Australian Maths Competition (1 Viewer)

kurt.physics

Member
Re: Australian Mathematics Competition

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

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kurt.physics

Member
I have posted the solutions here

Solutions

appl3licious

New Member
Re: Australian Mathematics Competition

Here they are =)
thank you for sharing~ u really saved me ^^

Lukybear

Active Member
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.

Lukybear

Active Member
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

study-freak

Bored of
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.

study-freak

Bored of
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
111=37x3
222=37x6
333=37x9
444=37x12
555=37x15
666=37x18
777=37x21
888=37x24=74x12
999=37x27
(C) 7.

study-freak

Bored of
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?
$\bg_white \\x+y+z=4 \\x^2+y^2+z^2=10 \\x^3+y^3+z^3=22 \\ \\x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx) \\10=16-2(xy+yz+zx) \\xy+yz+zx=3 \\ \\x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) \\22-3xyz=4(10-3) \\xyz=-2 \\ \\x^4+y^4+z^4=(x^2+y^2+z^2)^2-2xyz(x+y+z) \\=100+4(4) \\=116 \\ \\LOL... This way is too complicated.$

The A-Team
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?
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

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Lukybear

Active Member
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)

gurmies

Drover
Re: Westpac Maths Comp marathon

no rule, just a handy factorisation

lolokay

Active 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?
is the answer to this 50?

figure out my notation; )

$\bg_white \\ \alpha =4\\ \alpha ^2 = 10\\ \alpha ^3 = 22\\ \alpha ^4=???\\ \\ 1) (\alpha ^2)^2=\alpha ^4+2\alpha ^2\beta ^2=100\\ 2) (\alpha )(\alpha ^2)=\alpha ^3+\alpha \beta ^2=40\\ 3) (\alpha )(\alpha ^3)=\alpha ^4+\alpha\beta ^3=88\\ 4) (\alpha)^3=\alpha ^3+3\alpha \beta ^2+6\alpha \beta \gamma =64\\ 5) (\alpha )(\alpha \beta ^2)=2(\alpha \beta \gamma)(\alpha )+\alpha\beta ^3+2\alpha ^2\beta ^2 \\ \\ 2)->\alpha \beta ^2=18\\ 4)->\alpha \beta \gamma =-2\\ 1)+3),5) ->2\alpha ^4+(\alpha )(\alpha \beta ^2)-2(\alpha \beta \gamma)(\alpha )=188\\ 2\alpha ^4 = 188 - 4*18+2*-2*4\\ 2\alpha ^4 = 100\\ \alpha ^4 = 50$

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kurt.physics

Member
Re: Westpac Maths Comp marathon

is the answer to this 50?

figure out my notation; )

$\bg_white \\ \alpha =4\\ \alpha ^2 = 10\\ \alpha ^3 = 22\\ \alpha ^4=???\\ \\ 1) (\alpha ^2)^2=\alpha ^4+2\alpha ^2\beta ^2=100\\ 2) (\alpha )(\alpha ^2)=\alpha ^3+\alpha \beta ^2=40\\ 3) (\alpha )(\alpha ^3)=\alpha ^4+\alpha\beta ^3=88\\ 4) (\alpha)^3=\alpha ^3+3\alpha \beta ^2+6\alpha \beta \gamma =64\\ 5) (\alpha )(\alpha \beta ^2)=2(\alpha \beta \gamma)(\alpha )+\alpha\beta ^3+2\alpha ^2\beta ^2 \\ \\ 2)->\alpha \beta ^2=18\\ 4)->\alpha \beta \gamma =-2\\ 1)+3),5) ->2\alpha ^4+(\alpha )(\alpha \beta ^2)-2(\alpha \beta \gamma)(\alpha )=188\\ 2\alpha ^4 = 188 - 4*18+2*-2*4\\ 2\alpha ^4 = 100\\ \alpha ^4 = 50$
Yes, you are right =)

Would you like to post a new question?

lolokay

Active Member
Re: Westpac Maths Comp marathon

A rectangular prism 77 x 81 x 100 is cut into cubes of edge 1 by
planes parallel to the faces of the prism. A given internal diagonal
pierces how many of these cubes?
(A) 255 (B) 256 (C) 257 (D) 258 (E) 259

+ty kurt for the link to the papers. I'm gonna do them all eventually. just tried one, made a few silly errors and one question i wasn't sure about

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Dragonmaster262

Unorthodox top student
Re: Australian Mathematics Competition

Does the Australian Maths Competition only have stuff from the 4 unit Maths course or does it have any stuff from Uni maths?

kurt.physics

Member
Re: Australian Mathematics Competition

Does the Australian Maths Competition only have stuff from the 4 unit Maths course or does it have any stuff from Uni maths?
The AMC is generally a problem solving exam. You will see that the first 10 questions are rudimentary. The next 5 will be more problem solving, the last 15 will be problem solving although knowledge of both number theory and combinatorics and a solid understanding of geometry and algebra would certainly make the last few questions 5 times easier.

sharin_3

New Member
The number 2008! (factorial 2008) means the product of all the integers 1,2,3,4,...,2007,2008. With how many zeroes does 2008! end?

Tomorrow's the competition, and I need to know the solution of this wuestions, please? Thanks

Aerath

Retired
Re: Australian Mathematics Competition

Yeah, I usually bomb these Maths Comps. Usually a combination of me not giving a damn and I also suck at problem solving. Never ever get HDs, but I usually do ok in school maths, so heh.

kurt.physics

Member
The number 2008! (factorial 2008) means the product of all the integers 1,2,3,4,...,2007,2008. With how many zeroes does 2008! end?

Tomorrow's the competition, and I need to know the solution of this wuestions, please? Thanks
The amount of zeros a number ends in is the amount of 10s that are in the prime factorization. Now 10 = 2 x 5. And so when we prime factorize 2008! we need to count the amount of 2s and the amount of 5s.

Now there is clearly going to be more 2s than 5s, so to determine how many factors of 10 there are in the prime factorization of 2008! we must only find how many 5s are in the prime factorization.

Now clearly finding the prime factorization of 2008! would be a torturous task, so we will "look" at each term of 2008! ie 1, 2, 3, 4, 5, 6, . . . ,2007, 2008.

So how many multiples 5s are there between 1 and 2008, well

$\bg_white 5m < 2008$

$\bg_white m <401.6$

As m is an integer then there are 401.

Now some terms in the prime factorization will have 25 as a factor, and hence that terms will have 2 factors of 5. So we must count how many terms are multiples of 25. Now we already have counted 1 of those factors of 5 (out of those numbers that are multiples of 5^2), and so when we determine how many factors are multiples of 5^2, they will add 1 extra multiple of 5 into our count.

$\bg_white 25n < 2008$

$\bg_white \therefore n < 80.32$

As n is an integer then there are 80.

But some of the factors of 2008! will be multiples of 5^3 = 125, so when we count these by the same logic as before they will contribute an extra 1 multiple of 5 into our count, so

$\bg_white 125p < 2008$

$\bg_white p< 16.064$

As p is an integer then there are 16.

But some of the factors of 2008! will be multiples of 5^4 = 625, and so these will add an extra 1 multiple of 5 into our count. So,

$\bg_white 625j < 2008$

$\bg_white \therefore j < 3.2128$

As j is an integer there are 3 of these.

As 5^5 is greater than 2008, then there are no multiples of 5^5 up to 2008.

So in total the amount of factors of 5 in 2008! are

$\bg_white 401 + 80 + 16 + 3 = 500$

Therefore there are 500 factors of 10 in 2008! and hence 2008! ends in 500 zeros.

Good luck everyone for the competition today =)

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maths94

Member
How did u go in the Australian Maths Competition?

today (6-8-09) i did my intermediate AMC test.This is just a thread to see which question u were stuck on ...q26-30 i need help on dem,, and ye just a general chat on how u went in the AMC test.