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Lateral Thinking Logic Problems More Very Easy - Easy - Difficult - Very Difficult Print Problems 1. The Camels Print Solutions Four tasmanian camels traveling on a very narrow ledge encounter four Hide All tasmanian camels coming the other way. Show All As everyone knows, tasmanian camels never go backwards, especially Site Map when on a precarious ledge. The camels will climb over each other, but only if there is a camel sized space on the other side. Send Feedback The camels didn't see each other until there was only exactly one camel's width between the two groups. Discussion Board How can all camels pass, allowing both groups to go on their way, without any camel reversing? Show Hint Show Solution View the Solutions Hint: Use match sticks or coins to simulate the puzzle. Answer: etc.... 2. The Waiter Three men in a cafe order a meal the total cost of which is $15. They each contribute $5. The waiter takes the money to the chef who recognizes the three as friends and asks the waiter to return $5 to the men. The waiter is not only poor at mathematics but dishonest and instead of going to the trouble of splitting the $5 between the three he simply gives them $1 each and pockets the remaining $2 for himself. Now, each of the men effectively paid $4, the total paid is therefore $12. Add the $2 in the waiters pocket and this comes to $14.....where has the other $1 gone from the original $15? Show Solution View the Solutions Answer: The payments should equal the receipts. It does not make sense to add what was paid by the men ($12) to what was received from that payment by the waiter ($2) Although the initial bill was $15 dollars, one of the five dollar notes gets changed into five ones. The total the three men ultimately paid is $12, as they get three ones back. So from the $12 the men paid, the owner receives $10 and the waiter receives the $2 difference. $15 - $3 = $10 + $2 3. The Boxes There are three boxes. One is labeled "APPLES" another is labeled "ORANGES". The last one is labeled "APPLES AND ORANGES". You know that each is labeled incorrectly. You may ask me to pick one fruit from one box which you choose. How can you label the boxes correctly? Show Solution View the Solutions Answer: Pick from the one labeled "Apples & Oranges". This box must contain either only apples or only oranges. E.g. if you find an Orange, label the box Orange, then change the Oranges box to Apples, and the Apples box to "Apples & Oranges" 4. The Cannibals Three cannibals and three anthropologists have to cross a river. The boat they have is only big enough for two people. The cannibals will do as requested, even if they are on the other side of the river, with one exception. If at any point in time there are more cannibals on one side of the river than anthropologists, the cannibals will eat them. What plan can the anthropologists use for crossing the river so they don't get eaten? Note: One anthropologist can not control two cannibals on land, nor can one anthropologist on land control two cannibals on the boat if they are all on the same side of the river. This means an anthropologist will not survive being rowed across the river by a cannibal if there is one cannibal on the other side. Show Solution View the Solutions Answer: First, two cannibals go across to the other side of the river, then the rower gets called back. Next, the rowing cannibal takes the second across and then gets called back, so now there are two cannibals on the far side. Two anthropologists go over, then one anthropologist accompanies one cannibal back, so now there is one anthropologist and one cannibal on the far side. The last two anthropologists go over to the far side, so now all the anthropologists are across the other side, along with the boat and one cannibal. In two trips, the cannibal on the far side takes the boat and ferries the other two cannibals across the river. 5. The Father A mother is 21 years older than her child. In exactly 6 years from now, the mother will be exactly 5 times as old as the child. Where's the father? Show Solution View the Solutions Answer: With the mother. If you do the math, you find out the child will be born in 9 months. 6. The Double Jeopardy Doors You are trapped in a room with two doors. One leads to certain death and the other leads to freedom. You don't know which is which. There are two robots guarding the doors. They will let you choose one door but upon doing so you must go through it. You can, however, ask one robot one question. The problem is one robot always tells the truth ,the other always lies and you don't know which is which. What is the question you ask? Show Hint Show Solution View the Solutions Hint: The two robots know each others personality. That they talk when they're bored, lonely, etc. Try to get the two robots to cancel their evil & good ways out. Answer: Ask one robot what the other robot would say, if it was asked which door was safe. Then go through the other door. 7. The Frog A frog is at the bottom of a 30 meter well. Each day he summons enough energy for one 3 meter leap up the well. Exhausted, he then hangs there for the rest of the day. At night, while he is asleep, he slips 2 meters backwards. How many days does it take him to escape from the well? Note: Assume after the first leap that his hind legs are exactly three meters up the well. His hind legs must clear the well for him to escape. Show Hint Show Solution View the Solutions Hint: Try to think the problem through for a five meter well. Now what is the solution for the 30 meter well? Answer: 28 Each day he makes it up another meter, and then on the twenty seventh day he can leap three meters and climb out. 8. The Bobber You can paddle your canoe seven miles per hour through any placid lake. The stream flows at three miles per hour. The moment you start to paddle up stream a fisherman looses one of his bobbers in the water fourteen miles up stream of you. How many hours does it take for you and the bobber to meet? Show Solution View the Solutions Answer: 2 Ignore the speed of the stream, as the cork will be carried along at three miles per hour as will you. It takes two hours to travel fourteen miles, at a rate of seven miles per hour. 10. The Socks Cathy has six pairs of black socks and six pairs of white socks in her drawer. In complete darkness, and without looking, how many socks must she take from the drawer in order to be sure to get a pair that match? Show Solution View the Solutions Answer: 3 Socks do not come in in left and right, so any black will pair with any other black and any white will pair with any other white. If you have three socks and they are either colored black or white, then you will have at least two socks of the same color, giving you one matching pair. 11. There is something about Mary Mary's mum has four children. The first child is called April. The second May. The third June. What is the name of the fourth child? Show Solution View the Solutions Answer: Mary. Mary's mothers fourth child was Mary herself. 12. Petals Around the Rose The name of the game is Petals Around the Rose, and that name is significant. Newcomers to the game can be told that much. They can also be told that every answer is zero or an even number. They can also be told the answer for every throw of the dice that are used in the game. And that's all the information they get. The person who has the dice and knows the game, rolls five dice and remarks almost instantly on the answer. For example: in Roll #1 the answer is two. Roll #1. "The answer is what?" says the new player. "Two." "On that roll?" "Yes." "Would it still be two if I moved the dice without turning any of them over, just rearranging the pattern?" "I can tell you only three things: the name of the game, the fact that the answer is always even, and the answer for any particular throw. In this case the answer is two." "So that's how it is. What am I supposed to do?" "You're supposed to tell me the answer before I tell you. I'll give you all the time you want, but don't tell me your theory, just the answer. If you figure it out, you don't want to give the idea away to these other jokers around you. Make them work for the answers, too. If you get the answer right on six successive rolls, I'll take that as prima facie evidence that you understand the game." "OK, roll again." Roll #2. "I give up. What's the answer?" "The answer is eight." "Roll again." Roll #3. The answer is fourteen. Roll #4. The answer is zero. Roll #5. The answer is four. Roll #6. The answer is... An integral part of the puzzle is that those who have solved it are urged to keep the solution a secret, so there is no solution posted here. It is not a hard puzzle to figure out however. A claim that often accompanies these instructions is that the smarter an individual, the greater amount of difficulty the individual will have in solving it. If such a statement is true, it may be attributed to the fact that "smarter" people tend to be more knowledgeable in a wide range of information which they may unnecessarily attempt to draw upon to solve the puzzle. Very Easy - Easy - Difficult - Very Difficult More Very Easy Logic Puzzles >> Lateral Thinking Logic Problems More Very Easy - Easy - Difficult - Very Difficult 1. The Camels Four tasmanian camels traveling on a very narrow ledge encounter four tasmanian camels coming the other way. As everyone knows, tasmanian camels never go backwards, especially when on a precarious ledge. The camels will climb over each other, but only if there is a camel sized space on the other side. The camels didn't see each other until there was only exactly one camel's width between the two groups. How can all camels pass, allowing both groups to go on their way, without any camel reversing? Show Hint Show Solution View the Solutions Hint: Use match sticks or coins to simulate the puzzle. Answer: etc.... 2. The Waiter Three men in a cafe order a meal the total cost of which is $15. They each contribute $5. The waiter takes the money to the chef who recognizes the three as friends and asks the waiter to return $5 to the men. The waiter is not only poor at mathematics but dishonest and instead of going to the trouble of splitting the $5 between the three he simply gives them $1 each and pockets the remaining $2 for himself. Now, each of the men effectively paid $4, the total paid is therefore $12. Add the $2 in the waiters pocket and this comes to $14.....where has the other $1 gone from the original $15? Show Solution View the Solutions Answer: The payments should equal the receipts. It does not make sense to add what was paid by the men ($12) to what was received from that payment by the waiter ($2) Although the initial bill was $15 dollars, one of the five dollar notes gets changed into five ones. The total the three men ultimately paid is $12, as they get three ones back. So from the $12 the men paid, the owner receives $10 and the waiter receives the $2 difference. $15 - $3 = $10 + $2 3. The Boxes There are three boxes. One is labeled "APPLES" another is labeled "ORANGES". The last one is labeled "APPLES AND ORANGES". You know that each is labeled incorrectly. You may ask me to pick one fruit from one box which you choose. How can you label the boxes correctly? Show Solution View the Solutions Answer: Pick from the one labeled "Apples & Oranges". This box must contain either only apples or only oranges. E.g. if you find an Orange, label the box Orange, then change the Oranges box to Apples, and the Apples box to "Apples & Oranges" 4. The Cannibals Three cannibals and three anthropologists have to cross a river. The boat they have is only big enough for two people. The cannibals will do as requested, even if they are on the other side of the river, with one exception. If at any point in time there are more cannibals on one side of the river than anthropologists, the cannibals will eat them. What plan can the anthropologists use for crossing the river so they don't get eaten? Note: One anthropologist can not control two cannibals on land, nor can one anthropologist on land control two cannibals on the boat if they are all on the same side of the river. This means an anthropologist will not survive being rowed across the river by a cannibal if there is one cannibal on the other side. Show Solution View the Solutions Answer: First, two cannibals go across to the other side of the river, then the rower gets called back. Next, the rowing cannibal takes the second across and then gets called back, so now there are two cannibals on the far side. Two anthropologists go over, then one anthropologist accompanies one cannibal back, so now there is one anthropologist and one cannibal on the far side. The last two anthropologists go over to the far side, so now all the anthropologists are across the other side, along with the boat and one cannibal. In two trips, the cannibal on the far side takes the boat and ferries the other two cannibals across the river. 5. The Father A mother is 21 years older than her child. In exactly 6 years from now, the mother will be exactly 5 times as old as the child. Where's the father? Show Solution View the Solutions Answer: With the mother. If you do the math, you find out the child will be born in 9 months. 6. The Double Jeopardy Doors You are trapped in a room with two doors. One leads to certain death and the other leads to freedom. You don't know which is which. There are two robots guarding the doors. They will let you choose one door but upon doing so you must go through it. You can, however, ask one robot one question. The problem is one robot always tells the truth ,the other always lies and you don't know which is which. What is the question you ask? Show Hint Show Solution View the Solutions Hint: The two robots know each others personality. That they talk when they're bored, lonely, etc. Try to get the two robots to cancel their evil & good ways out. Answer: Ask one robot what the other robot would say, if it was asked which door was safe. Then go through the other door. 7. The Frog A frog is at the bottom of a 30 meter well. Each day he summons enough energy for one 3 meter leap up the well. Exhausted, he then hangs there for the rest of the day. At night, while he is asleep, he slips 2 meters backwards. How many days does it take him to escape from the well? Note: Assume after the first leap that his hind legs are exactly three meters up the well. His hind legs must clear the well for him to escape. Show Hint Show Solution View the Solutions Hint: Try to think the problem through for a five meter well. Now what is the solution for the 30 meter well? Answer: 28 Each day he makes it up another meter, and then on the twenty seventh day he can leap three meters and climb out. 8. The Bobber You can paddle your canoe seven miles per hour through any placid lake. The stream flows at three miles per hour. The moment you start to paddle up stream a fisherman looses one of his bobbers in the water fourteen miles up stream of you. How many hours does it take for you and the bobber to meet? Show Solution View the Solutions Answer: 2 Ignore the speed of the stream, as the cork will be carried along at three miles per hour as will you. It takes two hours to travel fourteen miles, at a rate of seven miles per hour. 10. The Socks Cathy has six pairs of black socks and six pairs of white socks in her drawer. In complete darkness, and without looking, how many socks must she take from the drawer in order to be sure to get a pair that match? Show Solution View the Solutions Answer: 3 Socks do not come in in left and right, so any black will pair with any other black and any white will pair with any other white. If you have three socks and they are either colored black or white, then you will have at least two socks of the same color, giving you one matching pair. 11. There is something about Mary Mary's mum has four children. The first child is called April. The second May. The third June. What is the name of the fourth child? Show Solution View the Solutions Answer: Mary. Mary's mothers fourth child was Mary herself. 12. Petals Around the Rose The name of the game is Petals Around the Rose, and that name is significant. Newcomers to the game can be told that much. They can also be told that every answer is zero or an even number. They can also be told the answer for every throw of the dice that are used in the game. And that's all the information they get. The person who has the dice and knows the game, rolls five dice and remarks almost instantly on the answer. For example: in Roll #1 the answer is two. Roll #1. "The answer is what?" says the new player. "Two." "On that roll?" "Yes." "Would it still be two if I moved the dice without turning any of them over, just rearranging the pattern?" "I can tell you only three things: the name of the game, the fact that the answer is always even, and the answer for any particular throw. In this case the answer is two." "So that's how it is. What am I supposed to do?" "You're supposed to tell me the answer before I tell you. I'll give you all the time you want, but don't tell me your theory, just the answer. If you figure it out, you don't want to give the idea away to these other jokers around you. Make them work for the answers, too. If you get the answer right on six successive rolls, I'll take that as prima facie evidence that you understand the game." "OK, roll again." Roll #2. "I give up. What's the answer?" "The answer is eight." "Roll again." Roll #3. The answer is fourteen. Roll #4. The answer is zero. Roll #5. The answer is four. Roll #6. The answer is... An integral part of the puzzle is that those who have solved it are urged to keep the solution a secret, so there is no solution posted here. It is not a hard puzzle to figure out however. A claim that often accompanies these instructions is that the smarter an individual, the greater amount of difficulty the individual will have in solving it. If such a statement is true, it may be attributed to the fact that "smarter" people tend to be more knowledgeable in a wide range of information which they may unnecessarily attempt to draw upon to solve the puzzle. Four people meet in a room. Each person shakes hands once with each other person. How many hand shakes are there in all? Message Posted: Thu Dec 23, 2010 5:44 am by Rag Post subject: Answer :) THE ANSWER IS 10 Message Posted: Wed Mar 16, 2011 5:49 pm by wesmode12 Post subject: The answer is 6. Message Posted: Fri May 27, 2011 5:57 am by ychick89c Post subject: the answer is 12... ex: there is guy 1,2,3, and 4 guy one shakes hands with: 1-2, 1-3, 1-4 thats 3 hand shakes continue with guy 2,3, and 4. or do 3 times 4. either way you find 12 Message Posted: Sun Jun 05, 2011 8:34 am by aamil4u Post subject: I agree with wesmode. The answer is 6 _________________ "Live as if you will die tomorrow, Learn as if you will live forever.." -Mahatama Gandhi there is the entrance of the place where two guards are there outside,one guard is facing north and another south, the one on the north smiles and the another one asks why ? how ? Message Posted: Thu Jan 06, 2011 1:42 pm by pollpo Post subject: there facing each other, one north and one south. I am very useful. I can be picked up. I can carry water. I have holes in me. I am very common. Who am I? (ok I know this is easy but I just want to share this with you) Message Posted: Sat Mar 22, 2008 2:28 am by Eontios Post subject: Not that easy....a watering hole...a bucket....a toilet? lol _________________ Hi. Message Posted: Sun Mar 23, 2008 2:04 pm by Minnie Post subject: hey You are a SPONGE Message Posted: Tue Dec 16, 2008 4:16 pm by leakybeak Post subject: could be a person Message Posted: Mon Dec 22, 2008 7:58 pm by alexonfyre Post subject: A Watering Can? Message Posted: Sat Nov 07, 2009 3:54 pm by shobainfo Post subject: Re: What am I? sakura wrote: I am very useful. I can be picked up. I can carry water. I have holes in me. I am very common. Who am I? (ok I know this is easy but I just want to share this with you) is this bucket... Message Posted: Sat Dec 05, 2009 12:08 pm by mdbbarilla Post subject: This is a watering can. Message Posted: Thu Dec 23, 2010 5:46 am by Rag Post subject: Answer SPONGE i have a bag, with some balls in it. all but 4 are blue, all but 4 are green, and all but 4 are red. how many balls do i have in total? Message Posted: Mon Nov 23, 2009 5:24 pm by rj82330 Post subject: 2 by 2 by 2 Six _________________ Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking? Message Posted: Sat Sep 25, 2010 4:39 pm by Legolas_Acer Post subject: answer 5. right? Message Posted: Thu Dec 23, 2010 5:45 am by Rag Post subject: answer 12 two fathers gave their two sons some money.one gave his son 150 pesos and other father gave 100 pesos to his son.when the two sons counted their money,they found that all together they had become richer by only 150 pesos.how come? Message Posted: Thu May 20, 2010 6:41 am by rezaru2000 Post subject: They are Father -> Son -> Grand Son A farmer has to sail three, a tiger , a goat and grass across the river and there is only one boat. The boat can carry only two of the above four at a time. How Farmer manages all this. Hint. Neither tiger will consume the goat nor the goat will consume the grass, if farmer is present. Farmer can have more than one trip. Message Posted: Sun Dec 27, 2009 6:53 am by applescript Post subject: Farmer + goat. Farmer returns. Farmer + tiger. Farmer + goat return. Farmer + grass. Farmer returns. Farmer + goat. You are on your way to Salalalah and it is midnight and you are in the middle of the dessert, when your tire goes flat. You remove the four bolts of your tire and keep them on the side of the road. Suddenly, a truck passes by at a very high speed and your bolts just go flying all over and you can't find them. What do you do to replace the tyre? You know there is a Tyre Repair Shop (equipment supplied by TTC ) about 20 KM away. As it is midnight no one will stop to help you. There are no camels around that you can ride and if you are thinking of using a flash light or wait till the sun rises hmmm its not the answer . Message Posted: Sun Aug 02, 2009 2:54 am by porcelina Post subject: Salahlahlah Would you remove one bolt from each of the other tires and then use them to put on your spare? Message Posted: Sun Dec 27, 2009 6:48 am by applescript Post subject: erm I thought tires only need nuts to be fixed onto a car? So yeah..its a trick question? During the lunch hour at school, a group of five boys from Miss Jones home room visited a nearby lunch wagon. one of the five boys took a candy bar without paying for it. When the boys were questioned by the school principal, they made the following statements in respective order: 1. Rex: "Neither Earl nor I did it." 2. Jack: "It was Rex or Abe." 3. Abe: "Both Rex and Jack are lying." 4. Dan: "Abe's statement is not true; one of them is lying and the other is speaking the truth." 5. Earl: "What Dan said is wrong." When Miss Jones was consulted, she said, "Three of these boys are always truthful, but everything that two of them say will be a lie." Assuming that Miss Jones is correct, can you determine who took the candy bar? Message Posted: Thu Dec 13, 2007 2:14 pm by Absentee Post subject: Abe? Message Posted: Mon Dec 22, 2008 8:27 pm by alexonfyre Post subject: Since Dan and Earl contradict each other one of them must be lying. If Earl is lying then Abe is lying and so is one of Rex and Jack, which is not possible (Earl + Abe + one of Jack and Rex = 3 liars). If Earl is telling the truth, then Dan is lying. However, only one part of Dan's statement has to be false for him to be lying, if the first part is false then that means that both Jack and Rex are lying, which is impossible again (Dan + Jack & Rex = 3 liars), which means the first part is definitely true and the second part is surely false. This means that not only is the statement (one of [jack and rex] is lying and the other telling the truth) false, but so is (both jack and rex are lying) leaving only one possibility, that they are both telling the truth. So: Truth-tellers: Rex, Jack, Earl Lairs: Dan & Abe So Rex says that neither Earl nor he did it, which we know is true Jack says it was Rex or Abe, but because what Rex said was true, it cannot be Rex so Abe is the culprit. =) In short, Absentee is correct you are an oil mogul considering the purchase of drilling rights to an as yet unexplored tract of land. the well's expected value to its current owners is uniformly distributed over [$1..$100]. (i.e., a 1% chance it's worth each value b/w $1..$100, inclusive). bcause you have greater economies of scale than the current owners, the well will actually be worth 50% more to you than to them (but they don't know this). the catch: although you must bid on the well before drilling starts (and hence, before the actual yield of the well is known), the current owner can wait until *after* the well's actual value is ascertained before accepting your bid or not. what should you bid? Message Posted: Sat Mar 22, 2008 2:29 am by Eontios Post subject: You should bid a cent more than the others....i think XD _________________ Hi. Message Posted: Mon Dec 22, 2008 8:03 pm by alexonfyre Post subject: I would buy a tenth of the land for 15 bucks (enough to put my wells on and roads or a pipeline to get out), and then get all of the oil on the whole tract through drainage. There was this clown. He saw the dust and die. Why did he die? Like I mean elaborate. Hints: He is a migit His size does not matter. Message Posted: Mon Jun 30, 2008 11:45 am by skintc Post subject: He could see a dust storm coming, but cos he's a clown everyone thought he was joking. So they all died. Message Posted: Thu Jul 03, 2008 1:05 am by smartgirl23 Post subject: Death Was the dust sawdust? If so, did he die because he killed him self as someone was cutting sawdust from his cane or such and he died because he thought that he was growing taller? :| _________________ Intelligence is the key to excellence. Message Posted: Mon Dec 22, 2008 7:55 pm by alexonfyre Post subject: Even if it isn't correct (though I think it should be) I like skintc's answer the best =) Anyway, here is my ridiculous answer (though this riddle is lateral thinking, not logic) Midget clowns are perfect for shooting out of cannons, so he was the cannon clown, however one day the cannon missed its target (the safety net) and he saw the dust on the circus tent floor (which is generally a dirt floor or covered in sawdust for a variety of reasons) before he landed and died. 1. You are driving down the road in your car on a wild, stormy night, when you pass by a bus stop and you see three people waiting for the bus An old lady who looks as if she is about to die. An old friend who once saved your life. The perfect partner you have been dreaming about. Knowing that there can only be one passenger in your car, whom would you choose? Show Hint Show Solution Hint: You can make everyone happy. Your car can only contain one passenger, so whom should it be? Answer: The old lady of course! After helping the old lady into the car, you can give your keys to your friend, and wait with your perfect partner for the bus 2. Acting on an anonymous phone call, the police raid a house to arrest a suspected murderer. They don't know what he looks like but they know his name is John and that he is inside the house. The police bust in on a carpenter, a lorry driver, a mechanic and a fireman all playing poker. Without hesitation or communication of any kind, they immediately arrest the fireman. How do they know they've got their man? Show Hint Show Solution Hint: The police only know two things, that the criminal's name is John and that he is in a particular house. Answer: The fireman is the only man in the room. The rest of the poker players are women. 3. A man lives in the penthouse of an apartment building. Every morning he takes the elevator down to the lobby and leaves the building. Upon his return, however, he can only travel halfway up in the lift and has to walk the rest of the way - unless it's raining. What is the explanation for this? Show Hint Show Solution Hint: He is very proud, so refuses to ever ask for help. Answer: The man is a dwarf. He can't reach the upper elevator buttons, but he can ask people to push them for him. He can also push them with his umbrella. 4. How could a baby fall out of a twenty-story building onto the ground and live? Show Hint Show Solution Hint: It does not matter what the baby lands on, and it has nothing to do with luck. Answer: The baby fell out of a ground floor window. 5. Bad Boy Bubby was warned by his mother never to open the cellar door or he would see things that he was not meant to see. One day while his mother was out he did open the cellar door. What did he see? Show Hint Show Solution Hint: His mother was an odd woman. Answer: When Bad Boy Bubby opened the cellar door he saw the living room and, through its windows, the garden. He had never seen these before because his mother had kept him all his life in the cellar. 6. A man and his son are in a car crash. The father is killed and the child is taken to hospital gravely injured. When he gets there, the surgeon says, 'I can't operate on this boy - for he is my son!!!' How can this possibly be? Show Hint Show Solution Hint: This has nothing to do with adoption or time travel. Answer: The surgeon can not operate on her own son; she is his mother. 7. There are six eggs in the basket. Six people each take one of the eggs. How can it be that one egg is left in the basket? Show Hint Show Solution Hint: A alternate version of the problem is... Saradhi, Nick & Ted win a raffle contest. The prize is three hard boiled eggs in a basket. After discussing how to divide the prize, each take one egg. Nick & Ted get hungry and so eat their eggs. One of the original eggs is still left in original basket. Answer: The last person took the basket with the last egg still inside. brown,jones & smith are suspected of income tax evasion.They testify,under oath as follows Brown: jone is guilty & smith is innocent Jones: if Brown is guilty,then so is Smith Smith: I am innocent but at least one of the other's is guilty. a)assuming every body told the truth who is or are innocent or guilty? b)assuming the innocent told the truth & guilty lied who is or are innocent or guilty? Message Posted: Thu May 19, 2011 4:28 pm by Verglas Post subject: a) Jones is guilty and the other 2 are innocent b) Brown and Smith are guilty and Jones is innocent I think all sane people are cachers and one third of all cachers are sane but half of all whackos are cachers with only one whacko that s sane If eight whackos are cachers and ninety are attending my ball how many cachers are neither sane nor whacko at all? Message Posted: Fri Mar 18, 2011 11:47 pm by ChibiHoshi Post subject: 48 90 total - 8 wackos = 82 cachers 82- 27 sane (third of 82) = 55 1 of the 8 cacher-wackos is one of these sane folk so - 7 cacher-wackos = 48 plain cachers Message Posted: Tue May 17, 2011 8:18 am by Dalton Mcmickle Post subject: hiiiii............... thanks dear to share knowledge about role of forums... :lol: __________________________ ps3 wireless controller, wenzel tents, tents, tent, ps3 games Achmed and Ali are camel-drivers and on one day they decided to quit their job. They wanted to become shepherds. So they went to the market and sold all their camels. The amount of money(dinars) they received for each camel is the same as the total of camels they owned. For that money they bought as many sheep as possible at 10 dinars a sheep. For the money that was left they bought a goat. On their way home they got in a fight and decided to split up. When they divided the sheep there was one sheep left. So Ali said to Achmed "I take the last sheep and you can get the goat". "That's not fair" said Achmed, "a goat costs lesser than a sheep". "Ok", Ali said "then I will give you one of my dogs and then we are even". And Achmed agreed. What costs a dog? Message Posted: Tue Jan 08, 2008 5:53 am by aatifahmed Post subject: Solution to what costs a dog?! The dog costs 4 dinar. The solution I am submitting below is encrypted so as not to take the fun out of your efforts! To decrypt use: www.rot-n.com The cipher used is ROT18-ROT5 & ROT13. Abgr gung Nyv naq Npuzrq unq rdhny ahzore bs pnzryf (yrgf pnyy a). Urapr; Gbgny ahzore bs pnzryf = 7a. Gbgny zbarl gurl erprvir sbe pnzryf = 9a^7 Gur furrcf pbhyq abg or qvfgevohgrq rirayl nzbat gurz, urapr vgf na bqq ahzore (yrgf fnl 7z+6). Naq yrgf fnl gung pbfg bs bar tbng vf t naq bar qbt vf q. Gurersber, 65*(7z+6)+p = 9a^7 Abgr p unf gb or rira nf gur EUF vf rira. Nyfb abgr gung p vf gur havg qvtvg bs YUF, jurernf gur EUF vf n fdhner, yrnivat p gb or bayl 9 be ; (fvapr 7 naq = pna arire or havg qvtvg bs n fdhner ahzore, naq 5 vf abg cbffvoyr). Abj, :z + (65+p)/9 = a^7 Guhf p unf gb qvivfvoyr ol 9, juvpu yrnir p gb or bayl ;. Fvapr p vf ; naq gur qvfgevohgvba vf rdhny fb q unf gb or 9!! Message Posted: Mon Dec 22, 2008 9:48 pm by alexonfyre Post subject: I don't think it ever said they have equal number of camels aatifahmed...though it works if they do...I am pretty sure it is still true if they aren't equal numbers, however. Message Posted: Sat Mar 19, 2011 12:55 am by ChibiHoshi Post subject: 4 interestingly enough, if you look at a table of perfect squares, those with the tens place as an odd number (the only way there could be odd sheep) are always followed by a 6 (cost of a goat) 16,36,196,256,576,676,1156,1296,1936, etc. So sheep-goat = dog 10-6 is 4 Two friends, whom we will call Arthur and Robert, were curators at the Museum of American History. Both were born in the month of May, one in 1932 and the other a year later. Each was in charge of a beautiful antique clock. Both of the clocks worked pretty well, considering their ages, but one of them lost ten seconds an hour and the other gained ten seconds an hour. On one bright day in January, the two friends set both clocks right at exactly 12 noon. ``You realize,'' said Arthur, ``that the clocks will start drifting apart, and they won't be together again until---let's see---why, on the very day you will be 47 years old. Am I right?'' Robert then made a short calculation. ``That's right!'' he said. Who is older, Arthur or Robert? Message Posted: Sat Mar 19, 2011 12:29 am by ChibiHoshi Post subject: Robert The clock would have been set in either 1979 or 1980 for one of them to turn 47. (1932 or 1933 birth year) The clocks will resync in 90 days. (Simple conversion calculation of 10s/hr would take 90 days to lose/gain 6 hours which is when both would resync) The Jan-May gap could only work if it was set 31st January and a non-leap year (1980 was a leap year and therefore the clock would have synced on April 30th) So Robert, the one reaching 47 on May 1st of 1979 was the one born in 1932 and older A person crosses a river on a paddle boat. He took 4 hours while paddling against the water flow and 3 hours while paddling in the direction of the water flow. On still waters, how long would he take? Message Posted: Fri Mar 18, 2011 5:15 pm by ChibiHoshi Post subject: Assuming the distances are all equal and they cancel each other out Rower+River=3 hours Rower-River=4 hours Basic algebra 2xRower=7 Rower alone takes 3.5 hours The Island of Baal Of all the islands of knights and knaves, the island of Baal is the weirdest and most remarkable. This island is inhabited exclusively by humans and monkeys. The monkey, as well as every human, is either a kngith or a knave. In the dead center of this islands stands the Temple of Baal, one of the most remarkable temples in the entire universe. The high priests are metaphysicians, and in the Inner Sanctum of the temple can be found a priest who is rumored to know the answer to the ultimate mystery of the universe; why there is something instead of nothing. Aspirants to the Sacred Knowledge are allowed to visit the Inner Sanctum, provided that they prove themselves worthy by passing three series of tests. I learned all these secrets, incidentally, by stealth: I had to enter the temple disguised as a monkey. I did this at great personal risk. Had I been caught, the penalty would have been unimaginable. Instead of merely annihilating me, the priests would have changed the very laws of the universe in such a way that I could never have been born! Well, a philosopher who was searching for the answer to the question, Wy is there something rather than nothing?" arrived on the island of Baal and agreed to try the tests. The first series took place on three consecutive days in a huge room called the Outer Sanctum. In the center of the rrom a cowled figure was seated on a golden throne. He was either a human or a monkey, and also a knight or a knave. He uttered a sacred sentence, and from this sentence the philosopher had to deduce exactly what he was whether a knight or a knave, and whether a human or a monkey. The First Test The speaker said, "I am either a knave or a monkey." Exactly what is he? The Second Test The speaker said, "I am a knave and a monkey." Exactly what is he? The Thrird Test The speaker said, "I am not both a monkey and a knight." What is he? Message Posted: Mon Nov 23, 2009 5:31 pm by rj82330 Post subject: First Test: Can never be false - or else it becomes contradictory. So it has to be true, so the speaker cannot be a knave. So FIRST Subject = MONKEY KNIGHT Second Test: Can never be true, since the speaker would then be a knave and a liar. So it must be false, in which can the speaker must be a knave, but not a monkey So SECOND Subject = HUMAN KNAVE Third Test: Can never be false, since the speaker would become a lying knight. So it must be true. In this case, the speaker must be a knight, so cannot be a monkey. So THIRD Subject = HUMAN KNIGHT _________________ Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking? Message Posted: Mon Jan 03, 2011 11:23 am by s.b. Post subject: it would have reeli helped if ud just mentioned that knaves always lie and knights are always truthful...maybe its obvious to u but to amateurs lyk me it would really help.....thanks! :) This question is from my geometry class, it only involve logic not probability. please help me solve this! it only middle school so i don't think it would be TOO hard anyways here it is. reply as soon as possible please --------------------------------------------------------------------------------------- Three geniuses stand in a file (one behind the other). Each can see only to the front, so the rear person can see the middle and the front, the middle person can see the front, and the genius in the front cannot see anyone. You have five hats. Two are white, and three are red. You blindfold the three geniuses, who are utterly truthful, and put a hat--at random--on the head of each. Then you hide the other two hats and remove the blindfolds. You then ask each genius to name the color of his hat (which he cannot see). The rear one says "I don't know." The middle one says, "I don't know." Then the front one says, "I know." WHAT COLOR HAT IS THE FRONT GENIUS WEARING?????? HOW DOES HE KNOW WITH 100% CERTAINTY WHAT COLOR HAT IT IS????? Message Posted: Fri Feb 08, 2008 11:50 pm by toothslooth Post subject: A bit of logic works wonders! The front genius is wearing a red hat. Here's how he knows... The front 2 were NOT both wearing white or the back 1 would know he was red. So the front 2 must either be wearing both red or 1 red 1 white. If the front was wearing white the middle 1 MUST be wearing red. As he does not know, the front 1 MUST THEREFORE be wearing RED. Enjoy! Message Posted: Tue Aug 11, 2009 12:42 pm by mdbbarilla Post subject: list the possibilities. then cancel out. man 1: red / white man 2: red / white man 3: red / white if man 1 : red then man 2 : red man 3 will not be able to know his hat because there is more than one possibility then man 2: white man 3 will still not know but if man 1: white man 2 : white man 3 will know that his hat is red. if man 2 was red man3 will not know his hat a man needs to go through a train tunnel. he starts through the tunnel and when he gets 1/4 the way through the tunnel, he hears the train whistle behind him. you don't know how far away the train is, or how fast it is going, (or how fast he is going). all you know is that if the man turns around and runs back the way he came, he will just barely make it out of the tunnel alive before the train hits him. if the man keeps running through the tunnel, he will also just barely make it out of the tunnel alive before the train hits him. assume the man runs the same speed whether he goes back to the start or continues on through the tunnel. also assume that he accelerates to his top speed instantaneously. assume the train misses him by an infintisimal amount and all those other reasonable assumptions that go along with puzzles like this so that some wanker doesn't say the problem isn't well defined. how fast is the train going compared to the man? Message Posted: Tue Jan 08, 2008 7:28 am by aatifahmed Post subject: Solution to A man needs to go through a train tunnel! The train is moving at 2 times the speed of man. Use www.rot-13.com to decipher the solution if need be. Yrg gur genva or ng n qvfgnapr f sebz gur ghaary. Yrg gur ghaary or bs yratgu q naq gur fcrrq bs zna i, juvyr bs gur genva ki. Gurersber: Gvzr gnxra ol gur zna naq gur znpuvar vf fnzr ng gur ragenapr => (f/ki)=(q/4i)=((f+q)/(ki+4i)) -- -- (v) (hfvat onfvp cebcregvrf bs engvb cebcbegvbaf). Gvzr gnxra ol gur zna naq gur znpuvar vf fnzr ng gur rkvg => ((f+q)/ki)=(3q/4i) ---- (vv) hfr (v) & (vv): (ki+4i)*q/4i=3q*ki/4i fbyir gb trg k=2. Message Posted: Tue Nov 04, 2008 11:03 pm by Axomomma Post subject: Sorry, but the correct answer is: The train is moving at 1.5 times the speed of the man. Assume that the tunnel is 400 yards long. The man hears the train when he is 1/4 of the way inside, or 100 yards in. If the man travels back 100 yards the train will just miss him. If he goes forward 300 yards the train will just miss him. The train was 150 yards away from the entrance to the tunnel when the man first heard it. If the man goes forward 300 yards the train will travel 450 yards. 300 x 1.5 = 450 If the man goes back 100 yards the train will travel 150 yards. 100 x 1.5 = 150 _________________ I'm not telling you because "I think", I'm telling you because "I know" Message Posted: Mon Dec 22, 2008 9:00 pm by alexonfyre Post subject: Actually the answer is that the train is moving twice the speed of the man, this forum doesn't have spoiler tags, so I will go ahead and post my work, hope that is okay. First to Axomomma, for your hypothetical to be true, then the train would have to travel 550 yards while the man traveled 300 (since it has to go the distance to the tunnel, then all the way through it, 400 + 150 = 550) and then your two ratios wouldn't match, so that cannot be the answer. It was a decent guess, it took me a while to realise the error, but I think the solution to this problem is a system of equations). There are 4 unknowns here: X (the length of the Tunnel), Y(the distance between the train and the tunnel), M (the speed of the man) and T (the speed of the train). The ultimate goal here is to eliminate x and y and leave a relationship between m and t. We have these two conditions given to us: The time it takes the man to move 1/4 of x is equal to the time it takes the train to go all of y which gives us: x/(4m) = y/t also, the time it takes the man to go 3/4 x is equal to the time it takes the train to go all of y and x or: 3x/(4m) = (y+x)/t For simplicity lets call the time it takes for the man to run through all of x, z or: z = x/m so z/4 = y/t and 3z/4 = (x+y)/t now we can solve for z in terms of y and t: z = 4y/t make another substitution and we will eliminate t (don't worry it will be back): (4y/t)*3/4 = (x+y)/t 3y/t=(x+y)/t 3y = x+y x = 2y (right here you know that the tunnel is twice the length of the distance between the train and tunnel, which is enough to solve it, but let's finish the elegant, mathematical way) now with x in terms of y we can eliminate them both from our initial equation: x/(4m) = y/t becomes 2y/4m = y/t y/2m = y/t (cross multiply) t = 2m or the speed of the train is twice the speed of the man. I couldn't get aatifahmed's decipher link to get his solution, it is broken now, so I don't know how similar his method was to mine, but I do know that his answer is correct. A famous restaurant starts a special marriage anniversary promo to the couples on Thursdays who carry their proof of marriage...50% discount on meals only on Thursdays. A couple enters the restaurant on Thursday, tells the manager that they got married on a fine Sunday 28years back hearing this the manager chased away the couple without being served. Why? Message Posted: Fri Sep 19, 2008 11:36 am by skintc Post subject: Cos he worked out that if they were married on a Sunday 28 years ago, that their anniversary could not be on a Thrusday that year. Message Posted: Fri Oct 03, 2008 12:36 pm by rj82330 Post subject: Further Isn't it the case that, every 28 years, dates will fall on the same day? It would be every 7 years, but the leap years throw it out. This is regulated every four cycles of 7 years. So, what happened on a Sunday 28 years ago, should be a Sunday this year. (The exceptions are if the 28 year period falls over 1700, 1800 or 1900, which are not divisible by 400 and therefore not leap years) _________________ Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking? You are standing outside a basement door, Down the stairs, there are 3 light bulbs. Next to you, are 3 switches. You cannot see the lights from upstairs, and you must decide which switch controls which lightbulb by going downstairs once. How do you do this? Message Posted: Mon Jun 30, 2008 11:39 am by skintc Post subject: Turn on first switch, leave it on for a few minutes. Then turn it off and flip second switch. When you go into the basement, the first switch is the bulb that is off, but hot. The second switch is the bulb that's on. The thrid switch is the bulb that's cold and off. you have two identical crystal orbs. you need to figure out how high an orb can fall from a 100 story building before it breaks. you know nothing about the toughness of the orbs: they may be very fragile and break when dropped from the first floor, or they may be so tough that dropping them from the 100th floor doesn't even harm them. what is the largest number of orb-drops you would ever have to do in order to find the right floor? (i.e. what's the most efficient way you could drop the orbs to find your answer?) you are allowed to break both orbs, provided that in doing so you uniquely identify the correct floor. Message Posted: Tue Dec 18, 2007 3:24 am by Rick Post subject: It may be less but I think I could do it in at most 14 drops. Message Posted: Tue Jan 08, 2008 5:02 am by aatifahmed Post subject: Solution to Falling Orbs! The answer is 7 drops. The solution I am submitting below is encrypted so as not to take the fun out of your efforts! To decrypt use: www.rot-n.com The cipher used is ROT18-ROT5 & ROT13. Guvax bs vg nf n ovanel ceboyrz, naq pbaireg vg vagb ovanel. Fgneg sebz bayl 65 fgbel uvtu ohvyqvat. Guhf lbh unir 6655 sbe 65. Rnpu cbfvgvba bs gur qvtvgf va 6655 qrfpevorf gur fgbel sebz juvpu gur beo vf gb or qebccrq, va gur qbjajneq snfuvba. Fb sbe rknzcyr bs 65, vg fubhyq or qebccrq sebz fgbel ahzore =, 9, 7 naq 6 erfcrpgviryl. Abgr - O zrnaf Oernx, AO zrnaf Ab Oernx. Guhf sbe 65: Q=-AO-Q65-AO-65 vf gur znk sybbe Q=-AO-Q65-O-Q>-AO-> vf gur znk Q=-AO-Q65-O-Q>-O-= vf gur znk sybbe Q=-O-Q9-AO-Q;-AO-Q<-O-; vf gur znk sybbe Q=-O-Q9-AO-Q;-AO-Q<-AO-< vf gur znk sybbe Q=-O-Q9-AO-Q;-O-Q:-AO-: vf gur znk sybbe Q=-O-Q9-AO-Q;-O-Q:-O-9 vf gur znk sybbe Q=-O-Q9-O-Q7-AO-Q8-AO-8 vf gur znk sybbe Q=-O-Q9-O-Q7-AO-Q8-O-7 vf gur znk sybbe Q=-O-Q9-O-Q7-O-Q6-AO-6 vf gur znk sybbe Q=-O-Q9-O-Q7-O-Q6-O-5 vf gur znk sybbe Nf h pna frr sebz nobir, abar bs gurz gnxr zber guna 9 qebcf! Fvzvyneyl vg pna or jbexrq bhg sbe 655, juvpu va ovanel vf 6655655, univat < qvtvgf naq urapr < qebcf!! Message Posted: Tue Jan 08, 2008 6:52 am by aatifahmed Post subject: Correction My solution holds true if infinite number of Orbs are available, for minimum number of drops. The correct answer to the problem is 14. Message Posted: Mon Dec 22, 2008 10:24 pm by alexonfyre Post subject: yeah, radically different math for the two solutions as well ;) Message Posted: Mon Jun 28, 2010 3:26 pm by shrek619 Post subject: i think the anwser is 7 (minimum no.). Message Posted: Thu Jun 02, 2011 5:26 pm by Unni Post subject: Correct answer Correct answer is 10 Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which. " It could be that some god gets asked more than one question (and hence that some god is not asked any question at all). What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.) Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely. Random will answer 'da' or 'ja' when asked any yes-no question.[2] Message Posted: Mon Dec 22, 2008 10:30 pm by alexonfyre Post subject: how would the truth telling god answer if one were to ask him "Would he tell me that he is the False God?" referring to the Random God? Message Posted: Sat Dec 18, 2010 2:30 pm by bugbustter Post subject: Three Gods, A,B and C called Truth, False and Random The Truth God would keep silent, because he would not be able to give a truthfull answer. _________________ bugbustter Message Posted: Tue May 17, 2011 4:08 pm by randompattern Post subject: I imagine that the false good also would remain silent if you asked what the random god would say bec the false god must be sure that his answer is false. With that in mind, I would start with A and ask: "What would B say if I asked him if the sky were blue?" (or any definite truth) If A will remains quiet: I know B is Random. I then ask A what C would say if I asked him if the sky were blue. Whatever the reply, that is the word for "no", because if A is True, then C is False, so A will truthfully tell me that C will lie and if A is False and C is true, then A will lie about C's truthful response. Now that I know which word means "no", I simply ask either A or C if the sky is blue. Let's Say I ask C: If the Answer is the word for "no", C is False and A is True. If it is "yes", C is True and A is False. If after my first question, A gives me an answer: I know that B is not Random. I ask B what C would say if I asked if the sky were blue. If B gives an answer, then I know A must be Random, and the Answer that B gave means "No". I then ask B or C if the sky is Blue to determine which one is True and which one is False. If A answers, but B remains silent: I know that C is Random, and that what A answered must have meant "No". I then ask A or B if the sky is blue to determine which one is True and which one is False. Say you have one string of alphabetic characters, and say you have another, guaranteed smaller string of alphabetic characters. Algorithmically speaking, what's the fastest way to find out if all the characters in the smaller string are in the larger string? For example, if the two strings were: String 1: ABCDEFGHLMNOPQRS String 2: DCGSRQPOM You'd get true as every character in string2 is in string1. If the two strings were: String 1: ABCDEFGHLMNOPQRS String 2: DCGSRQPOZ you'd get false as Z isn't in the first string. The naive way to do this operation would be to iterate over the 2nd string once for each character in the 1st string. That'd be O(n*m) in algorithm parlance where n is the length of string1 and m is the length of string2. Given the strings in our above example, thats 16*8 = 128 operations in the worst case. A slightly better way would be to sort each string and then do a stepwise iteration of both sorted strings simultaneously. Sorting both strings would be (in the general case) O(m log m) + O(n log n) and the linear scan after that is O(m+n). Again for our strings above, that would be 16*4 + 8*3 = 88 plus a linear scan of both strings at a cost of 16 + 8 = 24. Thats 88 + 24 = 112 total operations. Slightly better. (As the size of the strings grow, this method would start to look better and better) Finally, the best method would simply be O(n+m). That is, iterate through the first string and put each character in a hashtable (cost of O(n) or 16). Then iterate the 2nd string and query the hashtable for each character you find. If its not found, you don't have a match. That would cost 8 operations - so both operations together is a total of 24 operations. Not bad and way better than the other solutions. Is there another solution? What if - given that we have a limited range of possible characters - I assigned each character of the alphabet to a prime number starting with 2 and going up from there. So A would be 2, and B would be 3, and C would be 5, etc. And then I went through the first string and 'multiplied' each character's prime number together. You'd end up with some big number right? And then - what if I iterated through the 2nd string and 'divided' by every character in there. If any division gave a remainder - you knew you didn't have a match. If there was no remainders through the whole process, you knew you had a subset. Would that work? Message Posted: Mon Jan 10, 2011 1:30 am by harryaskham Post subject: The primes solution fails if the second string contains, say, 2 Ds when the original string has only 1 - you would try and divide by the corresponding prime twice when it exists only once as a prime factor, giving a remainder. For what it's worth, I think a hashing solution is overkill, given the space of only 26 characters - collisions would be very unlikely. Instead, a similar solution would be to simply keep a size-26 array of booleans - iterate over the first string, and set booleans[letterCode] to true for each. Then, you can simply iterate over the second string, returning false if booleans[letterCode] is false for any character, and true after the iteration has completed without returning. So basically a hash table, but since the key is the value, it itself can be used as the index (I tried both in Java and although both were fast, the array-backed solution used about 1/3 the time). Out of 52 cards 13 are up and 39 are down and are mixed . How to arrange them in two decks of equal numbers of up. Condition You are sitting in a dark room and there is no light. Message Posted: Mon Jun 28, 2010 3:17 pm by shrek619 Post subject: can you give a small hint ? Message Posted: Tue Sep 21, 2010 8:55 am by lonewolf Post subject: Saperate the piles into 2 equal piles of 26 cards each. Flip all the card of on pile. I think this question is simmilar to the 20 coin question. Message Posted: Wed Dec 29, 2010 11:33 am by moshu123 Post subject: Correct solution The correct solution is like this: Saperate the piles into 2 piles of 13 & 39 cards each. Flip all the card of smaller pile. Say you have one string of alphabetic characters, and say you have another, guaranteed smaller string of alphabetic characters. Algorithmically speaking, what's the fastest way to find out if all the characters in the smaller string are in the larger string? For example, if the two strings were: String 1: ABCDEFGHLMNOPQRS String 2: DCGSRQPOM You'd get true as every character in string2 is in string1. If the two strings were: String 1: ABCDEFGHLMNOPQRS String 2: DCGSRQPOZ you'd get false as Z isn't in the first string. The naive way to do this operation would be to iterate over the 2nd string once for each character in the 1st string. That'd be O(n*m) in algorithm parlance where n is the length of string1 and m is the length of string2. Given the strings in our above example, thats 16*8 = 128 operations in the worst case. A slightly better way would be to sort each string and then do a stepwise iteration of both sorted strings simultaneously. Sorting both strings would be (in the general case) O(m log m) + O(n log n) and the linear scan after that is O(m+n). Again for our strings above, that would be 16*4 + 8*3 = 88 plus a linear scan of both strings at a cost of 16 + 8 = 24. Thats 88 + 24 = 112 total operations. Slightly better. (As the size of the strings grow, this method would start to look better and better) Finally, the best method would simply be O(n+m). That is, iterate through the first string and put each character in a hashtable (cost of O(n) or 16). Then iterate the 2nd string and query the hashtable for each character you find. If its not found, you don't have a match. That would cost 8 operations - so both operations together is a total of 24 operations. Not bad and way better than the other solutions. Is there another solution? What if - given that we have a limited range of possible characters - I assigned each character of the alphabet to a prime number starting with 2 and going up from there. So A would be 2, and B would be 3, and C would be 5, etc. And then I went through the first string and 'multiplied' each character's prime number together. You'd end up with some big number right? And then - what if I iterated through the 2nd string and 'divided' by every character in there. If any division gave a remainder - you knew you didn't have a match. If there was no remainders through the whole process, you knew you had a subset. Would that work? A wily young man wishes to marry the princess Arlena. Unfortunately, her father, the Sultan, is opposed to the marriage and is willing to buy the young man off. The Sultan, therefore makes the young man the following offer--"You get to make one statement. If the statement is false, you will be put to death and receive none of the following three things. If the statement is true, you may have one of the following three things, but I get to choose! The three things are: (1) Arlena's hand in marriage; (2) A cup filled with extremely valuable diamonds; (3) A magic lamp with a genie (who, unfortunately, cannot do marriages)." The wily young man, however, is too wily for his future father-in-law. He makes a statement such that the only way the Sultan can keep his promise is by giving the young man Arlena's hand in marriage. What was the statement? Message Posted: Thu Nov 27, 2008 8:38 am by rj82330 Post subject: A variant Good one, this - only heard it will TWO options in the past. He says: "You will not give me the diamonds or the genie". _________________ Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking? Message Posted: Mon Dec 22, 2008 10:08 pm by alexonfyre Post subject: If that were correct RJ couldn't the Sultan just say "False, I chose the diamonds for you." and death. I think the actual answer is: "If you do not let me marry your daughter, I will be put to death" Meaning that if the king chooses diamond or genie and lets him live, he will be telling a lie and the king will have to put him to death, which would in turn create a paradox as he would be telling the truth at that point. The only resolution to the paradox is to allow the prince to marry his daughter. If the Sultan chooses to put him to death later, that would be another story, but in the mean time the Sultan would have to give up his daughter's hand in marriage. Message Posted: Tue Dec 23, 2008 11:39 am by rj82330 Post subject: My thinking was that with the three variable given, there is only one way to prevent creating either a paradox, or a contradiction in the question. I had the young man say "you will not give me certain presents" not "you would not have given me certain presents". (You have the sultan saying "I chose"). In this case, if the sultan wants to say "False, I will give you the diamonds", he has to actually fork over a present to prove the young man's statement false: but the scenario above doesn't allow for the young man to be given a present in the event of his making a false statement and ultimately being executed. - Contradiction If the sultan acknowledges that the young man's statement is true - because he is going to be executed rather than receiving diamonds or genie, then our hero will receive the punishment for making a false statement - Paradox. Surely, the only way to make sure that the integrity of the puzzle remains, and to avoid a paradox, is to make the statement true, for the one remaining true variable - i.e. giving away his daughter. ___ This is similar to the Pirate King of the Logical Positivists, who decrees that a certain empiricist is to be executed via the yardarm or the plank, depending on whether a statement he makes is true (he'll hang) or false (he'll drown). In that puzzle, the empiricist has to render the entire problem logically impossible (by saying "I will drown"). In this conundrum, he has to render most of the problem paradoxical, yet still leave one option within the puzzle open to him. _________________ Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking? Message Posted: Mon Mar 09, 2009 5:17 pm by Jeffkins Post subject: Similarly could he not say "You will give me Arlena's hand in marriage or you will put me to death." Essentially the reverse of rj82330's answer. Message Posted: Tue Nov 30, 2010 9:13 pm by Trenin Post subject: Optimal Solution How about this: "You will give me none of those three things, or you will give me your throne and I will forfeit my right to ask for a prize." To analyze this, it is basically two clauses - A or B - where: A is "You will give me none of those three things" B is "You will give me your throne and I will forfeit my right to ask for a prize" By making any one of them true, the entire statement is true. Lets suppose the sultan decides this statement is false. Then he gives the young man nothing and kills him. This makes clause A true, thus the entire statement is true, which is a contradiction. Thus the statement must be true. The first clause A can't be the only one that is true because then the sultan must give the young man a prize of the sultan's choosing if the young man wishes it (which he would) which makes clause A false - another contradiction. Thus the second clause B must be true - the sultan is forced to give up his throne. Some of you may say this is still a contradiction because if the statement is true, then the sultan must give the young man a prize. However, the wording of the puzzle is: If the statement is true, you may have one of the following three things, but I get to choose! Thus, it is up to the young man if he wants to get a prize, but it is up to the sultan which prize. Thus, the young man can refuse the prize and not have a contradiction. He does this explicitly in his statement, thus making the way clear for what the sultan must do. When the young man takes the throne, he then takes not only the princess' hand in marriage, but everything else in the power of the sultan as well. The leader of a group of dwarfs wants to test his group. So at night he paints on the back of each dwarf a dot, either red of blue. The next morning he let's the dwarfs gather at a meeting point and tell's them the rules of the game: You may never know what the color of your own dot is. (This is accomplished by saying they may not communicate in any way or use mirrors and such) He will give them a few minutes to discuss a tactic to solve the puzzle. The goal is that the number of dwarfs with a red dot, assume this number is X, on their back gather at the same meetingpoint the number of days, X, after the start of the game. for example if there are 10 red dots, after 10 days they must gather Any other entry upon the meetingpoint of any dwarf results in failure. Now, how must every dwarf think to solve the puzzle? Message Posted: Thu May 20, 2010 2:25 pm by sinb Post subject: well, the group selects a single person as a leader. The leader can see, let's say Y dots. either X = Y (if his own dot is green) or X = Y+1 (if his own dot is red) The plan is that the group meets at a place every day at a pre-decided point, but the leader is supposed to show up only on (Y-1)'th day. So everyone except for the leader know X, that is they know Y and depending on the leader's dot they can decide on X. When they decide on X (the total number of red dots), then they can decide whether their own dot is red or blue. Now on X'th day, every red dot dwarf meets at the same place if the leader's dot was red, otherwise they all meet at some other pre-decided place to which leader wont come. I dont know if the guy is allowed to indicate X value to others by going to the meetingplace, (it would be indirectly telling everyone which dots they have), but this is the solution I came up with. Let me know if you have some other solution. Message Posted: Mon Jun 28, 2010 2:58 pm by shrek619 Post subject: solution I assume there are at least 2 red dots. Consider one dwarf with either red or blue dot, let X be total no. of dots out there. If he is red dotted he sees X-1 dots let it be Y, if he is blue dotted he sees X dots let it be Y. Y is no. of red dots as he sees on other dwarfs. All he need to do is go on to the location on Y+1 day. Each one of them should follow this rule. If he is red dotted he goes to the location on correct day, if he is blue dotted the game is already over on Y th day so there is no need for going to the meeting place. am i right ? is there any other solution Message Posted: Mon Nov 29, 2010 9:12 pm by sudheerkuttiyat Post subject: Think Red dot Every dwarf must think he has a red dot on his back. Then every one with red dots in the back will meet X days from now, where X-1 is the number of red dots a dwarf with a red dot behind him can see. Anyone with blue dots on the back will see X red dots and will come to meet his mates X+1 days from now. Supposedly a true story: A 5'6" man was found hanged (suicide) in an empty cellar room in a huge mansion in England. He was a prominent man of good stature (lawyer I believe). It was found he was bi/sexual and couldn't handle the social pressure and mostly his devastated wife, so he decided to go out in style and baffle everyone. He did for many months. The situation was this: He was found hanging from a sprinkler system pipe in the middle of the room just 1' from the ceiling from a 4' piece of one stranded rope. The room (cellar) was 25' square (wall to wall, ceiling to floor). There were no windows, the room was totally bare, no furnishings whatsoever, no shelves, nothing. The walls and ceiling were smooth, the floor was smooth with the exception of the typical drain hole and it had only one entrance which was just a normal sized door and the door was locked from the inside with two dead bolts and a hinged padlock. It had only one light bulb in the center of the ceiling, light switch was outside the door. He was found about a month later decomposing. How did he do it? _________________ Just a human being having a spiritual experience! Message Posted: Fri Jul 17, 2009 12:30 pm by rackner Post subject: Could he not simply have stood on a block of ice to hang himself? The ice would then have melted and gone through the drain hole. Message Posted: Sun Jul 26, 2009 12:54 am by Callyst Post subject: maybe he blocked up the drain with either ice or the rope, and then took the rope and tied himself to the pipe. _________________ "whoever said the pen was mightier than the sword has obviously never encountered automatic weapons" -Douglas MacArthur- Message Posted: Tue Aug 18, 2009 12:15 am by Callyst Post subject: i forgot to add and busted the pipe open _________________ "whoever said the pen was mightier than the sword has obviously never encountered automatic weapons" -Douglas MacArthur- This was the Brain Teaser: A pack of 9 cards is numbered 1 to 9. One card is placed on the forehead of each of four logicians. Each of them can see the other three numbers, but not their own. In turn, each states whether or not he knows his own number. If not, he announces the sum of two of the numbers he can see. One game went as follows. Alf: “No, 14”. Bert: “Yes”. Chris: “No, 7. Dave: “No”- but before Dave could continue, Alf had worked out his own number. What was Dave’s number? And this was my proof that it can't be done: Definitions: There are two (and only two) possible “pairs” that make up the 14 that A can see: 6-8 and 5-9 These will be called the pairs, and 5,6,8,9 will be referred to as pair numbers. All the other numbers (i.e: 1,2,3,4,7) will be called “anonymous numbers”. A preliminary general observation: It is not possible that all four pair cards (5,6,8,9) are on the four foreheads, for there would clearly be no way that C could nominate a 2-card total of 7. Logic: A can see a 14-pair somewhere on B, C & D. B states that he knows his own number. So B must have half of a pair – if a whole pair was on C & D then B couldn’t possibly state that he knew his own number unless he was clairvoyant, it could be anything. So the other half of the pair will be on C or D, and, importantly, the person who doesn’t have the other half must have an anonymous number. For if they had one of the other pair then B couldn’t make his definitive statement. (Unless A conveniently had a third pair card, in which case B would know that he had to have the missing fourth one, but this is precluded – see general observation above). Just to recap by way of example – say B can see C 8 and D 3. He knows his own number, it must be a 6, because it’s the only way to make the14 that A has called. But if what he saw was C 8 and D 9, he wouldn’t have known his own number – it could be a 6 or a 5. So we have established: B has one of a pair; C or D has the other half of that pair, and the other has an anonymous number. So where is the other half of B’s pair? – is it on C or D? Lets say it’s on C. So C is looking at half a pair on B and an anonymous number on D, and would therefore immediately know his own number, which makes a 14-pair with B. So that’s not it, because C was unable to nominate his own number. So is it on D? No, for the same reason. D is looking at half a pair on B and an anonymous number on C, and would therefore know his own number. But he passed, so that’s not it. So the whole thing is impossible Message Posted: Tue Aug 11, 2009 1:10 pm by mdbbarilla Post subject: they only state the sum of the two numbers they can see. So A can announce the sum of C & D rather than B&C or B&D A sees 14 ( 6/8 or 5/9) C sees 7 ( 6/1 or 5/2 or 4/3) Let say they saw 6. A can see a 14-pair somewhere on B, C & D. B states that he knows his own number. So B must have half of a pair B has 6. B = 6 So the only pair that could have a sum of 7 is 6 and 1 If C had 1, C can't say that he sees a 7. C =/= 1 If D had the 1, A wouldn't be able to know his number. Because A doesn't know anything about his number. D =/= 1 If D didn't have the 1, A would now he has the number 1! Let say they saw 5. A can see a 14-pair somewhere on B, C & D. B states that he knows his own number. So B must have half of a pair. B has 5. So the only pair that has a sum of 7 is 5 and 2. If C had 2, C can't say he sees a 7 If D had 2, A won't know his number. But if D didn't get the 2, A would know his number is 2. So Dave's number would 3,4 or 7. Doesn't really matter if you use this approach. Message Posted: Fri Jun 03, 2011 7:22 pm by Unni Post subject: Correct answer What about placing the card (6 or 9) upside down? At a recent trial in Justiceville, the prosecutor, Mr. Jones, called five witnesses to the stand, after which the defense attorney, Ms. Smith, called five more witnesses. From this information and the clues, can you determine the first name (one is Diane) and last name (one is Anderson) and occupation (one is an accountant) of each of the ten witnesses and the order in which they testified? * The five called to the stand by Mr. Jones were, in no particular order: Kathy, Mark, Mr. Ducklow, Ms. Olson, and the computer programmer. * Ms. Smith's five witnesses were, in no particular order: John, Sandra, Ms. Fuller, Mr. Simpson and the secretary. * Neither Mark nor Kathy is the pilot. * Mary testified just before Miller, who was before the author, but not just before. * John is not the teacher. * Thompson, who is not Mary, testified just before the dentist. * The mechanic, who was not on the stand first, testified before McNeil, but not just before. * Frank testified just before the teacher who was just before Ms. Fuller. * The bank teller testified before Zimmer, but not just before. * Anne, who is a musician, testified just before Williams and just after the pilot. * The computer programmer was not just before the secretary. * Ms. Smith did not call Glenn or Betty. * Mark testified before Betty, but not just before. Message Posted: Sat Mar 19, 2011 4:59 pm by ChibiHoshi Post subject: Mark Thompson Bank Teller Kathy Anderson Dentist Betty Zimmer Computer Programmer Glenn Ducklow Pilot Anne Olson Musician Mary William Secretary John Miller Mechanic Frank Simpson Accountant Sandra McNeil Teacher Diane Fuller Author in a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. the king called a logician an a chemist to make the strongest poison for him. the logician went straight to work, but the chemist knew that he had no chance, for the logician was wiser than him, so instead he made up a plan to survive and make sure the logician dies. on the last day the logician suddenly realized that the chemist would know he had no chance, so he must have a plan. after a little thought the logician realized what was the chemist's plan must be, and he conducted a counter plan , to make sure he survives and the chemist die. when the time came, the king summoned both of them. they drank the poison as planned, the chemist died, the logician survived and the king got what he wanted. what exactly happen Message Posted: Wed Feb 16, 2011 6:15 pm by nomad Post subject: The chemist made the strongest poison available till time.And he had to drink it to prove it.So he died. But the logician made a weaker poison and drank it himselves.and to save his life he again took the poison prepared by the chemist and thats how he saved himselves Puzzle: There are a hundred coins sitting on the table, ten are currently heads and nintey are currently tails. You are sitting at the table with a blindfold and gloves on. You are able to feel where the coins are, but are unable to see or feel if they heads or tails. You must create two sets of coins. Each set must have the same number of heads and tails as the other group. You can only move or flip the coins, you are unable to determine their current state. How do you create two even groups of coins with the same number of heads and tails in each group? Solution: Create two sets of ten coins. Flip The coins in one of the sets over, and leave the coins in the other set alone. The first set of ten coins will have the same number of heads and tails as the other set of ten coins. Can somebody hep me better understand the solution? It would seem that if by chance I (blindfolded) create two sets of coins 0h:10t - 0h:10t and flip one set thus making 10h:0t - 0h:10t that I do not have the same number of heads and tails in each group. There are situations where this would work, but not 100% of the time. Am I missing a key point in the puzzle maybe? Many thanks! Message Posted: Mon Nov 29, 2010 9:26 pm by sudheerkuttiyat Post subject: Yes, The question should be ..."there are twenty coins, with 10 showing heads and 10 showing tails..... " The solution does not work for 100 coins with only 10 showing heads initially Splitting the twenty to two groups and flipping all coins in one group will make both groups look the same. ou have a chessboard with two opposite squares cut out. How can you cover the remaining 62 squares completely with 31 domino pieces ?(without breaking the dominoes or the board of course). Each domino covers exactly two squares. Message Posted: Tue Apr 15, 2008 11:22 am by tlmarjot Post subject: the chess board can be folded in half like a lot of chess boards. so half of the 62 squares is 31 same as the amount of dominos. each domino covers one square completely but when the board is folded over it covers all 62squares and remains true to the fact that each domino covers two squares Message Posted: Sat May 03, 2008 2:40 pm by dedo Post subject: tlmarjot, I didn't get it - can you cover it, or not? If yes - how? If not - why? _________________ In Nature - the only blemish is the mind None are to be called deform'd but the unkind Message Posted: Tue Jun 10, 2008 1:54 pm by steinjim Post subject: checkerboard If you cut out opposite corners, that will be two of the same color. Since each domino must cover one black and one white square, you will NOT be able to cover the remaining 62 squares with dominos (there will be 32 of one color and 30 of the other) Message Posted: Mon Dec 22, 2008 9:54 pm by alexonfyre Post subject: I took it to mean that one square of each color was cut out, at random...though it would be impossible to show a solution without the actual coordinates and/or a board showing such that we could draw on. Message Posted: Sun Aug 02, 2009 3:05 am by porcelina Post subject: If I'm not mistaken, the opposite squares could also be squares directly across the middle from one another. In which case, you'd simply lay the dominoes horizontally, filling up each row EXCEPT the two middle rows. There would be one square left uncovered on each. You would then place the last domino vertically, thereby covering both. A man goes out for a walk. He walks 1mile to the south, then 1 mile to the east and finally 1 mile to the north and ends up in the same place he started. now he dint started at north pole,for sure. . .so where did he :?: Message Posted: Tue Jul 07, 2009 7:59 am by rj82330 Post subject: Any number of points just over a mile NORTH of the SOUTH POLE. Provided the circumference of the earth at the point where he starts walking east is an exact factor of 1 mile (i.e. a circumference of 1/2 a mile, or 1/3 a mile, or 1/4 a mile etc), he can walk round and round on this latitude until he has walked a mile "east", whereupon he'll be at his point of origin, and can head 1 mile north to the original start point. Theoretically, he could start exactly 1 mile north of the South Pole, then "walk east" by standing on one spot and spinning in a circle until his body had arced 1760 yards, then walk back north: but this isn't really "walking". _________________ Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking? Message Posted: Tue Jul 07, 2009 8:31 am by v_jey Post subject: good job rj82330 a more simpler way to expln is by assuming a perfectly spherical Earth, somewhere one mile north of the latitude (in the southern hemisphere) that is one mile in circumference. The man walks south one mile to this latitude and walks one mile east, which takes him all the way around and back to where he started. The last step (one mile north) retraces the first step he took (one mile south). 8) A woman was being robbed and she could only run away with two of her golden balls they each were 10 pounds an she was 100 pounds in order to get away she had to cross a bridge and the bridge had a completely accurate sighn that said ony 110 pounds was aloud to cross. if all together they were 120 pounds how did she get across? answer:she jugled them across[/b][/i] _________________ Sylvia E Message Posted: Mon Mar 09, 2009 5:46 pm by Jeffkins Post subject: The act of throwing a 10 pound ball upwards generates greater than 10 pounds of force downwards. So the act of throwing one ball up in the air would cause the bridge to collapse. Juggling is not the answer. A possible answer would be to roll or throw the ball across the bridge. But of course that means this is a lateral thinking puzzle. Three coworkers would like to know their average salary. how can they do it, without disclosing their own salaries? Message Posted: Sat Mar 01, 2008 10:56 am by aggynd Post subject: ask each coworker total of other two's salary. total the three figures and divide by two. Message Posted: Sun Mar 30, 2008 12:42 pm by Jeebok Post subject: Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B, without showing C. B then adds his salary plus a random amount (BS + BR) and passes to C (at each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes to A. Now A subtracts her random number (AR), passes to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3. As has been noted already, there's no way to liar-proof the scheme. It's also worth noting that once they know the average, any of the three knows the sum of the other 2 salaries. Message Posted: Sat May 03, 2008 2:38 pm by dedo Post subject: Jeebok, you got it! _________________ In Nature - the only blemish is the mind None are to be called deform'd but the unkind