Australian Maths Competition (1 Viewer)

[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...

 

[MineTurtle]

Re: Australian Maths Competition 2013

Btw guys, in QLD most schools have received the results in the Australian Maths Competition.
My school has published our results and I got Prize Winner
Maybe we can share scores to figure out cut-offs in different states.
I'm soooo happy!!!

 

[obliviousninja]

Re: Australian Maths Competition 2013

Btw guys, in QLD most schools have received the results in the Australian Maths Competition.
My school has published our results and I got Prize Winner
Maybe we can share scores to figure out cut-offs in different states.
I'm soooo happy!!!

congrats.

 

[AhMcean]

Re: Australian Maths Competition 2013

wait isn it root(x+y) + root(y+z) = root(x+z), it probably works your way as well because when i did it i got the same answer as well, what i did was this:

root(x+y) + root(y+z) = root(x+z)

(x+y) + (y+z) + 2root((x+y)(y+z)) = z + x

cancelling out z+x:

2y=-2root((x+y)(y+z))

y^2=(x+y)(y+z)

Looking at this expression, it will only hold when x=z, so y^2=(x+y)^2
solving the quadratic, we get x=0 or y=-x/2

Now substituting x=z back to original equation we get root(x+y) + root(x+y)= root(2x)

so from the solution of the quadratic -x/2, we notice that -x=-2,-4,-6,-8....
However we see that from the root(2x) that x must be half of a even square number, ie. 2, 8, 18, 32...etc.

Picking -x=-2, we get y=-1. Therefore substituting x=1 and y=-1 to original equation, we get 1+1=2, which is true, so 1/x+1/y+1/z = 1/2 +1/2+-1

which equals 0.

 

[Squar3root]

Re: Australian Maths Competition 2013

wait isn it root(x+y) + root(y+z) = root(x+z), it probably works your way as well because when i did it i got the same answer as well, what i did was this:

root(x+y) + root(y+z) = root(x+z)

(x+y) + (y+z) + 2root((x+y)(y+z)) = z + x

cancelling out z+x:

2y=-2root((x+y)(y+z))

y^2=(x+y)(y+z)

Looking at this expression, it will only hold when x=z, so y^2=(x+y)^2
solving the quadratic, we get x=0 or y=-x/2

Now substituting x=z back to original equation we get root(x+y) + root(x+y)= root(2x)

so from the solution of the quadratic -x/2, we notice that -x=-2,-4,-6,-8....
However we see that from the root(2x) that x must be half of a even square number, ie. 2, 8, 18, 32...etc.

Picking -x=-2, we get y=-1. Therefore substituting x=1 and y=-1 to original equation, we get 1+1=2, which is true, so 1/x+1/y+1/z = 1/2 +1/2+-1

which equals 0.

you made an account for this lol

 

[Shadowdude]

Re: Australian Maths Competition 2013

maths is srs bsns

 

[MathsGuru]

Re: Australian Maths Competition 2013

y^2=(x+y)(y+z)

Looking at this expression, it will only hold when x=z

Unfortunately that doesn't follow - take e.g. y = 2, z = 2 and x = -1

 

[sohank]

Re: Australian Maths Competition 2013

Hope this actually makes sense, solution to question 22:

Interesting strategy.

I followed the same logic that the line CW cuts the shaded areas above and below the line equally. Hence it needs to pass through a point equidistant from the centers of two circles (1,1) and (4,1) ie at point coordinate (2.5, 1).

Any tilting of the horizontal line through (2.5, 1) will cut the circles symmetrically above and below the lines.

A line passing through (2.5, 1) and (6, 2) has a slope of 2/7 and y intercept of 2/7. Hence PW = 2/7

 

[sohank]

Re: Australian Maths Competition 2013

No.

Just solve these 4 equations in 4 unknowns:

p+q=r+s

p²+r²=53²

p²+s²=51²

q²+s²=25²

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.

Simple to you friend !! I have been trying to solve the above four equations for 2 hours w/o success. Can u pl send me solution working by pm or in the forum. Thanks heaps !! (I know my request is more than 1 year late)

 

[Kaido]

Re: Australian Maths Competition 2013

You must have a lot of time to solve 4 variable simultaneous equations

 

[InteGrand]

Re: Australian Maths Competition 2013

Simple to you friend !! I have been trying to solve the above four equations for 2 hours w/o success. Can u pl send me solution working by pm or in the forum. Thanks heaps !! (I know my request is more than 1 year late)

We have



where .

If you know your Pythagorean triples, you'll know that 7² + 24² = 25². So from Equation (4), (q, s) is a permutation of (7, 24).

We can see from Equations (2) and (3) that r > s. We can see from Equations (3) and (4) that p > q.

Subtracting Equation (3) from (2) gives









Now separately test s = 7 (i.e. s² = 49) and s = 24 (i.e. s² = 576 (if you didn't have this memorised, it's easy to work out using the expansion of (20 + 4)²)), and take the one that gives an RHS which is a perfect square. It turns out that this is using s = 24, which gives r² = 208 + 576 = 784 = 28², so that r = 28.

Also, as s = 24, we must have q = 7.

Now plug everything into Equation (1) and solve for the remaining variable.

It's probably good for these maths competitions to memorise the first thirty square numbers and some Pythagorean triples.

 

[tywebb]

Re: Australian Maths Competition 2013

You can also do it without necessarily using pythagorean triads.

 

[yo_yo]

Re: Australian Maths Competition 2013

Hello guys! Are you ready for Thursday ?