1. ## Re: Westpac Maths Comp marathon

Originally Posted by d3st1nyLiang
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.

2. ## Re: Westpac Maths Comp marathon

Originally Posted by shaon0
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 $100a + 10b + c = 19a + 19b + 19c$, we have:

$81a = 9b + 18c$

ie

$9a = b + 2c$

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

$0 \le b,c \le 9$

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

$9a = 9 + 2 \times 9$

$\implies a = 3$

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

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

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

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

3. ## 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

4. ## Re: Westpac Maths Comp marathon

or is it (c)?

5. ## Re: Westpac Maths Comp marathon

Originally Posted by kurt.physics
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 ten's digit and unit's digit respectively, then

$0 \le b,c \le 9$

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

$9a = 9 + 2 \times 9$

$\implies a = 3$

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

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

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

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.

6. ## Re: Australian Maths Competition

Originally Posted by kurt.physics
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.

7. ## Re: Westpac Maths Comp marathon

Originally Posted by nerdsforever
or is it (c)?

Here my solution:

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

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

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

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

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

8. ## Re: Australian Maths Competition

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!

10. ## 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

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

$u_{n} = u_{n-1}- u_{n-2}$ for n≥ 3.

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

Is it E) ?

12. ## Re: Westpac Maths Comp marathon

Originally Posted by nerdsforever
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

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

$u_{2} = \pi$

$u_{n} = u_{n-1}- u_{n-2}$ for n≥ 3.

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

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

$u_{2} = \pi$

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

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

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

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

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

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

We see the pattern emerging s.t.

$u_{1} = u_{7}$

$u_{2} = u_{8}$

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

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

$2008 \equiv 4 \mod 6$

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

13. ## 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?

14. ## Re: Australian Maths Competition

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

AMC papers (this is in Chinese ;P)

AMC papers (in English)

15. ## Australian Mathematics Competition

Hi everyone~
i was just wondering if anyone had the AMT solutions for year 2002 - 2006 (intermediate division)???

16. ## Re: Australian Mathematics Competition

Originally Posted by appl3licious
Hi everyone~
i was just wondering if anyone had the AMT solutions for year 2002 - 2006 (intermediate division)???
Here they are =)

17. ## Re: Australian Maths Competition

I have posted the solutions here

Solutions

18. ## Re: Australian Mathematics Competition

Originally Posted by kurt.physics
Here they are =)
thank you for sharing~ u really saved me ^^

19. ## Re: Westpac Maths Comp marathon

A train travelling at constant speed takes a quarter of a minute to pass a signpost
and takes three-quarters of a minute to pass completely through a tunnel which is
600m in length. The speed of the train, in kilometres per hour, is
(A) 50 (B) 56 (C) 64 (D) 72 (E) 80

I really dont get this question, even though it is extremely simple. Shouldnt it be 48 km/h?

BTW no idea to the above.

20. ## Re: Westpac Maths Comp marathon

How many different pairs of 2-digit numbers multiply to give a 3-digit number with
all digits the same?
(A) 5 (B) 6 (C) 7 (D) 8 (E) 9

21. ## Re: Westpac Maths Comp marathon

For the train Q, (D) is the answer.
Remember that the train takes 45 seconds to travel the distance of 600m+its own length.

22. ## Re: Westpac Maths Comp marathon

Originally Posted by Lukybear
How many different pairs of 2-digit numbers multiply to give a 3-digit number with
all digits the same?
(A) 5 (B) 6 (C) 7 (D) 8 (E) 9
111=37x3
222=37x6
333=37x9
444=37x12
555=37x15
666=37x18
777=37x21
888=37x24=74x12
999=37x27
(C) 7.

23. ## Re: Westpac Maths Comp marathon

Originally Posted by kurt.physics
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?
$\\x+y+z=4 \\x^2+y^2+z^2=10 \\x^3+y^3+z^3=22 \\ \\x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx) \\10=16-2(xy+yz+zx) \\xy+yz+zx=3 \\ \\x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) \\22-3xyz=4(10-3) \\xyz=-2 \\ \\x^4+y^4+z^4=(x^2+y^2+z^2)^2-2xyz(x+y+z) \\=100+4(4) \\=116 \\ \\LOL... This way is too complicated.$

24. ## Re: Westpac Maths Comp marathon

Originally Posted by kurt.physics
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?
Consider polynomial: P(x)=x^4-4x^3+8x^2+(8/3)x+a0

S1=a+b+c=4
--> S1-4=0

S2=a^2+b^2+c^2=10
-->S2-4(S1)+16=0

S3=a^3+b^3+c^3=22
--> S3-4(S2)+8(S1)+8=0

S4=a^4+b^4+c^4=x
--> S4-4(S3)+8(S2)+(8/3)S1+4(y)=0

S4=88-80-32/3-4a0

S4=-(8/3+4a0) Where a0 is the constant of P(x)

Thought this method would work but i would need to know a0 for it to work...fml

The method your looking for though i'll do tomoz if not answered

Update: Above was the method i was referring to

25. ## Re: Westpac Maths Comp marathon

*study freak:

what rule is applied when you say:
x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+yz+xz)

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