no, little tip: if there were 50 monks instead of 40, the answer would be the same

0?

24.
3 monks

Edited 6/18/2014 01:19:34

explanation?

the fist day no one leaves, that means >1 infected.
Day 2 no one leaves, that means >2 infected.
Day 3, they leave, so it is 3 infected.

The abbot let every monk know that all the other monks know they know etc. (common knowledge)

The only thing each monk does not know is, are they infected themselves? To figure out this they wait.
(Imagine yourselves as a sick guy) the fist day, you'll see all the other monks, if none of the are infected then you are(cause somebody is infected). If 1 other guy is infected (and he does not leave) then you know you are infected too. Thus you'd leave the 2nd day. If 2 other guys are infected (and they do not leave at day 2) you know you're infected. It goes on and on until they leave, as they left on day 3, 3 monks are infected. (2 other +"you").

Sorry if it's badly written, on my phone :p

Edited 6/18/2014 01:46:50

wut?

that doesn't explain why you wouldn't leave on day one, unless the abbot had the disease to begin with.
in any case, it is only 'possibly contagious' so I guess you would only leave after seeing evidence of it's contagious properties.

Let's say the abbot has it on day one. none of the monks leave as they believe themselves healthy and are unsure whether or not it is contagious.

Day 2. a monk gets the disease but is unaware of it. the other monks now know it is contagious, but stay because they don't know for certain that they have it (possibly spread to 2 people).

Day 3. the first monk to catch the disease sees another monk with the disease and deduces that it spreads to one person per day, therefore, he is ill. The other infected monk realises that he must be infected as he has seen the contagious properties of the illness and sees no new monks exhibiting the symptoms. they both leave. Abbot (crazy bugger that he is) stays on to spread the disease and create horrific riddles.

So. My answer is 2. :P

Edited 6/18/2014 07:58:30

Congratz to Hennns for having the right answer and the right explanation

His explanation seems flawed. Can you try to explain it in more detail?

This riddle was actually part of a math exam (in ~1980) for entering one of the best french engineering schools (Polythechnique). Although Hennns didn't prove it mathematically, he has the idea.I think the "contagious" part of the riddle is misleading though, i shouldn't have put it, but i just translated the original riddle.
To prove it mathematically it doesn't require heavy maths, a simple math induction method will do the trick.


base case:
assuming there is only 1 sick monk, he would have left on the first meeting. why? because we know for sure that there is a disease in the monastery, and therefore at least one monk is sick. The sick monk would see that there isn't any sick monk that he can see, and therefore he must be the one to be sick.

if there is 2 sick monks, both of them would see that the other is sick and no one else (number of sick monks <= 2). But there is no way for them to know if themselves are sick or not on the first meeting, so they wait for the second meeting. If no-one leave on the first meeting, it means that there can't be only 1 sick monk (base case), so there is exactly 2 sick monks and therefore if there is 2 sicks monks, they would figure it out after the 1st meeting and leave on the 2nd meeting.

Now lets make the hypothesis that if there is N sick monks, they would figure out they were sick and leave at the end of the Nth meeting


generalization:
if the hypothesis is true, and that there is N+1 sick monks

Each of these monks would see N sick monks in front of them. If we put ourselves in one of the sick monk's skin, there is only 2 possibilities: either there is N sick monks and we are healthy, or there is N+1 sick monks (N + me). So we must wait for the Nth meeting to see what happens. If no-one leaves on the Nth meeting, it can only mean that there is N+1 sick monks, and they can all leave on the N+1th meeting knowing that they are sick.

Since the base case and the generalization is true, the hypothesis is true for any N.

If only one monk would be infected on day 1, he would immediatedly leave. He would see 19 people that have no signs of a disease, making him the only one infected, since the fact that there is a disease is given.

If two would be infected, everybody would see that 2 people have signs of an infection, except for two, seeing only one. Now, because everybody is unsure of their own infection, they will have to wait for one day more than the number of infected they see, to clarify their own status.

So the one seeing only one infected would know by day 2, that he is infected himself, since if the other infected would have left on day one already if he would have seen no other people with infections.

Edited 6/18/2014 11:52:50

We posted at the same time :/

25. (classic)
Some guy just died and find himself at Hell and Heaven's doors. There is two doors, one for heaven and one for hell. Two giants, brothers, are guarding the doors (one giant for each door).The dead guy doesn't know wich leads to heaven and wich leads to hell, only god and the giant brothers knows that. God then tell the dead guy that one of the guard always tell the truth, and the other guard always lies (without telling him whom is whom ofc), and that he can ask only one question to one of the guards and then has to chose wich door he wants to open.
What question can he asks, so that he is assured to go to heaven?

I ask giant no1 :" Would your brother say that behind door A is heaven?"
or
I ask giant no2 :" Would your brother say that behind door A is heaven?"
or
I ask giant no1 :" Would your brother say that behind door A is hell?"
or
I ask giant no2 :" Would your brother say that behind door A is hell?"
or
I ask giant no1 :" Would your brother say that behind door B is heaven?"
or
I ask giant no2 :" Would your brother say that behind door B is heaven?"
or
I ask giant no1 :" Would your brother say that behind door B is hell?"
or
I ask giant no2 :" Would your brother say that behind door B is hell?"

Edited 6/18/2014 13:47:12

Yep, or "what would your brother say if i aked him where is the heaven's door?" :D

An archeologist in egypt unburies some ancient chambers. It is said that in holds either treasures or death. In the chambers he happens to find two enormous granite doors, one has the symbol of a King, the other the symbol of a snake on it.
Under the symbol of the Snake is written in glyphs: "Behind this door you will find death"
Under the symbol of the King the glyphs say: "One of the statements is true"

Which gate should the archeologist open?

Take the snake :) if the 2nd is true, the snake sentence is wrong -->he lives.
if the 2nd is wrong, both have to be wrong -->he lives again. Sorry for my bad explanation.

27. (headache garanteed)
One hundred dwarves were captured by a mean and hungry oger, but the oger wants to play with them and let them a chance to get out of his dungeon alive. He places them on a big staircase with a hundred steps, each darf with a hat on the head that can be either white or black.The dwarves are looking down, in a manner that one dwarf can only see the hats of the dwarves lower than him on the staircase.One by one, each dwarf then must announce the color of his hat, starting the dwarf on top of the staircase. If one shouts the right color he lives, if not the oger eats him immediatly. The oger, in his great clemency, allow the dwarves 5 min before starting the game to elaborate a strategy in order to save the maximum of dwarves as they can.
What would be the optimum strategy and how many dwarves would it statistically save?

Maybe i misunderstood the rules, but it didnt seem so heachachy.


They will save 99 dwarves and a 100 with a 50% chance.
First dwarf will go and say the color of the dwarf directly before him. So there is the 50% to die or live. Next dwarf will say the color he got told, but will put in some sort of delay, if that color is of the same color as the next dwarf before him or not. This way nobody else will have to die.

Edited 6/18/2014 16:12:44

#27: Define black hats as "1" and white hats as "0". Have the top dwarf count how many black hats there are and apply that number mod 2, which will be his guess.

Regardless of whether he is eaten or not, the next dwarf will know how many black hats the top dwarf saw (mod 2); since he can see every hat but his own (and the first guy's, but he doesn't matter), he should count the number of black hats he sees. If it matches (mod 2), then he has a white hat; if it doesn't match (mod 2), he has a black hat. He guesses correctly.

Everytime a dwarf (after the first one) says "Black Hat", the base guess (the one that the first dwarf said) should be flipped. Count the black hats, and again, if it matches (mod 2), then he has a white hat; if it doesn't match (mod 2), he has a black hat.

All dwarves after the first WILL guess correctly, thus 99 dwarves will always be saved. The first dwarf has a 50% chance of being saved. So, on average, 99.5 dwarves will live.

(This works if the dwarves have to go rapid-fire, and doesn't depend on wait times)

Edited 6/18/2014 16:16:55