Page 5 - Running for the Longest Thread

J1 crosses with J3 .............. 10 minutes
J1 returns ........................... 5 minutes
J2 and J4 cross ................... 25 minutes
J3 returns ........................... 10 minutes
J1 and J3 cross ................... 10 minutes
Total ................................... 60 minutes

That's the correct answer... Most people think since J1 is the fastest he should be the one doing all the running with the flashlight

Something similar (and probably quite simple) is a farmer who needs to cross a river in a small boat. With him he has a goat, a cabbage and a wulf (don't ask me why). Problem is, he can only take with him 1 'item' at a time and they all need to reach the opposite bank unharmed. If the wulf and the goat are left alone, well you can imagine... same goes for the goat and the cabbage. Can the farmer get all on the opposite bank?

Argh... >_<

With all those assumptions, the journey has to go like this:

Junkie with flashlight and a junkie without flashlight go over bridge.
Junkie with flashlight goes back and crosses with another junky.
Junkie with flashlight goes back and crosses with last junky.

And if it goes like that, then the fastest way to get all junkies over is to have junky 1, who only needs 5 minutes to cross, be the "guide" with the flashlight. And that takes 65 minutes!


Edit (saw rod k's answer) Ahhhhh... >_<

Ok, I have a riddle (got it from the movie "Die Hard 3")...


You have a 3 litre bottle and a 5 litre bottle (they are both exact).

There's a fountain next to you which you can use to fill the bottles.

How do you get exactly 4 litres into a bottle? You cannot try to fill 5 litre bottle 4/5ths full - that wouldn't be exact.

There's no trick in the wording, and you can't bring any other things, such as an additonal bottle, into the scenario.

Goat puzzle: (Be careful, Al might get turned on talking about goats...)

Cross with goat
return
cross with cabbage
return with goat
cross with wolf
return
cross with goat

Too much time spent alone with that goatl.. Al is definately gonna get excited

Bottles:

1) Fill the 5L bottle from the fountain
2) Fill the 3L bottle from the 5L bottle, leaves 2L is the 5L bottle
3) empty the 3L bottle
4) pour the 2L from the 5L bottle into the 3L bottle
5) fill the 5L bottle from the fountain
6) Fill the 3L bottle (already containing 3L) from the 5L bottle

This would leave 4L in the 5L bottle.

Hehe, yeah, that's right.

Or:

1) fill the 3L and pour it into the 5L
2) fill the 3L again and pour as much as possible into the 5L, there remains 1 litre of water in the 3L
3) empty the 5L bottle and pour the contents of the 3L into it
4) fill the 3L for the last time then pour it into the 5L, making it 4L

And rod, should that ..already containing 3L.. not be 2L?

Another one:

You have 8 ball-bearings and one is hollow. The weight difference is so slight that you can't feel it by hand. All you have is one of those balance beam scales. You know, the one that has two plates on either side. You have to use this scale to find the hollow ball bearing.

The challenge...
you can only use the scale twice.
How do you do it?

Take 6 balls, and put 3 on each side of the balance beam.

If they weigh equally much, then take all balls off, and weigh the two balls which you haven't weighed yet. The one which weighs less is the hollow one.

If they do not weigh equally much, you'll work with the 3 balls which were on the side of the balance beam which weighed less. Take two of those three and weigh them. If they weigh equally much, then the ball which you left out just now must be the hollow one. If the two balls do not weigh equally much, then the lighter one is the hollow one.

[QUOTE]Originally posted by CodE-E
[B]This is impossible since some of the statements could be lies and therefore we cannot really start anywhere.

It's possible...I'll post explanation soon, if nobody answers it.

This is one of the brain teasers in the algorithms section of this message board.

*EDITED SOLUTION --


Jake: "Both of my brothers are liars."
Dan: "Both of my brothers are honest."
Wendell: "Jake and Dan are both liars."
Nick: "Wendell and I are brothers."
Lito: "Dan and I are brothers."
Jay: "Lito is honest."
Jay; "Dan is one of the Brady's." [/B][/QUOTE]

1. It's easy to spot that Dan is a liar because neither set of brothers are exclusively liars or honest. DAN = LIAR!

Between Wendell and Jake is a liar and an honest. From this, we know that from the three examined brothers only one of them is honest, which leaves two spots for honest and one for liar.
2. Lito and Jay are honest (TWO of them) and Nick is a liar(One). LITO & JAY = HONEST; NICK = LIAR!

3. Now we know that Dan is a BRADY, so is Lito. Based on lying Nick, between Wendell and him is a BRADY and a BUNDY, leaving us with two spots for BUNDY - Jake and Jay fill in those spots.

RECAP =

LIAR = DAN, NICK, ?
HONEST = LITO, JAY, ?

BUNDY = JAKE, JAY, ?
BRADY = DAN, LITO, ?

4. JAKE and Jay are brothers which would make Jay a liar, which would make Wendell honest

5. Dan lied when he said both of his brothers are honest, if Lito is his brother and honest, the other one must be a liar, which leaves Nick.

There you go...

Nevermind... sorry, I see what you mean now. >_<

Edit: Actually, I don't.

I understand your first argument, and it's right.

What we know from the first argument is that Dan is a liar, and that one of his brothers is a liar, and the other one is honest. This must be so since not all three can be liars, and if his two brothers would be honest, then his statement would be true, which isn't possible (because it would make all three honest).



Now... I don't see how your second argument is valid.



Firstly, why to you place Jake in this? You don't yet know if what he says is true or false. If Lito's statement is is true, then he could be in group 1 (? (H)). If his statement is false, he would be the liar in group 2.

Sorry about that, you're right, fixed it.