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WL riddle thread reborn: 6/16/2014 23:53:34


Captain Bob
Level 44
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23. inkstand
WL riddle thread reborn: 6/17/2014 00:26:30


Genghis 
Level 54
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22. You make sure one of them is turned off. Then, you turn on the other 2. After 5 minutes, you turn off one of the 2 and leave the room. You can identify the one you left on easily. You figure out which one you left off because it will be colder than the other inactive bulb.
WL riddle thread reborn: 6/17/2014 16:57:26


Min34 
Level 63
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#21
A man and a woman


You never said it was his wife, so the stranger might have been in the car all the time :p

He made sure nobody was in the car

:O And her name was nobody!?!?!?!?!

...








I think I got it right
WL riddle thread reborn: 6/17/2014 17:05:04


Nightshade
Level 54
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The stranger was the taxi driver
WL riddle thread reborn: 6/17/2014 17:44:47


Genghis 
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The woman was dead? Well clearly something is missing from the answer.
WL riddle thread reborn: 6/17/2014 17:51:11


125ch209 
Level 58
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24. (hard one)

40 monks are living in a very strict monastery, they can only prey and can't communicate with each other by voice nor by sign. They can't even look themselve in a mirror. Every day, the Higher abbot, who is the only one allowed to speak, reunite the monks in a room to keep them informed of the daily news.

A rare disease very dangerous and possibly contagious is striking in the monastery, it is characterized by the presence of small red spots on the face, visible but not painful. It does not cause other symptoms at the start. Each monk therefore can not know if he is sick.

The abbot decides to informs all the monks about this disease. During the daily reunion, he told them about the danger of this disease, and asks that at the end of each reunion, when he says so, every sick monk aware of their own sickness leaves the monastery.

At the end of the first reunion (the one he told them about the disease), the abbot says :
" All monks aware of their own sickness, please leave at once" - But no one left

The next day, at the end of the reunion, the abbot said:
" All monks aware of their own sickness, please leave at once" - But no one left

The day after, at the end of the reunion, the abbot said:
" All monks aware of their own sickness, please leave at once" - and every single monk affected by the disease got up and left.

How many were they?

Edited 6/17/2014 20:09:00
WL riddle thread reborn: 6/17/2014 18:38:28


Daisuke Jigen
Level 56
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21. Think, you fools! Although badly worded, that is not where the answer lies. Lateral thinking. My guess is that the man's wife was pregnant, died in childbirth and the baby is the stranger. Have we met? gaaa!

Edit: 24. There were 40 monks. ;)

Edited 6/17/2014 18:40:57
WL riddle thread reborn: 6/17/2014 18:49:41


Genghis 
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Mmm that stranger has a piece of bacon.
WL riddle thread reborn: 6/17/2014 19:58:32


125ch209 
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24. There were 40 monks. ;)


nope ;)
WL riddle thread reborn: 6/17/2014 21:27:18


Timinator • apex 
Level 67
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i'd rather say 41 :P
WL riddle thread reborn: 6/17/2014 22:36:38


125ch209 
Level 58
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no, little tip: if there were 50 monks instead of 40, the answer would be the same
WL riddle thread reborn: 6/18/2014 00:41:02


Captain Bob
Level 44
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0?
WL riddle thread reborn: 6/18/2014 01:19:22

Hennns
Level 58
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24.
3 monks

Edited 6/18/2014 01:19:34
WL riddle thread reborn: 6/18/2014 01:23:44


125ch209 
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explanation?
WL riddle thread reborn: 6/18/2014 01:44:42

Hennns
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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
WL riddle thread reborn: 6/18/2014 02:02:46


Captain Bob
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wut?
WL riddle thread reborn: 6/18/2014 07:51:51


{rp} GeneralGror
Level 58
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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
WL riddle thread reborn: 6/18/2014 08:59:07


125ch209 
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Congratz to Hennns for having the right answer and the right explanation
WL riddle thread reborn: 6/18/2014 10:33:13


◄Atum►
Level 46
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His explanation seems flawed. Can you try to explain it in more detail?
WL riddle thread reborn: 6/18/2014 11:38:03


125ch209 
Level 58
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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.
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