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    Re: Australian Maths Competition 2013

    Quote Originally Posted by SimpletonPrime View Post
    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?
    Last edited by seanieg89; 1 Aug 2016 at 12:45 PM.

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    New Member SimpletonPrime's Avatar
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    Re: Australian Maths Competition 2013

    Quote Originally Posted by seanieg89 View Post
    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.

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    Re: Australian Maths Competition

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

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    Re: Australian Maths Competition

    Quote Originally Posted by Gana View Post
    i cannot see them
    About 7 years too late lol
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    Re: Australian Maths Competition

    Quote Originally Posted by Gana View Post
    i cannot see them

    here http://files.chiuchang.org.tw:8080/m...load/docu/AMC/

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    Re: Australian Maths Competition

    ***Renaming and stickying thread.***

    Feel free to ask any Australian MC questions in this thread.
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    Junior Member bujolover's Avatar
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    Question Re: Australian Maths Competition

    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.

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    Junior Member 1729's Avatar
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    Re: Australian Maths Competition

    Quote Originally Posted by bujolover View Post
    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.
    Last edited by 1729; 9 Jul 2017 at 11:54 PM.

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    Nomnomnomnomnom dan964's Avatar
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    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)
    Last edited by dan964; 7 Jun 2017 at 6:13 PM.
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    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!

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    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

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    Re: Australian Maths Competition

    Quote Originally Posted by si2136 View Post
    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.
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    Re: Australian Maths Competition

    Quote Originally Posted by si2136 View Post
    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.

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    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?

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    Re: Australian Maths Competition

    Quote Originally Posted by Mongose528 View Post
    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
    Last edited by 30june2016; 14 Jul 2017 at 5:58 PM.
    hello

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    Re: Australian Maths Competition

    Quote Originally Posted by 30june2016 View Post
    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.

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    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?

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    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?

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    Re: Australian Maths Competition

    The APMO papers are just freaking hard.

    American papers have so much probability it's not funny
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    Re: Australian Maths Competition

    Quote Originally Posted by si2136 View Post
    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...
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Re: Australian Maths Competition

    Quote Originally Posted by Mongose528 View Post
    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?
    Last edited by si2136; 20 Jul 2017 at 10:00 PM.
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    Re: Australian Maths Competition

    Quote Originally Posted by si2136 View Post
    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?
    Last edited by Mongose528; 20 Jul 2017 at 11:32 PM.

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    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?
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    Re: Australian Maths Competition

    Quote Originally Posted by si2136 View Post
    @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

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    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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