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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

There is a room with a light switch in it, which a mischevious ghost sees and decides to play with.

First he turns the light on. He waits for a minute, but no one reacts, so he turns it off again.

He is getting frustrated now, so he only waits for half a minute before turning the light back on.

Still, no one reacts and he is getting very angry now so he only waits a quarter of a minute before turning it off.

He continues to turn the light on and off, get no reaction and wait for half the time he waited previously until 2 minutes after he first started playing with the light, when he gives up and goes home. So, is the light on or off?

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,663

0:00 ON

1:00 OFF

1:30 ON

1:45 OFF

1:52½ ON

...

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,196

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + .............1/ ∞ = 2

I cannot think beyond that.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,663

Possibly the answer is "YES, the light IS on or off"

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

Five pupils, Amy, Brenda, Cathy, Darren and Elle stand in a row in a certain order. They hold 30 beanbags in all.

The person on the right of Cathy holds 23 beanbags.

The person on the left of Brenda holds 18 beanbags.

The person on the left of Darren holds 12 beanbags.

The person on the left of Elle holds 22 beanbags.

If Elle is the 1st person on the right and april is the 1st person on the left, how many beanbags is Amy holding.

*Last edited by Jenilia (2005-07-16 14:00:28)*

Ideas are funny little things, they won't work unless you do.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

You're too clever, MathsIsFun! I thought that would keep people confused for ages!

To Jenilia, I'm guessing that when you said person, you meant people, because in it's current form, the puzzle doesn't work.

If that's right, I worked out the answer to be the same as the last digit of 1x3x5x7x9x...x99.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,196

23+18+12+22 is already greater than 30! So, it should be 'people' as pointed out by Mathsy!:)

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

Do not forget that 2 person can carry 1 beanbag.

Ideas are funny little things, they won't work unless you do.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,663

Oh, wow, that is nice and tricky ... so, because they stand in a line, then 3 couldn't hold a beanbag, but 1 or 2 could.

BTW when you say: "If Elle is the 1st person on the right and april is the 1st person on the left, how many beanbags is Amy holding.", should that be Amy is the first person on the left (not april)?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

If I remember her puzzle correctly and all that she's changed is the names, then it should be Cathy who is on the left.

Why did the vector cross the road?

It wanted to be normal.

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**mad person****Member**- Registered: 2007-07-09
- Posts: 1

since the total sum is equal to 2 it indicates that there will be two cycles. hence the position after the completion of 2 minutes will be the same in the begining. since the ghost switch on first, the answer is the light is in off position now.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

The light switch would be broken and would become infinitely massive moving at the speed of light too, perhaps.

But neglecting any physical constraints... let's see.

I think the answer to this question is that if this scenario was repeated everyday, you would get different results, depending on whether infinity was odd or even at the end of those 2 minutes, on any given day.

**igloo** **myrtilles** **fourmis**

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

I don't think there is any possible answer. This question implies a hypertask, that is, an infinite number of tasks to be completed in a finite amount of time. It has been concluded by many that this is not possible.

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**Simon****Member**- Registered: 2006-06-06
- Posts: 41

If time were infinitely divisible, then neither the two minute threshold nor any other point in time could ever be reached by anyone. This is resolved by the notion that time is divided into instants that cannot be subdivided.

The question can therefore be answered if we know exactly how many of the ghost's waiting periods must occur before there is one small enough to include no more than a single instant of time.

Simon

*Last edited by Simon (2007-08-07 09:32:45)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

If time were infinitely divisible, then neither the two minute threshold nor any other point in time could ever be reached by anyone. This is resolved by the notion that time is divided into instants that cannot be subdivided.

Yes, but one does not have to take such drastic measures. Information, as far as we know, can not travel faster than the speed of light. Thus, the light being on or off can not, and so the ghost will eventually not be able to keep up with the time intervals as they get smaller and smaller.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

completely hypothetical: if we were to be able to measure where gravitational fluctuations arise from, that could be used for faster than list information, since moving a piece of mass in one part of the universe instantly changes the field of gravity somewhere else, or is there a 'speed' of gravity?

The Beginning Of All Things To End.

The End Of All Things To Come.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I find that very interesting. If the sun got stolen by aliens, we'd still orbit around where it was for around 8 minutes before starting to go in a line.

Why did the vector cross the road?

It wanted to be normal.

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

yeh, but infact it would be weird if it were instant, we would detect gravity coming from sometihng we can't yet see, and things we could see might suddenly lose their gravitation ;p

The Beginning Of All Things To End.

The End Of All Things To Come.

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