Brain teaser thread

[Ramo]

SD and SF, I was responding to jt's post (I crossposted).

[JohnM2433]

Originally posted by Jaguar Warrior
1 C and 1 M go over
1 C goes back
2 C go over (1 stays in boat)
1 M and 1 C go back
2 M go over

Danger! Danger, Will Robinson!

There are now 2 cannibals and 1 missionary on the side of the river they started out on, so clearly this answer does not work. (You can see this easily by using pennies and dimes to represent the cannibals and missionaries and moving them back and forth.)

And you can still solve the problem if you consider all the people on one side of the river to be "in the same place" whether they are in the boat or not. Your interpretation makes sense, of course, but it makes the problem too easy to solve.

C'mon, you guys, this one is easy if you think about it properly...

Oh, and regarding the labels and the jars, it seems to me again that you guys are engaging in a lot of unnecessary complex math. Each label has a 1/9 probability of being on the right jar, and there are 9 labels, so clearly the expected number of correct labels is 1, the chance of a label being correct times the number of labels. Or equivalently, each jar has a 1/9 probability of being correctly labeled, and there are 9 jars, so the expected number of correctly labeled jars is 1, the chance of a jar being correctly labeled times the number of jars.

[JohnM2433]

I got the number of bricks using some simple algebra, but I see that SnowFire posted the answer first. Since we got the same answer using different methods, I think we can say that it's the correct one even before Sagacious Dolphin's response.

My missionaries/cannibals problem is still waiting for a correct answer. Try doing it the hard way, where everyone on the same side of the river is always considered to be "in the same place". It's really not that difficult...

[Dauphin]

Originally posted by JohnM2433
And you can still solve the problem if you consider all the people on one side of the river to be "in the same place" whether they are in the boat or not.

2C ->
<- 1C
2C ->
<- 1C
2M ->
< - MC
2M ->
<- C
2C ->
<- M
MC ->

[JohnM2433]

Yep! If you realize that no outnumbering can occur in the boat, you can abstract the whole thing to one or two people crossing from each side every other turn. Then it's really fairly simple, since at most points there's only one possibility, and you can backtrack from what few dead ends there are. One just need keep in mind that if you have to make a move that undoes the previous move, you should backtrack, since the result is the same as doing nothing at all, and that can't help to lead to a solution. Soon, the remaining moves are obvious.

Personally, I would have sent a cannibal back for the remaining cannibal at the end, since the missionaries would presumably be even more pleased when they outnumber the cannibals. I don't know why I bother to mention that.

You've posted quite a few problems, so maybe you should give SnowFire a chance to post the next one.

[SnowFire]

Yup, can easly be done by algebra, but 'twas lazy. Being that I got one, here's another puzzle... as I'm sure you all know, there's an obvious answer from some calculation, a reverse psychology "but!" possibility if it seems too straightforward, and the reverse-reverse psychology strategy of "wait! the obvious thing was right after all!" So answers without reasons are pointless.

A man is bragging to his friends about his incredible speed as a runner. "No matter how fast I'm running, I can always double my speed if I want to!", he says. Then he adds modestly: "Well, actually, I cannot go faster than the speed of light, of course". Naturally, his friends don't believe him, and decide to put him to the test. They take him to a 60 km long, straight section of road at an abandoned place in the countryside, and tell him to start running, and that they will be right behind him in their car. They have a gun, and tell him that every time he hears the gun firing, he should double his speed, but not look back. Behind his back, his friends have agreed to treat him to a nice dinner if he runs the 60 kilometres in 17 seconds or less. The man starts running at a speed of 2 m/s, and his friends in the car behind him fire the gun once every second. Assuming the man really is capable of doubling his speed with only the speed of light as the limit, and that the car can somehow keep up with him, does he get the dinner?

[Urban Ranger]

60km/17s (3529.41m/s) is a lot faster than the speed of sound in air (330m/s). So at some point the man will no longer be able to hear the gun shots.

[JohnM2433]

Ooh! Good catch! That means that the man will never get up to 60km/17s, so of course his average speed will always be below that. Thus he will run the 60 km in more than 17 s, so he won't get the dinner.

I was about to post that his speed would be over 60 km/s during the 17th second, so he would have to have covered the entire distance after that second. But obviously that's not the case, since he stops hearing the gun.

[Urban Ranger]

Thanks

I'm digging up a question. If any of you has one handy, feel free.

[CapitanGarlic]

A man buys coconuts for 15$ (US, but that's just for the sake of argument) per dozen and sells them at $6 per dozen. (also $US) Doing this, he becomes a millionaire. How?

[Urban Ranger]

That's just a cover for his arms smuggling empire

[CyberGnu]

He started out a billionare?

[SnowFire]

How do you make a small fortune in general aviation? Start out with a large one.

I expect the answer's similar. He started out a billionaire or bigger.

Edit: This is what happens when you leave your reply-to window open for 20 minutes without refreshment before actually posting. A bit late on this one.

[CapitanGarlic]

yup.

[CyberGnu]

So, in the spirit of things:

A man walks into a hotel on Monday and stays two weeks, but he leaves on Friday. How is this possible?