• Help the next generation of students in the community for the new syllabus!
    Share your notes and exam papers on our Notes & Resources page
  • Like us on facebook here

Australian Maths Competition (1 Viewer)

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,468
Gender
Male
HSC
2010
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
Joined
May 6, 2008
Messages
1,468
Gender
Male
HSC
2010
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
Joined
Feb 8, 2008
Messages
1,143
Gender
Male
HSC
2009
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
Joined
Feb 8, 2008
Messages
1,143
Gender
Male
HSC
2009
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.
 

addikaye03

The A-Team
Joined
Nov 16, 2006
Messages
1,273
Location
Albury-Wodonga, NSW
Gender
Male
HSC
2008
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
 
Last edited:

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,468
Gender
Male
HSC
2010
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)
 

lolokay

Active Member
Joined
Mar 21, 2008
Messages
1,058
Gender
Undisclosed
HSC
2009
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; )

 
Last edited:

lolokay

Active Member
Joined
Mar 21, 2008
Messages
1,058
Gender
Undisclosed
HSC
2009
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
 
Last edited:

Dragonmaster262

Unorthodox top student
Joined
Jan 1, 2009
Messages
1,386
Location
Planet Earth
Gender
Male
HSC
2010
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
Joined
Jun 16, 2007
Messages
847
Gender
Undisclosed
HSC
N/A
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
Joined
Aug 6, 2009
Messages
2
Gender
Female
HSC
2011
How about the solution of:
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
Joined
May 10, 2007
Messages
10,178
Gender
Undisclosed
HSC
N/A
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
Joined
Jun 16, 2007
Messages
847
Gender
Undisclosed
HSC
N/A
How about the solution of:
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





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.





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





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,





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



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




Good luck everyone for the competition today =)
 
Last edited:

maths94

Member
Joined
Jul 14, 2009
Messages
448
Gender
Male
HSC
1998
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:mad: i need help on dem,, and ye just a general chat on how u went in the AMC test.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top