E=mc² Clan Forum

Another unsolved problem, which i heard on physics courses at school: noone before couldn`t made computer model of "trajectory of mutual movement of three bodies" (or four bodies, don`t remember at the moment...)

Ok, now a real accessible enigma, for everybody who want to solve it.

They are 100 monks who never speak or communicate to each other in anyway. During the day, the monk see all the others, but can't speak or communicate with the other one's. Monday, early in the morning, a badnews come at the monastery. At least one of the monks is sick. The sickness is not contagious. A monk which is certain that he is sick will kill himself during the night. The only visible sign of this sickness is that there is a red button (or spot, or something like that, don't know the exact translation) on his face. The next monday, all the sick monk(s) is/are dead, the other are still alive.

How many monks were sick?
How that happens??

Good luck in this new enigma...

well... I think only one of them was sick and he is the one, who is alive:)..
WHY:
Because news can only SPREAD, if they SEE the red spot (as they dont talk to each other).. So, 99 monks have SEEN the red spot on that guy's face, but, of course, could not tell him.. Supposing, that they didnt have mirrors - they all thought, that they could be sick, as well, so they killed themselves.. While the really sick monk DID NOT SEE any of the spots, so had no reasons to worry and kill himself... But I dont really understand why it is MONDAY ..

No, i don't think it is the right answer. And yes, I forgot to mention that those monk don't have any mirror.
But themonks kill themselves (or commit suicide, don't know exactly how to say that) only if they are certain that they are sick. So a monk seing a red point on another one can just think: "maybe I'm sick, maybe not" thus he is not sure about it's health.

But the monk who don't see any any red point know that he has one on his face, because there is a least one sick monk, and thus he will commit suicide during the next night.

That is the spirit of the enigma. But one sick person, it's to easy, and don't care about some information in the enigma... (I don't know the answer, i just found this enigma on the web, and while trying to solve it, i finally found a solution who take care of all information in the enigma, so i assume it is the right one...)

Try something else

well... Ok... but as they dont know the exact number, they can never be sure:).. because if there are two - they will see each others spots, but will not worry about themselves... as there might be just one sick monk..

OR... if there is one - they all die on the second night:).. because on the first night the sick one kills himself, as he doesnt see any other spots, and the others will kill themselves afterwards, as THEY will not see any other red spots...

it's pure logical problem...
But i have some trouble to explain my reasonnement in english lol.
Just I imagine the scene and the answer is logical for me.

A monk see if a monk is sick or not.

Imagine, one monk is sick :
a monk sees nobody who is sick, he's kill himself. (At least one monk is sick)
a monk see a monk who is sick, he knows he's not sick so he don't commit suicide. Find the sick monk takes one day.

Imagine, two monk are sick :
Sick monk sees one sick monk. So the next day, if a sick monk(1) see again the previous sick monk(2), he knows that he was sick the next day. Why? Because monk (1) and monk (2) are not commit suicide on day 1. because they saw one sick monk. So, on day 2, they commit suicide together...

We know that all sick monks are dead on the next monday. So, I think there are 8 sick monks...

Nice, Cynyx:).. I think it is right... not sure about the number (didnt count myself), but the logic seems to work..

7 or 8 ...
but i think 8...
Need a paper and a pen... I have only a computer at work LOL... hmmm at home also

hmmm, i need a paper and cannabis.....
have only tabacco and alco at home

lol Fede lol... 8)

Correct reasonment, but i personnlay think that the answer is 7. It's depends if we consider the monk kill themselves during the night between Sunday and Monday, or between Monday and Tuesday...

Nice work Cynyx

try thisgame.


Just click on the blue circle to start !

Aim : All people have to cross the river !

You have to follow the following rules

1. Only 2 people in the boat

2. The father can't be with the little girls WITHOUT the mother

3. The mother can't be with the little boys WITHOUT the father

4. The robber can't be with any familly member WITHOUT the cop

5. Only cop and parents can drive the boat

Click on the people for put them in or out the boat
Click on red button to cross the river

First to post a screenshot with all people at the next side of the river win the right to post an another enigma

Oh no, they all died...there was something wrong with cop....he killed the father with knife....mother smashed robber`s head to stone.....kids ran for survive....but they couldn`t swim....two girls died in water....tamahawk missle reached "the boat" with guys....mine aim bot killed all survived persons........
phoooooohhhh.......хороша афганка...

is it the right sollution....
and the screen with answer is right too?...

Weeehaaa..
http://img118.imageshack.us/my.php?imag ... aaagb3.jpg
..done..
I will think of a task now...

Ok.. there is a very very old one, so somebody might know the answer... and its very easy, anyway:)...
A man has 1000 coins and 10 bags, but once he put some money in the bag, he can not open it, so he can only pay with the full bags, and, of course, he wouldnt want to pay more than he needs. So, how put the money into the bags, so he can pay any amount using the bags, without opening them?..

Not sure it is the right way of doing it, but 1024=2^10
So

• bag 1 : 2^0=1 coin
  bag 2 : 2^1 = 2 coins
  bag 3 : 2^2 = 4 coins
  bag 4 : 2^3 = 8 coins
  bag 5 : 2^4 = 16 coins
  bag 6 : 2^5 = 32 coins
  bag 7 : 2^6 = 64 coins
  bag 8 : 2^7 = 128 coins
  bag 9 : 2^8 = 256 coins
  bag 10 : the remainings coins: 1000-511 = 489 coins

This coins repartition allow you to pay each amount of coins, as soon as you know how many coins are in each bags...

Yeah, thats right, Foug...

Ok, another very easy enigma:

Three poor guy's come to a cheap hotel and ask for a 3 rooms where they can sleep. The Motel owner ask them 10 € for each room. Thus, the three man paid a total of 30€. The guys come to their respective rooms, but soon the cleanng lady come to their doors. The price for tree room is only 25€, so teh owner asked to the cleaning lady to give back 5€. But 5 isn't a multiple of 3, so the cleaning lady decided to give back 3€, and she puts the 2 remaining s in his pocket.

So we can write (10-1)*3 +2 = 30. Thus 29 = 30.

WTF???

FougereDeGuerre wrote:Ok, another very easy enigma:

Three poor guy's come to a cheap hotel and ask for a 3 rooms where they can sleep. The Motel owner ask them 10 € for each room. Thus, the three man paid a total of 30€. The guys come to their respective rooms, but soon the cleanng lady come to their doors. The price for tree room is only 25€, so teh owner asked to the cleaning lady to give back 5€. But 5 isn't a multiple of 3, so the cleaning lady decided to give back 3€, and she puts the 2 remaining s in his pocket.

So we can write (10-1)*3 +2 = 30. Thus 29 = 30.

WTF???

I can't understand when we can write that formula. For me, there is no moment where (10-1)*3+2 describes the situation.

Total amount of money : 30
there are owner, guys and lady.

-> they pay the room
owner 30 guys 0 lady 0
-> the owner payback
owner 25 guys 0 lady 5
-> lady send money back and keep some
owner 25 guys 3 lady 2

At all the time, the sum is 30... and no situation describes (10-1)*3+2 ... this formula is wrong i think.

Yeah, finally someone got it

Ok here is mine

Try to find 4 mistakes, if u find them ill donate 20 Euro to u..


A rooster/cock/male chicken/kickirickooo (:P) sitting in the corner of a barrel laid Four 24 karat. gold eggs.

only one word to say : "too easy"(I know that are two words). 8)
Anyway I'll give you the answer but first send me the 20 euros

i meant i would donate 20 euros to the clan, not send em to u

eeenyway, lets see if anyone else knows the answer (anti u still havent said it so i suppose u didnt find it :P)

Be sure I know it^^.
I'm just waiting for the money :P

i want the money too:).. I know the answer (face it, it is not difficult):)..

it isnt difficult, but face it.. if u solve it then, haha, sorry

eeenyway i wanna hear the answers

ah cmon? no answers?

Think! (let me see brains exploding! or i can BOOM hEAdshot u :P)

Maxut wrote:Ok here is mine

Try to find 4 mistakes, if u find them ill donate 20 Euro to u..


A rooster/cock/male chicken/kickirickooo (:P) sitting in the corner of a barrel laid Four 24 karat. gold eggs.

Perhaps an answer :
-> a cock can't give eggs
-> a barrel has no corner
-> only one egg for a ckicken in one time
-> and the golden chicken don't exist !

wrong.. u were almost there..

oh comon guys ill explode in the end... here is the answer... :P


1st mistake: a cock cant lay eggs
2nd mistake: barrels are round, therefore no corners
3rd mistake: eggs cannot be gold

and last but not least..


I will not donate the 20 euros


(haha cynyx, couldnt think about it :P but because he found 3 out of 4, i give him a huge BANANA )