Australian Maths Competition (9 Viewers)

Mongoose528 · Jul 26, 2017

Jonothancroller said:
Hey, I've got some questions that I haven't been able to do. Would anyone care to do these questions as practice and offer an explanation? Thanks.

http://imgur.com/a/YfigO

Here's a solution to one of them:

Mongoose528 · Jul 26, 2017

Good luck for tomorrow everyone

He-Mann · Jul 27, 2017

Jonothancroller said:
Hey, I've got some questions that I haven't been able to do. Would anyone care to do these questions as practice and offer an explanation? Thanks.

http://imgur.com/a/YfigO

For the third question from the top,

$$Let $q(k) = p(k) - k^3.$ Then we wish to solve $q(k) = 0$, or to find the number of zeros $q(k)$ has. Sub in 100, $q(100) = -9900$. We can deduce the structure of $q$ by examining the prime factorization of -9900. \begin{align*}-9900 &= -3 \times 3 \times 11 \times 2 \times 2 \times 5 \times 5 \\ &= (-3) \times 3 \times 11 \times (-2) \times (2) \times (-5) \times (5). \end{align*} \\You could stop here but you wouldn't get be able to find the maximum number of roots of $q(k)$ because it isn't the 'worst-case'. The trick is to use the number $\textit{one}: \\\\ -9900 = (-3) \times 3 \times 11 \times (-2) \times (2) \times (-5) \times (5) \times (-1) \times 1 \times (-1). \\ \\ $Then $ q(100)$ becomes\\ \begin{align*}q(100) &= (100 - 103)(100 - 97)(100-89)(100-102)(100-98)(100-105)(100-95)(100-101)(100-99)(100-101) \end{align*}\\ Replace 100 with $x$ and multiply by some integer $m$ and you'll the 'worst-case' polynomial. Thus, maximum number of integer solutions for $q(k)$ is 9.$

_______________

For the reader to think about:

a) Why can't we use infinitely many 1s and -1s during the prime factorization?

b) Why does the 'worst-case' polynomial have 9 integer solutions?

si2136 · Jul 27, 2017

Senior was earlier than previous years I reckon this year. Couldn't do the last question though ran out of time. But the other 4 I think I did it right

Mongoose528 · Jul 27, 2017

si2136 said:
Senior was earlier than previous years I reckon this year. Couldn't do the last question though ran out of time. But the other 4 I think I did it right

You seem a great chance for a medal. I'm pretty confident I got 87 points and I'm hoping that one of the 3 I guessed were correct.

si2136 · Jul 27, 2017

Mongose528 said:
You seem a great chance for a medal. I'm pretty confident I got 87 points and I'm hoping that one of the 3 I guessed were correct.

Haahaha maybe just miss it and get HD

Mongoose528 · Jul 27, 2017

Surely you'd get a prize. I'm hoping for a prize, fed up of missing on them by 2/3 points.

si2136 · Jul 27, 2017

Mongose528 said:
Surely you'd get a prize. I'm hoping for a prize, fed up of missing on them by 2/3 points.

Depends on how the cohort goes. There were 3 others that I had to guess that were between 2 choices.

Mongoose528 · Jul 27, 2017

si2136 said:
Depends on how the cohort goes. There were 3 others that I had to guess that were between 2 choices.

Have questions about the area of a hexagon or a n-gon?

si2136 · Jul 27, 2017

Mongose528 said:
Have questions about the area of a hexagon or a n-gon?

A bit. It was liek a shuriken. You had to find the area of circle then use 180(n-2) to find the angles.

Mongoose528 · Jul 27, 2017

si2136 said:
Depends on how the cohort goes. There were 3 others that I had to guess that were between 2 choices.

Good luck, I hope you get at least a prize.

Mongoose528 · Jul 27, 2017

Anyone else complete the intermediate?

worldno17 · Jul 27, 2017

Mongose528 said:
Anyone else complete the intermediate?

Yeah, I did. Do you know what's the right answer for the toy soldier question?

Mongoose528 · Jul 27, 2017

worldno17 said:
Yeah, I did. Do you know what's the right answer for the toy soldier question?

Didn't get up to it :/, only had time to concentrate on the first 27 questions, I hate the short time limit. What did you get for 26/27?

Mongoose528 · Jul 29, 2017

Anyone from here doing the AIMO?

mathpie · Jul 30, 2017

Re: Westpac Maths Comp marathon

What did everyone get for the last 5 (free response) questions in the senior division? I couldn't get any.. lol

Mongoose528 · Jul 30, 2017

How would you solve this algebraically: A nude number is a natural number of whose digits are a factor of the number. Find all 3 digit nude numbers where no digits are repeated.

Can't seem to make in roads :/

I can solve it case by case, but I want to know a quicker way of how to do it.

Mongoose528 · Jul 30, 2017

_Anonymous said:
There was this question from the Intermediate section (the last one).

It was something like: find how many digits are in a googolplex (10^10 to the power of 100).

I'm getting 102, because 10^100 should be 1 followed by 100 zeros right? So if that's the case, shouldn't 10 to the power of 10 to the power of 100 be 102 digits since it's 10 to the power of 1 and 100 zeros? Some people are telling me it's 101 digits.

1 followed by 100 zeros is 101 digits

si2136 · Jul 31, 2017

si2136 · Jul 31, 2017

Re: Westpac Maths Comp marathon

mathpie said:
What did everyone get for the last 5 (free response) questions in the senior division? I couldn't get any.. lol

I'm trying to remember a q from there, can u recall all of them?

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