Rotwang Posted September 17, 2012 Share Posted September 17, 2012 Nobody? The next term is 312211. If you need an additional hint, try reading the sequence out loud. Quote Link to comment Share on other sites More sharing options...
Lemmiwinks Posted September 17, 2012 Author Share Posted September 17, 2012 Nobody? The next term is 312211. If you need an additional hint, try reading the sequence out loud. I still don't get it... Quote Link to comment Share on other sites More sharing options...
CoLoUr DoTZ Posted September 17, 2012 Share Posted September 17, 2012 ----------1 ---------11 ---------21 ------1211 ---111221 ---312211 13112211??? Quote Link to comment Share on other sites More sharing options...
Rotwang Posted September 17, 2012 Share Posted September 17, 2012 ----------1 ---------11 ---------21 ------1211 ---111221 ---312211 13112211??? Close, the correct answer is 13112221. Quote Link to comment Share on other sites More sharing options...
Lemmiwinks Posted September 17, 2012 Author Share Posted September 17, 2012 Close, the correct answer is 13112221. :huh: :huh: :huh: Quote Link to comment Share on other sites More sharing options...
Rotwang Posted September 17, 2012 Share Posted September 17, 2012 :huh: :huh: :huh: The first number is 1. It has one 1 in it -> 11. The second number has two 1's in it -> 21. The third number has one 2, followed by one 1 -> 1211. The fourth number has one 1 followed by one 2 followed by two 1's -> 111221 etc. Quote Link to comment Share on other sites More sharing options...
CoLoUr DoTZ Posted September 17, 2012 Share Posted September 17, 2012 Close, the correct answer is 13112221. For bonus points, what is the highest digit that appears in this sequence? The highest digit can only be 3 then....... Quote Link to comment Share on other sites More sharing options...
Rotwang Posted September 17, 2012 Share Posted September 17, 2012 The highest digit can only be 3 then....... Yep. Here's another fun riddle: a prison has 50 prisoners in it. One day at lunch time the prison warden tells the prisoners that, later that day, they will be placed on a staircase with 50 steps on it, facing downward. Each prisoner will then have a hat placed on his head, which may be black or white. The prisoners will not be able to see the colour of their own hat or the hat of anyone above them; they will only be able to see the hats of prisoners below them. The warden will then ask each prisoner in turn, starting with the prisoner at the top of the stairs and working his way down, to guess the colour of his hat. If he guesses wrong he will be shot, but otherwise he will be spared. Once on the stairs the prisoners will not be allowed to talk, or communicate in any way, except when prompted to guess the colour of their hat, and even then will only be allowed to say either "black" or "white". But they are allowed to talk freely for the duration of lunch, for the purpose of agreeing a strategy. Assuming they are smart enough, what is the maximum number of prisoners who will die? Quote Link to comment Share on other sites More sharing options...
CoLoUr DoTZ Posted September 17, 2012 Share Posted September 17, 2012 Yep. 1 from 3 tries......... is ok no?? Here's another fun riddle: a prison has 50 prisoners in it. One day at lunch time the prison warden tells the prisoners that, later that day, they will be placed on a staircase with 50 steps on it, facing downward. Each prisoner will then have a hat placed on his head, which may be black or white. The prisoners will not be able to see the colour of their own hat or the hat of anyone above them; they will only be able to see the hats of prisoners below them. The warden will then ask each prisoner in turn, starting with the prisoner at the top of the stairs and working his way down, to guess the colour of his hat. If he guesses wrong he will be shot, but otherwise he will be spared. Once on the stairs the prisoners will not be allowed to talk, or communicate in any way, except when prompted to guess the colour of their hat, and even then will only be allowed to say either "black" or "white". But they are allowed to talk freely for the duration of lunch, for the purpose of agreeing a strategy. Assuming they are smart enough, what is the maximum number of prisoners who will die? Do they know how many white and black hats are?? Quote Link to comment Share on other sites More sharing options...
Rotwang Posted September 17, 2012 Share Posted September 17, 2012 1 from 3 tries......... is ok no?? Yes. Do they know how many white and black hats are?? No. Quote Link to comment Share on other sites More sharing options...
CoLoUr DoTZ Posted September 17, 2012 Share Posted September 17, 2012 at the most 25 prisioners would die. First one says the color of the hat from the guy just below him, instead of trying to guess his. If he is lucky he lives, if he is not lucky he dies, but the guy below survives as he already knows the color of his hat. Following guy announce the color of the hat from the guy below him..... and so on... At the most 25 die, if no one of the odd numbers is lucky enough to have same hat color as his fellow even mate. Quote Link to comment Share on other sites More sharing options...
Rotwang Posted September 17, 2012 Share Posted September 17, 2012 That's good, but they can do better. Would you believe me if I said that the prisoners can agree a strategy that guarantees at most one of them would die? Quote Link to comment Share on other sites More sharing options...
Ormion Posted September 17, 2012 Share Posted September 17, 2012 The first number is 1. It has one 1 in it -> 11. The second number has two 1's in it -> 21. The third number has one 2, followed by one 1 -> 1211. The fourth number has one 1 followed by one 2 followed by two 1's -> 111221 etc. edit: Ignore me. I got that. That's a very simple, but very clever riddle. Nice! Quote Link to comment Share on other sites More sharing options...
Rotwang Posted September 18, 2012 Share Posted September 18, 2012 That's good, but they can do better. Would you believe me if I said that the prisoners can agree a strategy that guarantees at most one of them would die? Clue time: Consider the 49 prisoners below the first one. Each can see how many black hats the prisoners in front of him have, and by the time the warden gets to him he has heard how many black hats the prisoners behind him have. If he knew how many black hats there were among those 49 prisoners, he would know whether his hat were black or white. On the other hand, the first prisoner can see how many black hats there are among the 49 prisoners below him. But, since he can only say one of two words, he can't tell the other prisoners how many black hats there are - or can he? Quote Link to comment Share on other sites More sharing options...
Ormion Posted September 18, 2012 Share Posted September 18, 2012 Hey Rot, do oyu remember one about some monks that must not speak to each other and each of them had a cross on his back or something like that? That was a great one, but I can't remember it. Quote Link to comment Share on other sites More sharing options...
JISNEGRO Posted September 18, 2012 Share Posted September 18, 2012 They agree that if the first prisoner says black it means there is an odd number of black hats, but if he says white there is an even number of black hats. Knowing this, each other prisoner can count how many black hats are in front of him and because they have heard how many other hats have been told before, he can then know whether his own hat is black or not according to the parity initially set. The only prisoner guessing the color of his own hat is the first one. Quote Link to comment Share on other sites More sharing options...
Rotwang Posted September 18, 2012 Share Posted September 18, 2012 Hey Rot, do oyu remember one about some monks that must not speak to each other and each of them had a cross on his back or something like that? That was a great one, but I can't remember it. Yes, but I'm about to dash off. I'll be back tomorrow, probably, and I'll see if I can reproduce it then. They agree that if the first prisoner says black it means there is an odd number of black hats, but if he says white there is an even number of black hats. Knowing this, each other prisoner can count how many black hats are in front of him and because they have heard how many other hats have been told before, he can then know whether his own hat is black or not according to the parity initially set. The only prisoner guessing the color of his own hat is the first one. Bingo. Quote Link to comment Share on other sites More sharing options...
JISNEGRO Posted September 23, 2012 Share Posted September 23, 2012 A real-world mindbending problem is this: A man receives two invites for distinct dinners held at the same time, but in difference places in the same block. He leaves home to attend one of them. Next morning people that were in the two parties tell that they saw him at the party they were at too. Meaning he was in two different parties at the same time. How could this happen? If you find a solution, you have right now one million dollars waiting for you. Later, contracts worth billions of dollars. Plus, fame and a name in history like Eistein's. And, the situation above does happen in real world. Quote Link to comment Share on other sites More sharing options...
Rotwang Posted September 23, 2012 Share Posted September 23, 2012 Oh, I forgot to give the monk riddle. Here it is, IIRC: A monastery has 10 monks, each of whom has taken a vow of silence. They have no way to communicate with each other or the outside world. Each morning they go to the field to work, and each evening they go back to their own private rooms to pray. One night, God appears to each monk to inform them that he has placed a cross on some of their backs. He tells them that those monks, and only those monks, who have crosses on their backs must kill themselves; if they have not done so within 10 days, the world will end. They are not told how many crosses there are, except that there is at least one, and they are told that each monk has been told exactly the same thing. They cannot check their own backs for a cross, but they can see whether the other monks have crosses on their backs or not. WHAT DO? Quote Link to comment Share on other sites More sharing options...
Ormion Posted September 23, 2012 Share Posted September 23, 2012 IIRC this Is how it goes: Let's assume that there's 1 cross. That means that the monk with the cross on his back will check the other monks and see that none has one which means he has it then he kills himself. If there are 2 crosses: A cross-monk checks the others and sees 1 cross. He can't know if there's only this cross or has one on his back as well, so he does nothing. The next day if the other cross-monk is still alive then he realizes he has a cross as well since the other cross-monk should have killed himself like in example 1. If there are 3 crosses. A cross-monk would see 2 crosses on others backs. The next day he will do nothing like in example 2. The third day though if none of the other two cross-monks have killed themselves then he realizes he has a cross on his back (since if he didn't the two other would had ignored him and killed themselves in the first 2 days). And so on. More crosses, more days to wait. Since maximum crosses 10=10 days till earth gets destroyed Quote Link to comment Share on other sites More sharing options...
Ormion Posted September 23, 2012 Share Posted September 23, 2012 Ok one more. Replace a letter with a number so you can solve this equation. S E N D + M O R E _________ M O N E Y Quote Link to comment Share on other sites More sharing options...
Procyon Posted September 23, 2012 Share Posted September 23, 2012 @Jisnegro: I remembered "Sliding Doors" a 1998 movie with Gwyneth Paltrow exploring the 'division' of someone's reality in two states. In the movie this happens in the exact moment she is about to embark in a London subway car: she embarks and she misses the train at the same time (hence the title sliding doors), and the movie explores the two paralell worlds, until one of them collapses. A funny movie, worth watching. http://en.wikipedia.org/wiki/Sliding_Doors Quote Link to comment Share on other sites More sharing options...
Rotwang Posted September 24, 2012 Share Posted September 24, 2012 @ Ormion, if I understood what you're asking: Assuming that each line starts with a digit other than 0, and that no two letters correspond to the same digit, there is exactly one solution; namely s = 9, e = 5, n = 6, d = 7, m = 1, o = 0, r = 8, y = 2. If we drop the second requirement, there are loads of solutions. Quote Link to comment Share on other sites More sharing options...
JISNEGRO Posted September 24, 2012 Share Posted September 24, 2012 @Jisnegro: I remembered "Sliding Doors" a 1998 movie with Gwyneth Paltrow exploring the 'division' of someone's reality in two states. In the movie this happens in the exact moment she is about to embark in a London subway car: she embarks and she misses the train at the same time (hence the title sliding doors), and the movie explores the two paralell worlds, until one of them collapses. A funny movie, worth watching. http://en.wikipedia....i/Sliding_Doors Thanks for the recommendation, I will check it out. Quote Link to comment Share on other sites More sharing options...
Ormion Posted September 24, 2012 Share Posted September 24, 2012 @ Ormion, if I understood what you're asking: Assuming that each line starts with a digit other than 0, and that no two letters correspond to the same digit, there is exactly one solution; namely s = 9, e = 5, n = 6, d = 7, m = 1, o = 0, r = 8, y = 2. If we drop the second requirement, there are loads of solutions. Correct. Quote Link to comment Share on other sites More sharing options...
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