# Australian Maths Competition (1 Viewer)

#### JayNoob

##### zzz
Oh sweet thanks Kurt. Lol chinese website xD interesting.

#### kurt.physics

##### Member
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 it gets to post up another westpac question and then who ever solves that posts up another westpac question etc. We should start off fairly easy and increase the difficulty of the questions as the marathon progresses.

If you dont have any of the papers i have posted them here:

http://community.boredofstudies.org...australian-maths-competition.html#post4478693

Or if you have some of the questions from an earlier year feel free to post them =)

QUESTION 1

A rectangle has its length 25 times its width. What is the ratio of
its perimeter to the perimeter of the square of the same area?

-(A) 13 : 5 (B) 13 : 10 (C) 5 : 1 (D) 51 : 20 (E) 51 : 10

#### d3st1nyLiang

Re: Westpac Maths Comp marathon

is it a)?

A lucky number is a positive integer which is 19 times the sum of its digits. How
many different lucky numbers are there?

#### clue444

##### Member
The 2008 answers are in chinese damn lol. Thanks heaps Kurt!

#### gurmies

##### Drover
Re: Westpac Maths Comp marathon

Yes, (a)

#### shaon0

##### ...
Re: Westpac Maths Comp marathon

is it a)?

A lucky number is a positive integer which is 19 times the sum of its digits. How
many different lucky numbers are there?
I don't know if my methods right:
By using the conditions in the questions. It's easy to deduce that the number isn't a 1,2 or 4 digit number. So it's a 3 digit no.
Let the digits be as follow: 100 digits in hundreds place, 10 digits in 10s place and 10 digits in 1's place.

Alternatively, I think you can just go:
Z*19=ABC where {Z,A,B,C E Z}
So say for: a no. which has digits summing to 6 is:
6*19=114

Thus, Z is in between 5 and 16
So 10 numbers.

Last edited:

#### kurt.physics

##### Member
Re: Westpac Maths Comp marathon

I don't know if my methods right:
By using the conditions in the questions. It's easy to deduce that the number isn't a 1,2 or 4 digit number. So it's a 3 digit no.
Let the digits be as follow: 100 digits in hundreds place, 10 digits in 10s place and 10 digits in 1's place.

Alternatively, I think you can just go:
Z*19=ABC where {Z,A,B,C E Z}
So say for: a no. which has digits summing to 6 is:
6*19=114

Thus, Z is in between 5 and 16
So 10 numbers.
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 $\bg_white 100a + 10b + c = 19a + 19b + 19c$, we have:

$\bg_white 81a = 9b + 18c$

ie

$\bg_white 9a = b + 2c$

Note that as b and c are the ten's digit and unit's digit respectively, then

$\bg_white 0 \le b,c \le 9$

So the maximum $\bg_white (b,c)$ is $\bg_white (9,9)$ and therefore the maximum a is

$\bg_white 9a = 9 + 2 \times 9$

$\bg_white \implies a = 3$

So we must only check for $\bg_white a = 1,2,3$

For a = 1, we have $\bg_white (b, c) = (1, 4), (3,3), (5,2), (7,1), (9,0)$
For a = 2, we have $\bg_white (b, c) = (0, 9), (2,8), (4,7), (6,6), (8,5)$
For a = 3, we have $\bg_white (b, c) = (9, 9)$, and there are no other solutions.

Hence there are exactly 11 lucky numbers, namely, $\bg_white 114, 133, 152, 171, 190, 209, 228, 247, 266, 285, 399.$

We should post up easier questions first just to get the marathon going =)

#### kurt.physics

##### Member
Re: Westpac Maths Comp marathon

PROBLEM 3

Tina has a large number of 1 × 2 × 6 rectangular blocks. She wants
to make a solid cube out of the blocks. What is the smallest number
of blocks she needs?

(A) 6 (B) 12 (C) 18 (D) 36 (E) 144

#### nerdsforever

##### hsc zomg
Re: Westpac Maths Comp marathon

or is it (c)?

#### shaon0

##### ...
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 $\bg_white 100a + 10b + c = 19a + 19b + 19c$, we have:

$\bg_white 81a = 9b + 18c$

ie

$\bg_white 9a = b + 2c$

Note that as b and c are the ten's digit and unit's digit respectively, then

$\bg_white 0 \le b,c \le 9$

So the maximum $\bg_white (b,c)$ is $\bg_white (9,9)$ and therefore the maximum a is

$\bg_white 9a = 9 + 2 \times 9$

$\bg_white \implies a = 3$

So we must only check for $\bg_white a = 1,2,3$

For a = 1, we have $\bg_white (b, c) = (1, 4), (3,3), (5,2), (7,1), (9,0)$
For a = 2, we have $\bg_white (b, c) = (0, 9), (2,8), (4,7), (6,6), (8,5)$
For a = 3, we have $\bg_white (b, c) = (9, 9)$, and there are no other solutions.

Hence there are exactly 11 lucky numbers, namely, $\bg_white 114, 133, 152, 171, 190, 209, 228, 247, 266, 285, 399.$

We should post up easier questions first just to get the marathon going =)
Yeah it is 11. lol. I counted my solutions in between 5 and 16 incorrectly.

#### girish

##### New Member
Here is 2008 q + a and 2006 answers.
Hi Kurt,

Thanks for these papers and answers for AMC. For some years there was just the answers and not teh solution methods as such. Can you send these also if you have, say for 2008, 2006,2005,2004 and 2003? I am taking part inthe JUnior level. Thanks.

#### kurt.physics

##### Member
Re: Westpac Maths Comp marathon

or is it (c)?

Here my solution:

$\bg_white What ever way Tina arranges the rectangular boxes, each side of the box must evenly divide the side of the cube. This implies that the side of the cube is a multiple of\,1, 2\, and\, 6\, and therefore the side of the cube is a multiple of\, LCM[1,2,6] = 6$

$\bg_white The smallest multiple of\,6\,is\, 6\,, so the volume of the cube is\, 6^3$

$\bg_white The volume of each rectangular box is\,1 \times 2 \times 6 = 12$

$\bg_white So the smallest amount of boxes Tina will need is:$

$\bg_white \frac{6^3}{12} = 18 \implies \boxed{C}$

#### Aquawhite

##### Retiring
I think I am meant to be in this >_>... I'm really not in the mood for doing this since the Trials are just around the corner and all I've focused on is the syllabus and not really extra curricular (interesting maths ...).

Ima probably end up with just a participation or a credit if I'm lucky probably. Not that I care XD

thx kurt!

#### nerdsforever

##### hsc zomg
Re: Westpac Maths Comp marathon

Cheers kurt! you're awesome. what do you usually get in the maths comp?

Here's a new question.

A sequence {u1, u2, . . . , un} of real numbers is defined by

$\bg_white u_{1} = \sqrt{2}$
$\bg_white u_{2} = \Pi$

$\bg_white u_{n} = u_{n-1}- u_{n-2}$ for n≥ 3.

What is u2008?
(A) −√2
(B) 2008(√2 − 2008pi)
(C) 1003√2 − 1004pi
(D) pi
(E)√2

#### whackymac

##### New Member
Re: Westpac Maths Comp marathon

Is it E) ?

#### kurt.physics

##### Member
Re: Westpac Maths Comp marathon

Cheers kurt! you're awesome. what do you usually get in the maths comp?

Here's a new question.

A sequence {u1, u2, . . . , un} of real numbers is defined by

$\bg_white u_{1} = \sqrt{2}$

$\bg_white u_{2} = \pi$

$\bg_white u_{n} = u_{n-1}- u_{n-2}$ for n≥ 3.

What is u2008?
(A) −√2
(B) 2008(√2 − 2008pi)
(C) 1003√2 − 1004pi
(D) pi
(E)√2
$\bg_white u_{n} = u_{n-1} - u_{n-2}$

$\bg_white u_{1} = \sqrt{2}$

$\bg_white u_{2} = \pi$

$\bg_white u_{3} = \pi - \sqrt{2}$

$\bg_white u_{4} = (\pi - \sqrt{2}) - \pi = - \sqrt{2}$

$\bg_white u_{5} = - \sqrt{2} - \pi + \sqrt{2} = - \pi$

$\bg_white u_{6} = - \pi + \sqrt{2}$

$\bg_white u_{7} = - \pi + \sqrt{2} + \pi = \sqrt{2}$

$\bg_white u_{8} = \sqrt{2} + \pi - \sqrt{2} = \pi$

We see the pattern emerging s.t.

$\bg_white u_{1} = u_{7}$

$\bg_white u_{2} = u_{8}$

$\bg_white u_{n} = u_{n+6}$

So consider the pattern in modulo 6.

The 2008th term will be equivalent a specific residue modulo 6 and hence we can determine which term the 2008th term is equivalent to

$\bg_white 2008 \equiv 4 \mod 6$

$\bg_white u_{2008} = u_{4} = - \sqrt{2} \implies \boxed{A}$

#### kurt.physics

##### Member
Re: Westpac Maths Comp marathon

Problem 5

The sum of three numbers is 4, the sum of their squares is 10 and the sum of their
cubes is 22. What is the sum of their fourth powers?

#### kurt.physics

##### Member
I re-found the website which has the papers =)

AMC papers (this is in Chinese ;P)

AMC papers (in English)