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March 29, 2024, 06:25:37 am

Author Topic: Victoria University Problem Solving Competition  (Read 5837 times)  Share 

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Mao

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Re: Victoria University Problem Solving Competition
« Reply #15 on: August 10, 2008, 12:20:17 pm »
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lolol, i took the privelage to not show up. 6 hours wasted in travel is not worth it considering the prize is a scholarship I will not even consider accepting :P

I feel like I have done a good deed today, haha, I helped others by increasing the chances of them being awarded a place in VU :P

Lol mao... I was trying to guess which asian you were
You woulda won for sure

no way
I only scored 6/8 for the online thing [quite relieved that I will not be obligated to attend second round]
and then an email popped through =S
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015

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Re: Victoria University Problem Solving Competition
« Reply #16 on: August 10, 2008, 12:25:20 pm »
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Here's what I can remember if you're interested

1, 2 (cbf drawing, just geometry anyway)

3. Two cars race around a circular track at constant speeds. If they go in opposite directions they will cross every 30 seconds. If they go in the same direction one will overtake the other every 120 seconds. The track is 1800m. Find the speeds of the cars.

4. . Find the smallest value of .

5. Find the sum of all the possible values of x.

cara.mel

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Re: Victoria University Problem Solving Competition
« Reply #17 on: August 10, 2008, 12:43:30 pm »
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I've done question 3 before but with different numbers in it in spec!

I liked that, I actually got how to set up the question :P

But 4 and 5, no idea how you would do it. Apart from maybe factor question 5 as first step :P And maybe make a all over 1 denominator. I dunno

dcc

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Re: Victoria University Problem Solving Competition
« Reply #18 on: August 10, 2008, 12:55:04 pm »
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4. . Find the smallest value of .

Question 4:

Rewrite as:

Since we know that , then we take note that

So to ensure that both sides of the equations are Natural numbers, let (since this is the first number which lets the LHS be divisible by the right.  b = 8 Would also work, however this would not be the <lowest> value of b)



Question 5:

5. Find the sum of all the possible values of x.

Now the only reason why a number would be if or or and a gives a positive 1 (for example )

First:

Second:

Now notice that the discriminant of is so this quadratic has no real solutions.

Adding these solutions together, we find that the answer (hopefully) is
« Last Edit: August 10, 2008, 12:58:42 pm by dcc »

ingramjack48

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Re: Victoria University Problem Solving Competition
« Reply #19 on: August 10, 2008, 01:07:19 pm »
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Prove (for x) that: 1 + 4  = 5

Define the y intercept for the parabola foruma of 3x^2

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Re: Victoria University Problem Solving Competition
« Reply #20 on: August 10, 2008, 01:11:11 pm »
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Second:


Noooooooo lol I forgot about that