• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

HSC 2013 MX2 Marathon (archive) (1 Viewer)

Status
Not open for further replies.

psychotropic

Member
Joined
Feb 24, 2013
Messages
42
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

Three distinct numbers a,b,c are chosen at random from the numbers 1,2,...100. Find the probability that a,b and c are in a geometric progression
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

Three distinct numbers a,b,c are chosen at random from the numbers 1,2,...100. Find the probability that a,b and c are in a geometric progression
If its in geometric progression, then for integers a and k, it must be in the form:



Essentially, we are finding how many integer solutions there are for:



Where (a,k) are positive integers with

So lets sketch the graph:

for positive y, therefore:



When we sketch this graph, we have to find all integer points lying on or under the curve.
To measure this we sketch the lines, y=1, y=2, y=3, y=4 ... y=10

The intersection of y=1 and the curve is (100,1), that means (99,1) and (98,1) and so on are under the curve.
That is 100 solutions EDIT: oops

y=2, (25,2), that is another 25 solutions

y=3, (100/9, 3), that is 11 solutions

y=4, (6.25, 4) that is 6 soln

y=5, (4, 5), that is 4 soln

y=6, (2.7, 6), that is 2 soln

y=7, (2.0, 7) that is 2

y=8 (1. , 8) that is 1

y=9 (1.,, , 9) that is 1

y=10 (1, 10) that is 1.

Add them all together to get 153 integer pair solutions, that means the probability of picking a geometric progression is:

???
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon



















 
Last edited:

psychotropic

Member
Joined
Feb 24, 2013
Messages
42
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

pretty much except u have to discard those 100 solutions since we have to have distinct a,b,c. Otherwise you are correct.
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: HSC 2013 4U Marathon

A bit flawed, the radius of convergence of a Taylor series can be 0. Eg consider the function defined piecewise as e^{-1/x} for x > 0 and 0 for x < 0.

sine just happens to have an infinite radius of convergence.
 

harryharper

New Member
Joined
May 22, 2012
Messages
22
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

If its in geometric progression, then for integers a and k, it must be in the form:



Essentially, we are finding how many integer solutions there are for:



Where (a,k) are positive integers with
Isn't this not quite right?

It's not specified that the common ratio must be an integer. Here we're missing out on solutions like (4,6,9), (25,30,36), (27,36,48), (44,66,99)...In fact, there are nearly twice as many solutions (I think) if we allow non-integer ratios.

That is, if we solve instead for , where

A very similar approach gives 53 solutions for p=1 (the case that Sy solved), and 52 more for other values of p. Instead of solving for #values of k under , we're solving for the number of coprime pairs.

31 for a=1, 17 for a=2, 9 for each of a=3 and a=4, 5 for each of a=5,6, 3 for each of a=7,8,9,10,11 and 1 for each of a=12,13,14,...,25

31 + 17 + 2x9 + 2x5 + 5x3 + 14x1 = 105
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

Isn't this not quite right?

It's not specified that the common ratio must be an integer. Here we're missing out on solutions like (4,6,9), (25,30,36), (27,36,48), (44,66,99)...In fact, there are nearly twice as many solutions (I think) if we allow non-integer ratios.

That is, if we solve instead for , where

A very similar approach gives 53 solutions for p=1 (the case that Sy solved), and 52 more for other values of p. Instead of solving for #values of k under , we're solving for the number of coprime pairs.

31 for a=1, 17 for a=2, 9 for each of a=3 and a=4, 5 for each of a=5,6, 3 for each of a=7,8,9,10,11 and 1 for each of a=12,13,14,...,25

31 + 17 + 2x9 + 2x5 + 5x3 + 14x1 = 105
Very true, my bad.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon







 
Last edited:

JJ345

Member
Joined
Apr 27, 2013
Messages
78
Gender
Female
HSC
N/A
Re: HSC 2013 4U Marathon

The letters in the square root are supposed to be gd right?
t = d/(Vcosa) --> where a is the varying angle
Sub into y = -1/2 gt^2 + Vtsina
And let a=45 (minimum conditions for the projectile to hit the wall as range is max at this angle so velocity will the be min required--> I've worded this terribly :()
y=d-gd^2/v^2
y>0 for it to hit the wall
So rearranging, v>sqrt(gd)
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

The letters in the square root are supposed to be gd right?
t = d/(Vcosa) --> where a is the varying angle
Sub into y = -1/2 gt^2 + Vtsina
And let a=45 (minimum conditions for the projectile to hit the wall as range is max at this angle so velocity will the be min required--> I've worded this terribly :()
y=d-gd^2/v^2
y>0 for it to hit the wall
So rearranging, v>sqrt(gd)
Yeah I think that makes sense, well done.

My way was more straightforward, by subbing in t=d/Vcos a, and then y > 0
We arrive at:



Since (a + 1/a) > 2, put in a = than theta to get,
 

JJ345

Member
Joined
Apr 27, 2013
Messages
78
Gender
Female
HSC
N/A
Re: HSC 2013 4U Marathon

Yeah I think that makes sense, well done.

My way was more straightforward, by subbing in t=d/Vcos a, and then y > 0
We arrive at:



Since (a + 1/a) > 2, put in a = than theta to get,
From your second last step if we rearrange in terms of tan theta then solve for discriminant we get the required expression aswell--> But what that means physically doesn't really make sense to me--> Is it just that yes there's a real value for tan theta so its hitting the wall at some angle??
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

From your second last step if we rearrange in terms of tan theta then solve for discriminant we get the required expression aswell--> But what that means physically doesn't really make sense to me--> Is it just that yes there's a real value for tan theta so its hitting the wall at some angle??
The inequality came about due to the condition that y > 0 at x = d, thus if we take a quadratic in terms of tan theta, it will describe all possible tan theta in terms of V such that y > 0 at x=d (i.e. the conditions we need for the question), so yeah that's another method
 

VBN2470

Well-Known Member
Joined
Mar 13, 2012
Messages
440
Location
Sydney
Gender
Male
HSC
2013
Uni Grad
2017
Re: HSC 2013 4U Marathon

Yep well done.

Can anyone post a question please? (preferably probability)
An unbiased die is thrown six times. Find the probability that the six scores obtained will be such that a 6 occurs only on the last throw and exactly three of the first five throws result in odd numbers.
 

VBN2470

Well-Known Member
Joined
Mar 13, 2012
Messages
440
Location
Sydney
Gender
Male
HSC
2013
Uni Grad
2017
Re: HSC 2013 4U Marathon

Prove that is an integer for all positive integer values of .
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

Prove that is an integer for all positive integer values of .
Proving by mathematical induction on n, for integer m





















We know that

OR



Therefore, OR









For the case



 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top