Australian Maths Comp

[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.

[Mulan]

omygsh whose the 3 times medallist?! :O

[InitialDRulz]

I have the paper, but i don't have a scanner, so i'll do the next best thing, take photos of it and hope that you guys can read it lol
*Image Below*
If thats alright with you guys, then yeah i'll do the whole thing lol

[simonhu81292]

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~~~

[InitialDRulz]

EDIT: Question 10 is fuzzy, another photo

[Andiio]

Could someone go through the method to complete Q26?

I got a quadratic which when factorised gave a surd answer so I didn't know what to do..

[kamil9876]

so

thus





edit: lol i kinda miss high school... but meh uni problems are heaps better

[Martoman]

Given



We want so its reasonable to cube the first equation.




Crux move #1 -> -> crux move #2
                       

hmmmm we want mn somehow. Oh I know in the other equation looks like an expansion of with the devilishly absent.

so from the second equation:
                                       
                                       

[Andiio]

ahh i got mn = 11 as well >_>

[Mulan]

wats the answer for q. 27 on senior??? Intermediate had that q. too. I wrote 123 in a rush. totally wrong cause of the first no. having to be a four digit no.

oh and wats the answer to this q.

what is the remainder when 2^2010 is divided by 7?
a)1
b)2
c)3
d)4
e)5

[/0]

what is the remainder when 2^2010 is divided by 7?
a)1
b)2
c)3
d)4
e)5

If leaves remainder 1 upon division by then leaves remainder 1 upon division by .

Since:
, .
.

Now, leaves remainder 1.

So leaves remainder 1.

[Martoman]

what is the remainder when 2^2010 is divided by 7?
a)1
b)2
c)3
d)4
e)5

If leaves remainder 1 upon division by then leaves remainder 1 upon division by .

Since:
, .
.

Now, leaves remainder 1.

So leaves remainder 1.

why so pro?

LOL I just found that it repeats remainder of 1 for every power of 2 which is a multiple of 3. that is 2^3 = 8 remainder 1, 2^6 = 64 remainder 1.

Then 2010/3 = 670 so it will have a remainder 1.

[Mulan]

is it jst me but the maths comp papers were WAYYYY harder than last yrs! i guessed 15 q.s...half of it :S
who else did intermediate and found that most of teh answers were d and a? (8

Yeah, for the start. It was like DDDA IIRC
Couldn't remember the rest, but yeah, there were a handful of A's.
Did you do any of questions 26-30?

no. attempted and failed. for some reason most of my answers weren't even part of the multiple choice...
wat did you get for q.26 and 28 btw for short answer (8

I've got 76 and 789 respectively.
Lol, I just went back and looked at the paper, question 11. And now I'm wondering if I actually even answered that question.
FML

I FULL ON GUESSED Q.11. spent like 10 minutes on it :S wat a waste of time...
ohhhs i got 22 for q. 26  and not 76 :S

what is the remainder when 2^2010 is divided by 7?
a)1
b)2
c)3
d)4
e)5

If leaves remainder 1 upon division by then leaves remainder 1 upon division by .

Since:
, .
.

Now, leaves remainder 1.

So leaves remainder 1.

OMYGSH I GOT IT RIGHT! even though it was a lucky mid-guess.
thats a very smart way to do it (only understood the last two lines of your working out XD). first i worked out that the last digit of 2^2010 had to be 4. Then for some reason i went 64/7= something remainder 1. then i jst wrote 1. (8
hehee what luck!

[/0]

what is the remainder when 2^2010 is divided by 7?
a)1
b)2
c)3
d)4
e)5

If leaves remainder 1 upon division by then leaves remainder 1 upon division by .

Since:
, .
.

Now, leaves remainder 1.

So leaves remainder 1.

why so pro?

LOL I just found that it repeats remainder of 1 for every power of 2 which is a multiple of 3. that is 2^3 = 8 remainder 1, 2^6 = 64 remainder 1.

Then 2010/3 = 670 so it will have a remainder 1.

lol I guess we basically had the same idea though. If you want to harness the full power of remainders you should investigate Modular arithmetic

[Mulan]

wats the answer for this question no. 26 on intermediate paper? O:

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

i got 22 btw. no idea if this is right