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# Current Year 12 Thread Mark VI Tweet

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1. Re: Current Year 12 Thread Mark VI
(Original post by Blutooth)
2)[***]Determine all sets of non-negative integers x, y and z which
satisfy the equation
I remember that question. When my brother first introduced me to pell equations he used that as an example... I wish I was as smart as him
2. Re: Current Year 12 Thread Mark VI
(Original post by wcp100)
Prove:

Given xyz =1
I cba with latex right now - too tired.

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I really really don't want to be doing lots of annoying algebraic manipulation, so I'll see if I can get there by reasoning alone. So...

None of x,y,z is equal to zero as xyz =1.

The fractions imply that none of x,y,z is ever equal to one because otherwise the fractions would be undefined. This means that at least one of x,y,z < 1 and at most two of x,y,z <1. Also, this means that at least one of x,y,z >1.

Let x < 1 (one of them has to be 0 I'm just arbitrarily choosing x). Then either x <= 0.5 or x > 0.5.

x<= 0.5 implies |x-1| >= 0.5 and so the fraction x^2/(x-1)^2 will have a maximum value of 1 and a minimum value that tends to but can never be zero.

Let z >1. Then z^2 > (z-1)^2 for all such z and tends to it's maximum value as z tends to one. However, z^2/(z-1)^2 is always greater than one and the larger z gets, the closer the fraction tends to one.

It doesn't really matter whether y is less than or greater than 1 because it makes no difference: the properties of z will remain unchanged. If the fractions in x and y are so small that they are practically zero, then the value of z will be so big that the fraction associated with it will tend to one (but will always be greater than it).

The inequality is therefore true.
3. Re: Current Year 12 Thread Mark VI
(Original post by Maths_Lover)
I cba with latex right now - too tired.

Spoiler:
Show
I really really don't want to be doing lots of annoying algebraic manipulation, so I'll see if I can get there by reasoning alone. So...

None of x,y,z is equal to zero as xyz =1.

The fractions imply that none of x,y,z is ever equal to one because otherwise the fractions would be undefined. This means that at least one of x,y,z < 1 and at most two of x,y,z <1. Also, this means that at least one of x,y,z >1.

Let x < 1 (one of them has to be 0 I'm just arbitrarily choosing x). Then either x <= 0.5 or x > 0.5.

x<= 0.5 implies |x-1| >= 0.5 and so the fraction x^2/(x-1)^2 will have a maximum value of 1 and a minimum value that tends to but can never be zero.

Let z >1. Then z^2 > (z-1)^2 for all such z and tends to it's maximum value as z tends to one. However, z^2/(z-1)^2 is always greater than one and the larger z gets, the closer the fraction tends to one.

It doesn't really matter whether y is less than or greater than 1 because it makes no difference: the properties of z will remain unchanged. If the fractions in x and y are so small that they are practically zero, then the value of z will be so big that the fraction associated with it will tend to one (but will always be greater than it).

The inequality is therefore true.
hmm, I like how thorough you are
Spoiler:
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But surely just saying x,y,z =/= 1 and ergo at least one of x,y,z is greater than |1|, and ergo at least one fraction is greater than (or tends to) 1 as y^2 > (y-1)^2 for all y>|1|. and ergo the inequality must hold.

I suppose perhaps the reasoning may require some insight for 2 of x,y,z to be negative, but the square sort of deals with that anyway... and the 1-(var) implies a tendency towards 1... I'll stop nitpicking there though
Last edited by Llewellyn; 31-05-2012 at 21:35.
4. Re: Current Year 12 Thread Mark VI
(Original post by Blutooth)
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4)[*]There are 100 passengers waiting to board a plane. Passenger 1 is allocated seat 1, passenger 2 allocated seat 2, etc. The passengers board in numerical order. Every passeger remembers his seat, except passenger 1, who boards the plane and sits in a random seat.

Passenger 2 now boards. If his seat is free, he sits in it, otherwise he picks a random seat.

All remaining passengers do this, 1 by 1 boarding the plane, sitting in their intended seat if it is free, or picking randomly if not.

What is the probability that passenger 100 sits in seat 100?
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Passenger 1 has a 1/100 chance of sitting in his seat. If he does sit in his seat, then the other passengers will all sit in their correct places and hence the probability of passenger 100 sitting in seat 100 is 1/100.

If passenger 1 does not sit in his correct seat, then none of the other passengers will sit in their correct seats.

Therefore the probability of passenger 100 sitting in seat 100 is 1/100?
5. Re: Current Year 12 Thread Mark VI
(Original post by Llewellyn)
hmm, I like how thorough you are
Spoiler:
Show
But surely just saying x,y,z =/= 1 and ergo at least one of x,y,z is greater than |1|, and ergo at least one fraction is greater than (or tends to) 1 as y^2 > (y-1)^2 for all y>|1|. and ergo the inequality must hold.

I suppose perhaps the reasoning may require some insight for 2 of x,y,z to be negative, but the square sort of deals with that anyway... and the 1-(var) implies a tendency towards 1... I'll stop nitpicking there though
Thank you.

Spoiler:
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Yes. I do hve a problem with being concise.
6. Re: Current Year 12 Thread Mark VI
(Original post by Maths_Lover)
Spoiler:
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Passenger 1 has a 1/100 chance of sitting in his seat. If he does sit in his seat, then the other passengers will all sit in their correct places and hence the probability of passenger 100 sitting in seat 100 is 1/100.

If passenger 1 does not sit in his correct seat, then none of the other passengers will sit in their correct seats.

Therefore the probability of passenger 100 sitting in seat 100 is 1/100?
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Consider:

passenger 1 picks seat 2 at random
passenger 2 picks seat 1 at random
passenger 3 picks seat 3.
thus, it continues, and so the probability must be greater than 1/100

Edit: Sorry, I'm not sure if I interpreted your answer right. maybe it's better if I solve this before trying to give advice
Last edited by Llewellyn; 31-05-2012 at 21:56.
7. Re: Current Year 12 Thread Mark VI
(Original post by Llewellyn)
Spoiler:
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Consider:

passenger 1 picks seat 2 at random
passenger 2 picks seat 1 at random
passenger 3 picks seat 3.
thus, it continues, and so the probability must be greater than 1/100
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Oh damn...

New development:

Situation 1: If passenger 1 picks passenger 100's seat then passenger 100 can't sit there.

Situation 2: If passenger 1 picks a seat other than passenger 100's seat i.e. passenger n's seat, then as long as passenger n doesn't sit in passenger 100's seat then passenger 100 will get to sit in their seat.

There is a 99/100 x 98/99 chance of passenger 100 sitting in his seat.

If this is still wrong, then I am going to get some rest before attempting it again because I am mentally exhausted.
Last edited by Maths_Lover; 31-05-2012 at 21:56.
8. Re: Current Year 12 Thread Mark VI
Can we do something that isn't maths? I'm thoroughly naffed off that I haven't got any maths right today.
Last edited by bananarama2; 31-05-2012 at 21:59.
9. Re: Current Year 12 Thread Mark VI
Or go to bed because we have FP1 tomorrow
10. Re: Current Year 12 Thread Mark VI
(Original post by Maths_Lover)
Spoiler:
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Oh damn...

New development:

Situation 1: If passenger 1 picks passenger 100's seat then passenger 100 can't sit there.

Situation 2: If passenger 1 picks a seat other than passenger 100's seat i.e. passenger n's seat, then as long as passenger n doesn't sit in passenger 100's seat then passenger 100 will get to sit in their seat.

There is a 99/100 x 98/99 chance of passenger 100 sitting in his seat.

If this is still wrong, then I am going to get some rest before attempting it again because I am mentally exhausted.
You've confused me
11. Re: Current Year 12 Thread Mark VI
(Original post by wcp100)
Can we do something that isn't maths? I'm thoroughly naffed off that I haven't got any maths right today.
Did you watch the apprentice yesterday, I'm hoping Ricky will win...
12. Re: Current Year 12 Thread Mark VI
(Original post by Llewellyn)
You've confused me
That question is doable in about 2/3 lines. There is a very neat solution that requires very little algebra/ workings. the answer is surprisingly large.
Last edited by Blutooth; 31-05-2012 at 22:09.
13. Re: Current Year 12 Thread Mark VI
(Original post by Llewellyn)
Edit: Sorry, I'm not sure if I interpreted your answer right. maybe it's better if I solve this before trying to give advice
That's alright, I think we're all a bit tired by now.

(Original post by wcp100)
Can we do something that isn't maths? I'm thoroughly naffed off that I haven't got any maths right today.

I am going to bed soon as I am mentally exhausted.
14. Re: Current Year 12 Thread Mark VI
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Click the spoiler you know you want to...
15. Re: Current Year 12 Thread Mark VI
(Original post by Blutooth)
That question is doable in about 2/3 lines. There is a very neat solution that requires very little algebra/ workings. the answer is surprisingly large.
Spoiler:
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1 - sum from 1 to 99 of 1/100 ^x

which is 1 - 0.101010101010101 = 0.89898989898... (there are 199 d.p.s)

But I think ML inadvertently discounted that method.
Last edited by Llewellyn; 31-05-2012 at 22:14.
16. Re: Current Year 12 Thread Mark VI
(Original post by silentlife)
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Click the spoiler you know you want to...
17. Re: Current Year 12 Thread Mark VI
(Original post by Blutooth)
Hope that is going to come up in my interview
18. Re: Current Year 12 Thread Mark VI
(Original post by Llewellyn)
You've confused me
I've confused myself as well.

(Original post by Blutooth)
That question is doable in about 2/3 lines. There is a very neat solution that requires very little algebra/ workings. the answer is surprisingly large.
19. Re: Current Year 12 Thread Mark VI
(Original post by Maths_Lover)
I've confused myself as well.

Nope sorry. Silentlife has the answer but not the exposition.
20. Re: Current Year 12 Thread Mark VI
(Original post by wcp100)
Or go to bed because we have FP1 tomorrow
Well said... I', trying to go to bed but I keep getting distracted by all of the new maths questions that have cropped up.