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Re: Westpac Maths Comp marathon Close, the actual answer is 11 :| Here is the solution: As shaon showed, the lucky number must be a 3 digit number. Suppose the number is abc. Then 100a + 10b + c = 19a + 19b + 19c, we have: 81a = 9b + 18c ie 9a = b + 2c Note that as b and c are the...

The polynomial will be in the form f(x) = \left(x^2 - 9\right)(x^2 + d) with d>0 (2^2 - 9)(4 + d) = -45 (4-9)(4 + d) = -45 -5(4 + d) = -45 d + 4 = 9 \therefore d = 5 \therefore f(x) = (x^2 - 9)(x^2 + 5) edit: beaten ;)

Westpac Maths Comp marathon okay, so the Westpac maths comp (aka AMC) is coming up so for those who are doing it or just want do some hard maths problems this "marathon" is a kind of warm up =) So to get the ball rolling, i will post up a westpac maths question and who ever correctly solves...

http://community.boredofstudies.org/129/year-10-school-certificate/215824/australian-maths-competition.html#post4478728 This has all the worked solutions to the 2007 papers. And also on posts above the one in the link there is the papers from 2002-2006. The 2008 papers are a bit down the same...

wtf? lol which amc paper will you do in year 12? Even though im doing year 12 maths i still get to do the intermediate paper ;P Goodluck!

\triangle = 4\left(x^2 + c^2 + y^2\right) if x = y = c = 0 then \triangle = 4 \times 0 = 0 \therefore there is just one root. if any of x, y or c is >0 for all values then your statement "thus more than 1 root exists" would be correct. But if \triangle \ge 0 then there is either 1 or 2 roots.

Westpac is on the 6th of August, ie next Thursday :S Are you doing it?

The last question you asked involved a bit more problem solving. I had no idea what to do. So when you start a problem, instead of jumping in and jumbling up equations to try some how push a solution out, think about ways to go about solving the problem. Doing it graphically doesn't look to...

Your welcome. Good luck ;)

What makes this problem hard is that it is in terms of 2 variables. But we are given that x + y = 1, so y = 1 - x. Substitute this into the equation 7x^2 - 4x(1-x) + 9(1-x)^2 = 7x^2 - 4x + 4x^2 + 9 - 18x + 9x^2 = 7x^2 + 4x^2 + 9x^2 - 4x - 18x + 9 =20x^2 - 22x + 9 The minimum value that this...

To solve this problem use the second equation (the linear one) and make x the subject of the equation (or whatever), you will find it is y = \frac{-(2ax + c)}{2b} Squaring gives y^ = \frac{\left(2ax + c\right)^2}{4b^2} plug this into the first equation x^2 + \frac{\left(2ax +...

Consider the first equation: (a-b)^2 + (b-c)^2 + (c-a)^2 = 0 Then, (a-b)^2 \geq 0 (b-c)^2 \geq 0 (c-a)^2 \geq 0 (This is because they are all squares) So it can only equal zero when (a-b)^2 = 0 (b-c)^2 = 0 (c-a)^2 = 0 ie when a= b b= c c= a Combining these three statements...

lol about choosing senior =) Would you possibly be able to private message me the questions please =)

How come you did the senior exam? P.S. You wouldn't happen to still have the questions?

The formula we mentioned was derived from generalizing that particular function then differentiating with the chain rule.

f(x) = log_{e} \left(2x + 1\right) \frac{d}{dx}\left(log_{e} f(x) \right) = \frac{f'(x)}{f(x)} \therefore f'(x) = \frac{2}{2x+1} f'(1) = \frac{2}{3}

She is actually fairly good at mathematics, in the past year she has consistently been coming top 6 in all the national competitions. I think she would be dev because she was sick during the exam and so didn't do as well as she could have.

Here is a photo:

The Senior papers i posted on the forum (the senior ones are the ones with the "s" on the end. And the only solutions i have are the ones i posted. I do have all the National Papers from 2004 and 2005 and the AMO paper from 2006. Which national papers are you interested? AMOC, AIMO, AMO?

I got them from some chinese website lol =)