 10.
The question then comes down to a Chinese Remainder Theorem calculation.
You want a number which is a multiple of both 9 and 11, hence a multiple of 99, since 9 and 11 are coprime.
It then has to be the smallest multiple of 99 which is also 5 more than a multiple of 10.
I agree with both Q28: 495 and Q29: 506.
For Q28, a sum of 9 consecutive positive integers is always a multiple of 9 (it's 9 times the middle number).
Similarly a sum of 11 consecutive positive integers is always a multiple of 11.
For even numbers (when there's no middle number), it works slightly differently.
Try some and you'll see that the sum of 10 consecutive positive integers is always 5 more than a multiple of 10.
The question then comes down to a Chinese Remainder Theorem calculation.
You want a number which is a multiple of both 9 and 11, hence a multiple of 99, since 9 and 11 are coprime.
It then has to be the smallest multiple of 99 which is also 5 more than a multiple of 10.
Re: Australian Maths Competition 2013
A slight miscalculation I think... make that 161 for Q27
Re: Australian Maths Competition 2013
Here's another variation: Since the line divides the shaded area in half (this is only true because it also divides the whole rectangle in half, by the way), it must past through the midpoint of the line connecting the circles' centres. If we take P as (0,0) and C as (6,2), this midpoint is (2.5, 1). Then the gradient and y-intercept of the line can easily be shown to be both 2/7, pretty much as you did before.
Another way to get this gradient is to imagine 'sliding' the right hand circle up the line like a bead along a wire, to sit in the top right hand corner of the rectangle, with new centre (11,3) - this gives the same sort of nice symmetrical picture you used to choose XP = 1.
Looks good.Hope this actually makes sense, solution to question 22:
Here's another variation: Since the line divides the shaded area in half (this is only true because it also divides the whole rectangle in half, by the way), it must past through the midpoint of the line connecting the circles' centres. If we take P as (0,0) and C as (6,2), this midpoint is (2.5, 1). Then the gradient and y-intercept of the line can easily be shown to be both 2/7, pretty much as you did before.
Another way to get this gradient is to imagine 'sliding' the right hand circle up the line like a bead along a wire, to sit in the top right hand corner of the rectangle, with new centre (11,3) - this gives the same sort of nice symmetrical picture you used to choose XP = 1.
RealiseNothing
what is that?It is Cowpea
Re: Australian Maths Competition 2013
Very nice alternative methods as well.
Yep, I know the bolded.
Very nice alternative methods as well.
Re: Australian Maths Competition 2013
If you're doing something like the Melb Uni maths comp where you need to write out reasoning as well as the answer, you can use induction to rigorously prove various statements, e.g.:
there are 2 x (3^n) ways of the first 2n+1 goals being scored (at which time one team or the other will be 1 goal ahead)
or (what we really need here):
there are 4 x (3^n) ways of the first 2n+2 goals being scored.
You have the pattern and the reasoning already, all of this just serves to formalise it a bit more so you can be sure that the pattern does continue & you are correct
972 is correct, and a nice way to picture things is with a trellis showing possible scores and the numbers of ways of reaching them. Apologies for the crappy image, but since you already have the working I think you can get the picture - the numbers are added like in Pascal's triangle:
If you're doing something like the Melb Uni maths comp where you need to write out reasoning as well as the answer, you can use induction to rigorously prove various statements, e.g.:
there are 2 x (3^n) ways of the first 2n+1 goals being scored (at which time one team or the other will be 1 goal ahead)
or (what we really need here):
there are 4 x (3^n) ways of the first 2n+2 goals being scored.
You have the pattern and the reasoning already, all of this just serves to formalise it a bit more so you can be sure that the pattern does continue & you are correct
x3reme_c00l
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Re: Australian Maths Competition 2013
sorry to post here. but i've question to ask. having some difficulties in answering this question. hope any of you can help me with it. really appreciate it.
sorry to post here. but i've question to ask. having some difficulties in answering this question. hope any of you can help me with it. really appreciate it.
Re: Australian Maths Competition 2013
Whenever an odd number of goals have been scored, the teams must be one goal apart.
If team X leads team Y in such a situation, the next 2 goals can be scored in 3 ways: XY, YX, YY
The teams are then one goal apart again.
So there are 2 ways the first goal can be scored, 3 ways each subsequent pair of goals can be scored, and 2 ways the final goal can be scored.
In the case of this Q, i.e. a 12-goal game, we have 2 x 3^5 x 2, in general 2 x 3^(n/2 - 1) x 2 for a game with n goals (n even) or just 2 x 3^((n-1)/2) if n is odd
I've just hit on the short and neat way of doing this Q
Whenever an odd number of goals have been scored, the teams must be one goal apart.
If team X leads team Y in such a situation, the next 2 goals can be scored in 3 ways: XY, YX, YY
The teams are then one goal apart again.
So there are 2 ways the first goal can be scored, 3 ways each subsequent pair of goals can be scored, and 2 ways the final goal can be scored.
In the case of this Q, i.e. a 12-goal game, we have 2 x 3^5 x 2, in general 2 x 3^(n/2 - 1) x 2 for a game with n goals (n even) or just 2 x 3^((n-1)/2) if n is odd
HSC2014
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Re: Australian Maths Competition 2013
It's the holidays so hopefully I'll receive it early next term O:So the results are out, how did you guys do?
Makematics
Well-Known Member
Re: Australian Maths Competition 2013
sooo is there any way to view amc results online? i know what award i got, but id like to see what my score was
sooo is there any way to view amc results online? i know what award i got, but id like to see what my score was