Australian Maths Comp

[luken93]

Q1-Q15...
everything seems right except..
Q10.B
Q11.D
looks like you will get a distinction+ for sure... 

woops i meant
10 B (x^2)
and i think i may have treated 11 as 7x7 not 7 + 7

[/0]

Lol, best I can do is do a logical guess at the degree and fiddle around with numbers...but knowing theorems and stuff would significantly decrease the amount of time you spend on the question..

Definitely. Well, at least for the last digit thing Fermat's/Euler's Theorem would help though sometimes not essential. I have seen questions where Menelaus' Theorem has been applied and it saves a lot of effort and time. I've even seen Ptolemy's Theorem once I think! Though it must've been a while ago...

u dont rly do proof-y stuff in vce maths, if you are passionate about maths and love problem solving you can begin by reading art and craft of problem solving, rly good to get started etc
competition maths is quite different from vce maths.
basically just read ALOT

YES. I also recommend the Andreescu books and the Engel one, they are awesome! I personally have taken chapters/chunks from a lot of books to learn the theory

I also recommend a book with a similar title to Truetears' book. It is at a slightly lower level than olympiad but it is very well written:

"Art of Problem Solving" (Vol. 1 & 2) by Richard Rusczyk and Sandor Lehoczky

[Hutchoo]

i didnt do that q. looked too long and dint have time...
wat about this one?

How many whole no.s less than 2010 have exactly three factors?

THe numbers that have exactly three factors are exactly those that are the square of some prime.

e.g: 2^2=4, 3^2=9, 5^2=25...

In general one way to work out how many factors a number has is to use unique prime factorisation like this:

2^100 * 3^100 has how many factors?

well i can chose any number of the form 2^a * 3^b with a and b between 0 and 100. Thus there are 100*100 factors of this number. It follows from this method that only the squares of a prime have exactly 3 factors.


so i guess you just have to list all the prime numbers until u find one whose square is larger than 2010. (this last prime will be less than 50 because 2010<50^2, if that makes u feel any better).

this was exactly my thinking when i showed mulan through pm hahaha just cudn't be bothered finishing it lol

Oh from memory:

D
A
A
C
A
B
E
NO FUCKING IDEA
B
D
12,13 can't remember
A
C If 1 isn't a prime
16,17 can't remember
E
...
25 was B
I remember 26 and 28 to be what I posted. Rest I don't remember.

Q8..E
Q12.B
Q13.c
Q15...i got D . .
and WHAT....Q9 is B???damn ..chose A
my consecutive is gone~~~

http://en.wikipedia.org/wiki/List_of_prime_numbers
1 is not a prime number according to that...
Therefore 15 D

according to convention, 1 is not a prime. If it were, it would make a lot of theorems very annoying...

Terry Tao ftw

did someone just mentioned my god.....?

Keke yes, Terry Tao is a beast,  i like watching his lectures etc; prime number theory almost made me blow my pants.
He taught me that 1 isn't a prime number -I forgot (cough-.

[luken93]

Did anyone else get 166 for Q29?
I just remembered I got that one aswell how could i forget

[brightsky]

Question 30 (for both Intermediate and Senior papers) is 24 from two eminent sources. The solution seems very logical/simple...hate AMC questions for that. 

[Martoman]

Yeah now i looked at the paper again I remember what I did. I reduced it to a simple you had 6 routes and 4 ways of possibly doing so so i just did 6*4 = 24

[luken93]

Yeah now i looked at the paper again I remember what I did. I reduced it to a simple you had 6 routes and 4 ways of possibly doing so so i just did 6*4 = 24

noooooo i rubbed that out, thinking that it was too simple!!!
what did you do for Q29?
I ended up with the equation:

[Andiio]

How are the triangles are congruent in Q28?
And how did everyone do Q19?

[Martoman]

How are the triangles are congruent in Q28?
And how did everyone do Q19?

PRTQ is a cyclic quad. You can find two triangles there, RTQ and PQR. Through combination of sin rule and cos rule all angles are the same for both triangles.

Also, all sides are the same as PQ = PR and the traingles share base QR.

[EvangelionZeta]

It was on today, I did the past exams from 2007-2009.. Got the first 20 right.. last 10 = mega fail.
Does that math even compare to methods/spesh?

Nope, it's quite a different type of maths to VCE, so I don't think methods/spech helps much with the AMC. And also, the AMC tests your speed and intuition too.

To be fair, a lot of VCE maths techniques work on the AMC - I distinctively remember a Methods-esque substitution question at around #15 in last year's paper, for instance.

I did pretty badly comparatively (didn't get any in the last 10 this year, sigh, although I hear 26 was relatively easy), but from what a three-time medalist said, it was probably the hardest Senior paper in a long time.

LOL around 15? What particularly VCE-esque? There's no calculus in comps - trig is useful though, graphing functions occasionally helps you determine the number of solutions but this is rare.
The #26 in Intermediate this year was a joke 3-time medallist = Colin Lu I bet lol Apparently the senior paper last year was the hardest in a while.

It was something like 5^x+5^(2x)=10, solve for x or something.  Probably harder than that, but of the same general form - it's the same question as half of the Methods substitution questions.

[brightsky]

, so a logical guess at the degree of the polynomial would be 4 as is closest to 3438.

So the polynomial is in the form , where .





We have only limited possibilities. We know that a must be 1 as . We also know that or wouldn't be big enough. Try . Therefore 1 of the pronumerals must be 0.

If we keep the remaining pronumerals as 1, then we get , so we know that .

Hence we get the solutions: .

So the polynomial is .

Therefore .

[luken93]

, so a logical guess at the degree of the polynomial would be 4 as is closest to 3438.

So the polynomial is in the form , where .





We have only limited possibilities. We know that a must be 1 as . We also know that or wouldn't be big enough. Try . Therefore 1 of the pronumerals must be 0.

If we keep the remaining pronumerals as 1, then we get , so we know that .

Hence we get the solutions: .

So the polynomial is .

Therefore .

yay! 3 right from 26 - 30

[luken93]

So when do results come out usually?
From memory is it just before Sept School Holidays?

And roughly how do you think I'd go with a score ~95-100 / 135

[Martoman]

idk, im expecting Prize or better given i missed 29 and a few multi choice. damn i wanted a shiny medal.

[brightsky]

idk, im expecting Prize or better given i missed 29 and a few multi choice. damn i wanted a shiny medal.

LOL, wow, I'm expecting like a Distinction/High Distinction :L