Brain teaser thread

[Rogan Josh]

Edit: Sorry - the answer I just gave was bull****. Still thinking....

[Rogan Josh]

Ok - think I have it now. All the rooms numbered 1,4,16,64.... will have their lights on. ie. numbers 4^n with n=0,1,2,3.....

Not sure I have a rigorous argument why though.

[Dauphin]

Those will be on, but so will others.

Just in case I wasn't clear, every 3rd, then 4th, then 5th, then 6th then 7th ... then 998th, then 999th, then 1000th is switched.

[Rogan Josh]

Hmmm... well it really depends on how many numbers a given number is divisible by. For example 50 is divisible by 1,2,5,10,25 and 50, so it light number 50 will be switched 6 times so it will be off. Similarly 51 is divisible only by 1 and 51, so it will be flicked twice and also be off. Conservely, 64 is divisible by 1,2,4,8,16,32 and 64 - an odd number of numbers, so light 64 will be on.

So I can see how to see if any particular light is on or off, but I still can't see the general pattern...

You thought this one was easier than the last one? You must have a strange mind...

[Dauphin]

What property does a number have to have to have an odd number of factors?

[Dry]

Originally posted by Combat Ingrid
Here's another one:

It cannot be seen, cannot be felt.
Cannot be heard, cannot be smelt.
It lies behind stars and under hills,
And empty holes it fills.
It comes first and follows after,
Ends life, kills laughter.

Silence?

[Dry]

Originally posted by Sagacious Dolphin
What property does a number have to have to have an odd number of factors?

Not be a square number.

[monolith94]

I was very tired when I wrote my reasoning… still am… but my train of thought was basically the same as rogan josh's…
Just for the record…

[Richard Bruns]

This is a really great thread. I've had fun solving some of these.

The only lights that are on are the rooms with square numbers. But I didn't really solve that.

I have a question about the robbery liars, however. From Rogan Josh's correct answer:

Chris lying -> one of Brad or Dave must be telling the truth

I don't see how that can be inferred from the problem. All of them could be lying about the robbery, which means that all of them could be truthful when they say that others are lying about the robbery. You can't infer anything from that information.

I suppose that their statements are meant to refer to who is lying about who is lying, and not about who is lying about the robbery. But that creates a paradox. How can Alex know who is lying until he has heard all the statements of the others? How can any of them make such a statement until they have heard the others? It seems to be an impossible situation, a self-referencing paradox.

Can someone enlighten me?

Also, a couple of comments about the three-way duel. It is very clear that MLeonard gains by not firing. But consider the eventual outcome of the duel:

Once the initiative passes to Ming, he faces the same choice. There are three outcomes. If he plugs MLeonard, he dies when MtG shoots. If he frags MtG, he has a 1/3 chance of getting blasted by MLeonard next turn. If he misses or doesn't shoot, the choice of firing passes to MtG.

If MtG wastes MLeonard, then he has a 2/3 chance of getting inhumed by Ming. If he eliminates Ming, he has a 1/3 chance of kicking the bucket when MLeonard fires.

But MtG knows, at this point, that MLeonard will not fire if all three are left alive. So his best choice of action is to blow down a tree or something, so that everyone is left alive. That way nobody tries to gun him down next turn. If Ming knows that MtG knows this, his best chance of survival is also to not shoot at anybody, because that is the only way that nobody will fire at him next turn.

Basically, if all three are logical and they know that the other two are logical, then nobody will fire in the duel. They will all waste shots until they run out of ammo. Of course, if they were all logical they wouldn't have gotten into a duel in the first place. Three-power situations typically result in standoffs, since nobody gains by taking the forst shot.

[Dauphin]

Originally posted by Richard Bruns
I suppose that their statements are meant to refer to who is lying about who is lying, and not about who is lying about the robbery.

Correct.

But that creates a paradox. How can Alex know who is lying until he has heard all the statements of the others? How can any of them make such a statement until they have heard the others? It seems to be an impossible situation, a self-referencing paradox.

Can someone enlighten me?

Its only a paradox if you take it they know that they are a liar or not. They could all be guessing - then the paradox is removed

[Dauphin]

Originally posted by Richard Bruns
If MtG wastes MLeonard, then he has a 2/3 chance of getting inhumed by Ming. If he eliminates Ming, he has a 1/3 chance of kicking the bucket when MLeonard fires.

But MtG knows, at this point, that MLeonard will not fire if all three are left alive. So his best choice of action is to blow down a tree or something, so that everyone is left alive. That way nobody tries to gun him down next turn. If Ming knows that MtG knows this, his best chance of survival is also to not shoot at anybody, because that is the only way that nobody will fire at him next turn.

Basically, if all three are logical and they know that the other two are logical, then nobody will fire in the duel. They will all waste shots until they run out of ammo. Of course, if they were all logical they wouldn't have gotten into a duel in the first place. Three-power situations typically result in standoffs, since nobody gains by taking the forst shot.

Depends on if MtG is playing to win, or to not lose.

Ming and ML will never target each other - that would be tantamount to suicide. Therefore MtG's odds will never improve. So if he wants to win he has to shoot someone first.

[Zero-Tau]

Originally posted by Sagacious Dolphin
A hotel has 1,000 rooms numbered 1 to 1,000. Jack is looking after the hotel for the winter and is going slightly crazy. He decides to flick every roomlight switch in the hotel. He then decides to flick the switch in every other room (starting with room 2). He then flicks the switch in every third room (starting with room three). He then flicks the switch in every fourth room (starting in room 4).

If all lights were originally off, which lights remain on once Jack has finished switching every 1,000^th roomlight switch?

All rooms with square numbers have their lights flicked on. Their prime factorization would be a1^e1*a2^e2*a3^e3... with all exponents being even. Using the formula for the number of divisors (e1+1)(e2+1)(e3+1)... we see that the square numbers have an odd number of divisors, meaning that they'll end up with the lights on. Conversely, if the number of divisors is odd, then each of e1+1, e2+1, e3+1,... would be odd, meaning that all the exponents would be even, making the number a square.

[Dauphin]

I've got another one, if no-one posts one in the next hour.

[Ramo]

I'm off to a class in 5 minutes. Post it now!

[Dauphin]

You have a box with four balls of different colours. If you randomly draw two at a time, then paint the first ball to match the second. What is the expected number of drawings before all balls are the same colour?