Lateral Thinking VIII and IX

[tera]

Is the "continued for a number of days" part important? It seems to imply that the monks collect some additional information over these days.

I've been trying to do a mathematical strategy, but so far I can't figure out more than probabilities past the sick monks = 1 case.

[Sayumi]

The answer doesn't have much to do with math and nothing with probabilities.
The answer to your question will be another hint I guess:

Spoiler:

The number of days depends an the number of monks who are sick.

Follow ayw's hints step by step and keep this one in mind... good luck and have fun 

[sstimson]

Another Question. Do any of the monks know they might be sick. Look at it like this. The sick monk does not know by himself. If all the monks are doing the normal habits with not change, then the sick will never know he is sick!.What do the monks know about the illness? Do they know they can not catch it. If so then they might not even care that one has purple spots. Based on your question, it sound like the only way one might know they are sick is a change in behavor toward that monk. if so then if the sick monk see a different behavor towards him by at least two different monks he might think he was sick, if he knew he could be sick. This question needs the answer to the question. " How much do the monks know about the illness?"

Later

[sstimson]

To expand this puzzle same start but this time make 2 piles. one all heads and other all tails. You can still flip coins. How would you do this?

Interesting, but I think that's impossible. You need some form of feedback.

Let's say you could ask at any time whether there are more heads or tails or equal numbers facing up. How often would you have to ask and how would you manipulate the coins to get one pile of heads and one pile of tails? Assume of course that you can always distinguish two coins from their location when you're blindfolded.

How about this. You can ask four time about amount of heads in a pile.

when you figure that one out, then try the either more head ot more tails answer same four questions though I think three would be enough

Later

[ayw]

Another Question. Do any of the monks know they might be sick. Look at it like this. The sick monk does not know by himself. If all the monks are doing the normal habits with not change, then the sick will never know he is sick!.What do the monks know about the illness? Do they know they can not catch it. If so then they might not even care that one has purple spots. Based on your question, it sound like the only way one might know they are sick is a change in behavor toward that monk. if so then if the sick monk see a different behavor towards him by at least two different monks he might think he was sick, if he knew he could be sick. This question needs the answer to the question. " How much do the monks know about the illness?"

Later

Nope this has nothing to do with behaviour towards each other (that would count as communication). The monks can deduce whether they're sick or not purely from the logical deductions they can make at dinner, where they can see how many of other monks are sick.

Of course a monk would know whether another monk has been to the doctor because the spots would have disappeared. But the fact is that, due to their lateral thinking skills, every monk will know after the same number of dinners whether he is sick or not, and therefore all sick will go to the doctor together. Your task is to explain how this happens by outlining the deductions of the monks. (Note: your task is not to find out how many monks are ill. In fact, the number of sick remains a free parameter in the solution.)

Follow the hints already given in the thread and you will arrive at the solution.  ;)
 
 

[bash7353]

Do the monks know that there is at least on sick person there?
Because:

Spoiler:

say only one monk is sick, then this one sees 24 other monks with no illness so he can easily come to the conclusion that he is sick.
Now assuming there are two sick monks, at dinner these two monks see from their perspective one sick monk. Due to the fact that he does not go to the doctor they can conclude there has to be another sick monk, which has to be theirself.

Same with 3 monks. Let's say you are one of the 3 sick monks, at dinner you see two sick monks, which after two days still haven't gone to the doctor, from that you can deduce that there has to be one other sick person, and that must be you.

What I think was somethimes a bit confusing is to come th conclusions by the other monks non-actions, because in reality people - and even monks - make mistakes

Hope my answer is right, gotta go to school now, so I have time to think about the coin riddle
See ya...

[ayw]

@googleearth

Spoiler:

Your answer is correct! Test your mates at school with this one..

[sstimson]

Did the monks agree ahead of time it was all or nothing ( they all went or none went ). No, Right? Here another problem Every time they sit down we see the same number of sick monks whether it one time of 25 times. You did say say the illness would not spread.If 24 monks sit down sick and one is not he see 24 sick would likely think he sick too ( The idea of peer influence ) the other 24 see one well and 23 sick. But lets make it simpler. One monk hears a least one monk is sick . Since he the only one he nows it is him.
Two monks both sick.One of the two monks hopes it is olny one monk that is sick.So he waits for day two. Since told at least one monk is sick and the other monk is still there he knows he sick. the other monk thinks the same way.so both go
two monks one sick. one monk will see one well monk and know it is him so that monk will go that day.
Three monks get a little confusing so I wait and think on that one ( 3 has 6 possibilities {AAB, ABB, BAA, BAB, AAA, BBB} where b is sick and a is well )
but one of those possibilities AAA can not happen.
1 -1 day
2 monks either 1 or 2 days

Later

[bash7353]

@googleearth

Spoiler:

Your answer is correct! Test your mates at school with this one..

I did. They said they thought this through at home, I'm looking forward to speaking with them about that tomorrow.

Just as I said before I used my rather boring classes - and there's a lot of them here in Germany - to think about the coin problem. Here's what I've come up with:

Spoiler:

I think you just have to choose 5 of the coins for group A and flip them all, and the other 5 you just leave the way they are in group B. The sums of heads up coins and tail up coins is in the beginning 5 each, after just choosing 5 for group A and the other 5 for group B you will see that the sums in each group of head/tail up coins contain the same two numbers, just the other way around (I don't know if you understood this but I'm no English native speaker and I couldn't think of any different way to put this).
For instance: In group A you got 2 heads up and 3 tails up coins, that automaticly means you have 3 heads up and 2 tails up coins in group B because there's a total of 5 heads up coins, some goes for tails up coins. Now you just have to flip all 5 coins in one group in oder for the numbers to change places, so in both groups it is identical.

Hey, this is fun, you got other "Lateral thinking"s?

[Sayumi]

@sstimson

Spoiler:

Okay, your getting really close. Try not changing the total number of monks (25) and just vary the number of sick monks, starting with one, then two and so on. (It's not necessary to write down all the possibilities - with 25monks that would take ages...)

@googleearth

Spoiler:

Good job, both of your answers are right!!!  The monk one is actually quite tricky. Your answer for the coin one is perfectly fine and understandable.
I'm German as well so if really necessary you can ask/post in German as well, but let's try to stick to English whenever possible...

Glad you liked them! Try looking into some of the older posts (if you haven't already done so). My recommendation would be Lateral Thinking VI (really hard, but totally awesome  ), LT V (only if you've got some spare time - you can't solve this one just with logic, you'll have to go with trial and error) and LT IV (not that hard, but really nice  ) Try looking into the cases as well, they're really fun to do! I've got a few more LTs at hand and I will post them sometime this or next week... if you ask nicely 

[bash7353]

[...]

@googleearth

Spoiler:

Good job, both of your answers are right!!!  The monk one is actually quite tricky. Your answer for the coin one is perfectly fine and understandable.
I'm German as well so if really necessary you can ask/post in German as well, but let's try to stick to English whenever possible...

Glad you liked them! Try looking into some of the older posts (if you haven't already done so). My recommendation would be Lateral Thinking VI (really hard, but totally awesome  ), LT V (only if you've got some spare time - you can't solve this one just with logic, you'll have to go with trial and error) and LT IV (not that hard, but really nice  ) Try looking into the cases as well, they're really fun to do! I've got a few more LTs at hand and I will post them sometime this or next week... if you ask nicely 

In that one situation I just wasn't sure in what words to put it, so everyone will understand it. In that matter, I think the English language is way better than German one. To post in German wouldn't have helped me in that situation...
But still, good to know that you live in Germany, too.
Later

EDIT:
@Sayumi: It really would be nice if you (or anyone else) put up other riddles like these...

[ayw]

EDIT:
@Sayumi: It really would be nice if you (or anyone else) put up other riddles like these...

As Sayumi said, have a look at earlier threads. There are Lateral Thinking I - IX, Cases I - V, and lots of other Riddles. You may have to go up to 3 pages back to see the earliest.

We'll try and keep this up.   

[Sayumi]

Yep, I'll work on something as soon as I have the time and a few good ideas  . Here just some little things to keep the mind busy...
I got these ones of the internet, but I changed both of them a bit... the second one was with cement originally, but I like chocolate cake better .

The Cubes
Yuusaku Kudo once said to little Shinichi: "I've got two cubes (with a number on each face) on my desk. I arrange the them every day, so that the front faces show the current day of the month. What numbers are on the faces of the cubes?"
(You can't represent the day "5" with a single cube with a side that says 5 on it. You have to use both cubes all the time. So the 5th day would be "05".)

Five liters
You are baking a huuuge chocolate cake and the recipe calls for five liters of milk. You have a cow giving you all the milk you need. The problem is that you only have a four liters and a seven liters bucket and neither has graduation marks. Find a method to measure five liters. (You'll have to waste a lot of milk - but it's an amazing cow with a neverending supply of milk  )

[Holmes]

I´m back!(from Holidays, I don´t start school till the 20th  )(I know you all missed me  )  And Oh! my God we´re already on page 2!

Well I read all the comments so, I know the answers to the Lateral Thinking ones, I really didn´t mean to, but, I had to read them to actualize myself in this thread.

And googleearth, have the recommendations of Sayumi and ayw( they have the best reputation in solving problems, so...)

And finally,

My reasoning, bah, I don´t really call it reasoning, but, nevermind:

Five litres ---

Spoiler:

First, fill in 4 litres of water in the 4 litres bucket. After that, from the milk you just put, put it in the 7 litres bucket, so, now you have nothing in the 4 litres bucket and 4 litres in the 7 litres buckt. Then, fill in again the 4 litres bucket and re-put it in the 7 litres bucket. Now you have 1 litre in the 4 litres bucket and the 7 litres bucket is full. Finally, throw away all 7 litres and put in the 1 litre milk from the 4 litres milk in the 7 litres bucket, and then fill in the 4 litres bucket. If my maths are correct, 1+4=5

And please Sayumi, I don´t quite understand "The Cubes", can you explain it please?

[Sayumi]

@Holmes-Of course we all missed you! I started to think that you had retired  As for your answer:

Spoiler:

Yes, that would work and it's probably the easiest way. Well done!

Okay the cubes (I'll put this in a spoiler because I think it might be solvable without it. It doesn't tell the solution but it makes the answer a bit easier):

Spoiler:

You have two cubes.  Each of these cubes has six sides/faces (like a dice). Your task is to write a number on each side, so that with rearranging the two cubes you can express each day of a month (today it would be "04"-zero on one cube and four on the other one-  because it's 4thMarch (at least it's here). 21st of a month would be "2" and "1", and so on...)

This is only if it's still hard to understand after the first spoiler!!!

Spoiler:

Arrange the numbers from 0 to 9 on the 2x6 sides of the cubes so that you can form any number from 1 to 31 (no month has more than 31 days) with the two cubes.

Hope this explained it a bit. It's actually not that difficult and just to keep you guys busy, until someone comes up with something harder.