Australian Maths Competition (5 Viewers)

[Paradoxica]

Re: Australian Maths Competition 2013

Why do you imply n-m must be a perfect square?

Actually, m only less than n.

True, that's not necessarily the case.
but I must ask again, are a,b,c allowed to be negative? If not, then the problem jumps a few notches in complexity...

 

[SimpletonPrime]

Re: Australian Maths Competition 2013

True, that's not necessarily the case.
but I must ask again, are a,b,c allowed to be negative? If not, then the problem jumps a few notches in complexity...

No, a,b,c are positive integers.

I got 199 for the question , almost forgot the case a=1...

 

[Paradoxica]

Re: Australian Maths Competition 2013

Why do you imply n-m must be a perfect square?

Oh, in the test, m only less than n.

I already took that back...

Since a,b,c can be assumed to be any integers, then all we have to do is partition the factors of 2016² into two sides and use multiplicity arguments to remove repeated solutions from the set, and then double the answer because of negative factorisations.

 

[Paradoxica]

Re: Australian Maths Competition 2013

No, a,b,c are positive integers.

I got 199 for the question , almost forgot the case a=1...

... ok then that just ruined my day a bit, looks like we need a bit more cleverness to determine the possible values of n-m...

 

[fluffchuck]

Re: Australian Maths Competition 2013

No, a,b,c are positive integers.

I got 199 for the question , almost forgot the case a=1...

O: I got 199 too, is that the correct answer?

 

[SimpletonPrime]

Re: Australian Maths Competition 2013

O: I got 199 too, is that the correct answer?

I hope so ...

 

[RealiseNothing]

Re: Australian Maths Competition 2013

If its asking for the maximum number of chords without intersection, not number of ways, then I think its 125, idk though, 75% sure its wrong.

Yeah I got this.

 

[seanieg89]

Re: Australian Maths Competition 2013

OK. I remembered two questions:

Given positive integer and where . How many values of ?

Uh, if a,b,c,m are positive integers, how can we possibly have am^2+bm+c=f(m)=0?

 

[SimpletonPrime]

Re: Australian Maths Competition 2013

Uh, if a,b,c,m are positive integers, how can we possibly have am^2+bm+c=f(m)=0?

I think a,b,c are integers. I'm sorry, I don't really remember the problem.

 

[Gana]

Here is 2005/2006

i cannot see them

 

[RealiseNothing]

i cannot see them

About 7 years too late lol

 

[Mongoose528]

i cannot see them

here http://files.chiuchang.org.tw:8080/myweb/download/docu/AMC/

 

[dan964]

***Renaming and stickying thread.***

Feel free to ask any Australian MC questions in this thread.

 

[bujolover]

Thought I'd revive this thread...

The number of digits in the decimal expansion of 2²⁰⁰⁵ is closest to...

(A) 400 (B) 500 (C) 600 (D) 700 (E) 800

Please provide all working out, as I actually have no idea how to do this question (got it out of a past questions book- source: 2005 Senior Paper Q25). The solution in the book makes zero sense to me, as I'm not a super mathematically advanced person. :/

When you post working out, please post a new question as well (I want this to become a marathon) to either test us/ask help from us.

 

[1729]

Thought I'd revive this thread...

The number of digits in the decimal expansion of 2²⁰⁰⁵ is closest to...

(A) 400 (B) 500 (C) 600 (D) 700 (E) 800

Please provide all working out, as I actually have no idea how to do this question (got it out of a past questions book- source: 2005 Senior Paper Q25). The solution in the book makes zero sense to me, as I'm not a super mathematically advanced person. :/

When you post working out, please post a new question as well (I want this to become a marathon) to either test us/ask help from us.

 

[dan964]

Numerically
2^k=10^n =A that is n=k*log(2) by log laws.

Pick n=1, A=10
2^3=8, 2^4=16
1/4<log(2)<1/3

halving the interval:
7/24<log(2)<1/3

Pick n=2, A=100
2^6=64, 2^7=128
2/7<log(2)<1/3

halving the interval
7/24<log(2)<13/42

Pick n=3, A=1000
2^9=512, 2^10=1024
3/10<log(2)<1/3

From here we conclude that
0.3000 < log(2) < 0.3095

Let us approximate therefore log(2) = 0.3
So n=k*log(2)
k=2005

>> n=601.5

>> n approx = 600 so (C)

(it is possible to make these calculations by hand, I was just lazy and used a calculator to calculate some of the fractions)

 

[si2136]

Anyone doing it this year?

It's sad that it's the day after UMAT as I'll be burned out, but hopefully I get at least HD this year, and hopefully a prize!

 

[pikachu975]

Q: http://prntscr.com/fupd74

How would one approach this? I'm not sure how the answer works. I tried using sum of series, but it didn't help.

A: http://prntscr.com/fupdr6

Sum = n/2 (a+l)

So by observing the initial numbers e.g. 1,3,5,7,9 here there's 5 terms, so I just observed that to get 5 we do (9+1)/2 = 5 and tested it on 1,3,5,7 etc and it worked, so the number of terms is (k+1)/2.

Now, Sum = (k+1)/4 * (k+1) and solve from there.

 

[InteGrand]

Q: http://prntscr.com/fupd74

How would one approach this? I'm not sure how the answer works. I tried using sum of series, but it didn't help.

A: http://prntscr.com/fupdr6

The sum of the first n positive odd numbers is n^2, so let k = 2n-1 (so there are n terms in the sum), so n^2 = 1000000, so n = 1000. So k = 2n-1 = 1999.

 

[Mongoose528]

Would you need to know things such as modular arithmetic and geometric identities for the intermediate section? Especially for the last 5?