# Australian Maths Competition (1 Viewer)

#### Mongoose528

##### Member
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

##### Member
Good luck for tomorrow everyone

#### He-Mann

##### Vexed?
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,

\bg_white 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.

_______________

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?

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#### si2136

##### Well-Known Member
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

##### Member
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

##### Well-Known Member
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

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

#### si2136

##### Well-Known Member
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

##### Member
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

##### Well-Known Member
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

##### Member
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

##### Member
Anyone else complete the intermediate?

#### worldno17

##### Active Member
Anyone else complete the intermediate?
Yeah, I did. Do you know what's the right answer for the toy soldier question?

#### Mongoose528

##### Member
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

##### Member
Anyone from here doing the AIMO?

#### mathpie

##### New Member
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

##### Member
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

##### Member

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

101

#### si2136

##### Well-Known Member
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
I'm trying to remember a q from there, can u recall all of them?