i cannot see them
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Australian Maths Competition (1 Viewer)
- Thread starter maths94
- Start date
RealiseNothing
what is that?It is Cowpea
About 7 years too late lol
Mongoose528
Member
dan964
what
***Renaming and stickying thread.***
Feel free to ask any Australian MC questions in this thread.
Feel free to ask any Australian MC questions in this thread.
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.
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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dan964
what
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)
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)
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pikachu975
Premium Member
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.
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
Member
Would you need to know things such as modular arithmetic and geometric identities for the intermediate section? Especially for the last 5?
30june2016
Active Member
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
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Mongoose528
Member
Would you have any tips for the amc (especially last 5) and aimo? Completing them both this year.
Mongoose528
Member
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?
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?
Mongoose528
Member
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?
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?
Paradoxica
-insert title here-
hence why we are shit with our technology... everything is probability these days...
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?
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Mongoose528
Member
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?
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