1. ## Re: Australian Maths Competition 2013

Originally Posted by SimpletonPrime
OK. I remembered two questions:

Given positive integer $m and $f(m)=0,f(n)=2016^2$ where $f(x)=ax^2+bx+c$. How many values of $n-m$?
Uh, if a,b,c,m are positive integers, how can we possibly have am^2+bm+c=f(m)=0?

2. ## Re: Australian Maths Competition 2013

Originally Posted by seanieg89
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.

3. ## Re: Australian Maths Competition

Originally Posted by kurt.physics
Here is 2005/2006
i cannot see them

4. ## Re: Australian Maths Competition

Originally Posted by Gana
i cannot see them
About 7 years too late lol

5. ## Re: Australian Maths Competition

Originally Posted by Gana
i cannot see them

7. ## Re: Australian Maths Competition

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.

8. ## Re: Australian Maths Competition

Originally Posted by bujolover

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.
$\noindent The number of digits in N \in \mathbb{N} can be found using \log_{10}{N} (or, rather (\log_{10}{N} + 1)). The number of digits in 2^{2005} is \log_{10}{2^{2005}} = 2005\log_{10}{2} = 603.565 \ldots meaning the answer is (C)$

9. ## Re: Australian Maths Competition

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
halving the interval:
7/24
Pick n=2, A=100
2^6=64, 2^7=128
2/7
halving the interval
7/24
Pick n=3, A=1000
2^9=512, 2^10=1024
3/10
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)

10. ## Re: Australian Maths Competition

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!

11. ## Re: Australian Maths Competition

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

12. ## Re: Australian Maths Competition

Originally Posted by si2136
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.

13. ## Re: Australian Maths Competition

Originally Posted by si2136
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.

14. ## Re: Australian Maths Competition

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

15. ## Re: Australian Maths Competition

Originally Posted by Mongose528
Would you need to know things such as modular arithmetic and geometric identities for the intermediate section? Especially for the last 5?

You never know, but for the Australian Intermediate Mathematics Olympiad (similar to AMC, but harder) a few years ago, we got asked a question that could be solved using modular arithmetic/bases etc

16. ## Re: Australian Maths Competition

Originally Posted by 30june2016
You never know, but for the Australian Intermediate Mathematics Olympiad (similar to AMC, but harder) a few years ago, we got asked a question that could be solved using modular arithmetic/bases etc
Would you have any tips for the amc (especially last 5) and aimo? Completing them both this year.

17. ## Re: Australian Maths Competition

Terry has invented a new way to extend lists of numbers. To Terryfy a list such
as [1, 8] he creates two lists [2, 9] and [3, 10] where each term is one more than
the corresponding term in the previous list, and then joins the three lists together
to give [1, 8, 2, 9, 3, 10]. If he starts with a list containing one number [0] and
repeatedly Terryfies it he creates the list
[0, 1, 2, 1, 2, 3, 2, 3, 4, 1, 2, 3, 2, 3, 4, 3, 4, 5, 2, 3, 4, . . . ].

What is the 2012th number in this Terryfic list?

18. ## Re: Australian Maths Competition

Continuing the Marathon:

1. A high school marching band can be arranged in a rectangular formation with exactly
three boys in each row and exactly five girls in each column. There are several sizes
of marching band for which this is possible. What is the sum of all such possible
sizes?

2. Around a circle, I place 64 equally
spaced points, so that there are
64×63÷2 = 2016 possible chords
between these points.
I draw some of these chords, but
each chord cannot cut across more
than one other chord.
What is the maximum number of
chords I can draw?

3. A symmetrical cross with equal arms has an area of 2016 cm2 and all sides of integer
length in centimetres. What is the smallest perimeter the cross can have, in
centimetres?

19. ## Re: Australian Maths Competition

The APMO papers are just freaking hard.

American papers have so much probability it's not funny

20. ## Re: Australian Maths Competition

Originally Posted by si2136
The APMO papers are just freaking hard.

American papers have so much probability it's not funny
hence why we are shit with our technology... everything is probability these days...

21. ## Re: Australian Maths Competition

Originally Posted by Mongose528
3. A symmetrical cross with equal arms has an area of 2016 cm2 and all sides of integer
length in centimetres. What is the smallest perimeter the cross can have, in
centimetres?
Let x = Large Square
Let y = Small Square

Therefore 2016 = (x+2y)(x-2y)

Through trial and error, x+2y = 56, x-2y = 36.

Therefore 2x = 92.

Therefore the perimeter is 184cm.

NEW Q:

There are 42 Points P1, P2, P3, ....., P42, placed in order on a straight line so that each distance from Pi to P(i+1) is 1/i, where 1≤i≤41. What is the sum of the distances between every pair of these points?

22. ## Re: Australian Maths Competition

Originally Posted by si2136
Let x = Large Square
Let y = Small Square

Therefore 2016 = (x+2y)(x-2y)

Through trial and error, x+2y = 56, x-2y = 36.

Therefore 2x = 92.

Therefore the perimeter is 184cm.

NEW Q:

There are 42 Points P1, P2, P3, ....., P42, placed in order on a straight line so that each distance from Pi to P(i+1) is 1/i, where 1≤i≤41. What is the sum of the distances between every pair of these points?
Completed it for 1≤i≤2, 1≤i≤3 ... 1≤i≤5, and the answer was always a triangular number of n-1

For: 1≤i≤2, the sum of distances was 1

For: 1≤i≤3, the sum of distances was 3

For: 1≤i≤5, the sum of distances was 10

So for 1≤i≤42 the sum of distances will be 41*42/2 = 861

New Q: In a 3 × 3 grid of points, many triangles can be formed using 3 of the points as
vertices. Of all these possible triangles, how
many have all three sides of different lengths?

23. ## Re: Australian Maths Competition

@M: Is the answer 68 for the new q?

Is there a faster way to solve 3 system of lines with 4 unknowns than matrices?

24. ## Re: Australian Maths Competition

Originally Posted by si2136
@M: Is the answer 68 for the new q?

Is there a faster way to solve 3 system of lines with 4 unknowns than matrices?
Unfortuantely that's incorrect @si2136

The answer they've given is 40

25. ## Re: Australian Maths Competition

1 2 3
4 5 6
7 8 9

Note: you can only pick max of 2 points on any given line/diagonal. You must pick 3 points.

Pick (1) as a fixed point:
Pick (5) = 2,3,4,6,7,8 = 6
Remove (5), Pick (9) = 2,3,4,6,7,8 = 6
Remove (9), Pick (2) = 4,6,7,8 = 4 or Pick (3) = 4,6,7,8 = 4
Remove (2,3), Pick (4) = 6,8 = 2 or Pick (7) = 6,8 = 2
Remove (4,7), Pick (6) = 8, = 1
= 25

Remove (1), Pick (9) as a fixed point
Pick (5) = 2,3,4,6,7,8 = 6
Remove (5), Pick (3) = 2,4,7,8 = 4 or Pick (6) = 2,4,7,8 = 4
Remove (3,6), Pick (7) = 2,4 = 2 or Pick (8) = 2,4 = 2
Remove (7,8), Pick (2) = 4, = 1
= 19

Remove (9), Pick (3) as a fixed point
Pick (5) = 2,4,6,8 = 4
Remove (5), Pick (7) = 2,4,6,8 = 4
Remove (7), Pick (2) = 4,6,8 = 3
Remove (2), Pick (4) = 6,8 = 2
Remove (4), Pick (6) = 8, = 1
= 14

Remove (3), Pick (7) as a fixed point
Pick (5) = 2,4,6,8 = 4
Remove (5), Pick (2) = 4,6,8 = 3
Remove (2), Pick (4) = 6,8 = 2
Remove (4), Pick (6) = 8, = 1
= 10

Remove (7), Pick (2) as a fixed point
Pick (5) = 4,6 = 2
Remove (5), Pick (4) = 6,8 = 2
Remove (4), Pick (6) = 8, = 1
= 5

Remove (2), Pick (8) as a fixed point
Pick (5) = 4,6 = 2
Remove (5), Pick (4) = 6, = 1
= 3

Remove (8), note that only (4,5,6) remain which no valid triangles can be formed.

1 2 3
4 5 6
7 8 9

Pick (1) as a fixed point:
Pick (5) = 6,8 = 2
Remove (5), Pick (9) = 2,4,6,8 = 4
Remove (9), Pick (2) = 6,7,8 = 3 or Pick (3) = 4,6 = 2
Remove (2,3), Pick (4) = 6,8 = 2 or Pick (7) = 8, = 1
= 14

Remove (1), Pick (9) as a fixed point
Pick (5) = 2,4 = 2
Remove (5), Pick (3) = 2,8 = 2 or Pick (6) = 2,4,7 = 3
Remove (3,6), Pick (7) = 4, = 1 or Pick (8) = 2,4 = 2
= 10

Remove (9), Pick (3) as a fixed point
Pick (5) = 4,8 = 2
Remove (5), Pick (7) = 2,4,6,8 = 4
Remove (7), Pick (2) = 4,8 = 2
Remove (2), Pick (4) = 6, = 1
Remove (4), Pick (6) = 8, = 1
= 10

Remove (3), Pick (7) as a fixed point
Pick (5) = 2,6 = 2
Remove (5), Pick (2) = 4,8 = 2
Remove (2), Pick (4) = 6, = 1
Remove (4), Pick (6) = 8, = 1
= 6

Note no scalene triangles remain

Total of triangles = 76
Total of scalene triangles = 40

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