Australian Maths Competition (1 Viewer)

[jihad1234]

Re: Australian Maths Competition 2013

Thanks for 28 and 29, I assume the answers are 495 and 506 respectively?

 

[Gnowyhtac]

Re: Australian Maths Competition 2013

Thanks for 28 and 29, I assume the answers are 495 and 506 respectively?

Could you please show how to get it?

 

[MathsGuru]

Re: Australian Maths Competition 2013

Could you please show how to get it?

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.

 

[MathsGuru]

Re: Australian Maths Competition 2013

For 26, 27 and 30 I get 512, 162 and 765 respectively

A slight miscalculation I think... make that 161 for Q27

 

[MathsGuru]

Re: Australian Maths Competition 2013

Hope this actually makes sense, solution to question 22:

Looks good.

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]

Re: Australian Maths Competition 2013

Looks good.

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.

Yep, I know the bolded.

Very nice alternative methods as well.

 

[MathsGuru]

Re: Australian Maths Competition 2013

Consider the scores are equal at (x,x), then you have two possibilities (x+1,x) or (x,x+1).

If the scores are unequal at (x,x+1) you also have two possibilities (x+1,x+1) or (x,x+2).

However if the scores are 2 apart (x,x+2) you only have one possibility (x+1,x+2).

Start at (0,0) and apply this, you find the pattern is double then add half of what you have, i.e.:

1) 2
2) 4
3) 6
4) 12
5) 18
6) 36
7) 54
8) 108
9) 162
10) 324
11) 486
12) 972

I might add I don't know for sure if my answers are correct, I'm just posting my personal working out.

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]

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.

 

[MathsGuru]

Re: Australian Maths Competition 2013

Consider the scores are equal at (x,x), then you have two possibilities (x+1,x) or (x,x+1).

If the scores are unequal at (x,x+1) you also have two possibilities (x+1,x+1) or (x,x+2).

However if the scores are 2 apart (x,x+2) you only have one possibility (x+1,x+2).

Start at (0,0) and apply this, you find the pattern is double then add half of what you have, i.e.:

1) 2
2) 4
3) 6
4) 12
5) 18
6) 36
7) 54
8) 108
9) 162
10) 324
11) 486
12) 972

I might add I don't know for sure if my answers are correct, I'm just posting my personal working out.

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

 

[jihad1234]

Re: Australian Maths Competition 2013

So the results are out, how did you guys do?

 

[HSC2014]

Re: Australian Maths Competition 2013

So the results are out, how did you guys do?

It's the holidays so hopefully I'll receive it early next term O:

 

[Makematics]

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

 

[obliviousninja]

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

Did you get the D?

My teacher didn't bother to let us do AMC this year, she wanted to focus on hsc revision.
I think I got a distinction last year though.

 

[MathsGuru]

Re: Australian Maths Competition 2013

Prime factors of 2013 are 3, 11, 61.

They all have powers of 1, thus the amount of consecutive integers that add to 2013 is:



Take another example, say 72, it's prime factorisation is

So we take each power, add one, multiply them together, and minus one:

I think you've got the right answer for the wrong reason here...
I agree that there are 7 ways of writing 2013 as the sum of at least 2 consecutive positive integers, but there certainly aren't 11 ways of writing 72 as such a sum.
Also, your formula would say that any product of 2 primes could be written as such a sum in 3 ways.
But (for example) 6 can only be written as 1+2+3, and 15 can only be written as 7+8 or 4+5+6.

Your formula appears to be the right one if you want to count how many factors a number has, other than itself... but that's not what this questiom was about ;-)

The way I've done it (just now, better late than never lol) is based on these 2 observations:
- if n is odd, the sum of any n consecutive integers is a multiple of n
- if n is even, the sum of any n consecutive integers is n/2 more than a multiple of n

 

[MathsGuru]

Re: Australian Maths Competition 2013

My solutions to question 21 and 23:

Question: If , what is the value of ?

Multiply each side by



Now multiply each side by



Substitute this in gives:



If you repeat this a few times you get the answer of

Question 23:

Question: Given find

Square both sides:



Cancel out the on each side:





Squaring again gives:







We want to find the value of , but if we make this one fraction we get:



But so

I agree almost verbatim with what you did for Q23

For Q21, I started by writing x^5 = (x^2)^2 * x = (x+3)^2 * x = ..., maybe this gets to the answer a bit quicker??

 

[RealiseNothing]

Re: Australian Maths Competition 2013

I think you've got the right answer for the wrong reason here...
I agree that there are 7 ways of writing 2013 as the sum of at least 2 consecutive positive integers, but there certainly aren't 11 ways of writing 72 as such a sum.
Also, your formula would say that any product of 2 primes could be written as such a sum in 3 ways.
But (for example) 6 can only be written as 1+2+3, and 15 can only be written as 7+8 or 4+5+6.

Your formula appears to be the right one if you want to count how many factors a number has, other than itself... but that's not what this questiom was about ;-)

The way I've done it (just now, better late than never lol) is based on these 2 observations:
- if n is odd, the sum of any n consecutive integers is a multiple of n
- if n is even, the sum of any n consecutive integers is n/2 more than a multiple of n

http://en.wikipedia.org/wiki/Polite_number

 

[MathsGuru]

Re: Australian Maths Competition 2013

http://en.wikipedia.org/wiki/Polite_number

oops, I missed 15 = 1+2+3+4+5

Thanks for the very informative link... most importantly it defines:
"the politeness of x equals the number of odd divisors of x that are greater than one" (emphasis added)

So your answer to the original problem is right for the right reason (or at least one right reason, I did it a bit differently, and more clumsily without the concept of polite numbers )

But because of the restriction to odd divisors, your example with 72 needs revision down to just (3+1)-1 = 2, which seems right according to my calculations: 72 = 23+24+25 (3 #s with ave 24) = 4+5+6+...+12 (9 #s with ave 8)

 

[RealiseNothing]

Re: Australian Maths Competition 2013

oops, I missed 15 = 1+2+3+4+5

Thanks for the very informative link... most importantly it defines:
"the politeness of x equals the number of odd divisors of x that are greater than one" (emphasis added)

So your answer to the original problem is right for the right reason (or at least one right reason, I did it a bit differently, and more clumsily without the concept of polite numbers )

But because of the restriction to odd divisors, your example with 72 needs revision down to just (3+1)-1 = 2, which seems right according to my calculations: 72 = 23+24+25 (3 #s with ave 24) = 4+5+6+...+12 (9 #s with ave 8)

Ah, yes that is true. I completely skimmed over the "odd" unfortunately haha.

Also for the example of 72, would it not just be (3)-1= 2? Since (3+1)-1 = 3 not 2

 

[MathsGuru]

Re: Australian Maths Competition 2013

Ah, yes that is true. I completely skimmed over the "odd" unfortunately haha.

Also for the example of 72, would it not just be (3)-1= 2? Since (3+1)-1 = 3 not 2

Yes indeed, I think I was reading the wrong half of your factorisation and 72 and copied it without thinking, I meant to write the (2+1)-1 part and leave out the (3+1) coming from 2^3

Sometimes my brain goes to sleep recently... must be because the end of the year is coming

 

[MathsGuru]

Re: Australian Maths Competition 2013

Has anyone done Q29,30 on the senior paper by the way?

I get 176 and 870 for these two Qs respectively...