***Renaming and stickying thread.***
Feel free to ask any Australian MC questions in this thread.
2014 HSC
2015 Gap Year
2016-2018 UOW Bachelor of Maths/Maths Advanced.
Links:
View English Notes (no download)
Thought I'd revive this thread...
The number of digits in the decimal expansion of 2^{2005} 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.
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.
2014 HSC
2015 Gap Year
2016-2018 UOW Bachelor of Maths/Maths Advanced.
Links:
View English Notes (no download)
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!
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
2016 HSC (Accelerated)
// 2 Unit Maths // Studies of Religion 1 //
2017 HSC
// Biology // Physics // Maths Extension 1 // Maths Extension 2 // English Advanced //
Would you need to know things such as modular arithmetic and geometric identities for the intermediate section? Especially for the last 5?
Last edited by 30june2016; 14 Jul 2017 at 5:58 PM.
hello
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?
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?
The APMO papers are just freaking hard.
American papers have so much probability it's not funny
MX2 - ADV - ECO - BUS
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.
MX2 - ADV - ECO - BUS
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.
@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?
MX2 - ADV - ECO - BUS
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
2014 HSC
2015 Gap Year
2016-2018 UOW Bachelor of Maths/Maths Advanced.
Links:
View English Notes (no download)
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