Long story riddles

cleanstorylogicclever

After recent events, Question Mark is annoyed with his brother, Skid Mark. Skid thought it would be funny to hide Question's wallet. He told Question that he would get it back if he finds it. So, first off, Skid laid five colored keys in a row. One of them is a key to a room where Skid is hiding Question's wallet. Using the clues, can you determine the order of the keys and which is the right key? Red: This key is somewhere to the left of the key to the door. Blue: This key is not at one of the ends. Green: This key is three spaces away from the key to the door (2 between). Yellow: This key is next to the key to the door. Orange: This key is in the middle.
The order (from left to right) is Green, Red, Orange, Blue, Yellow. The blue key is the key to the door.
72.35 %
107 votes
logiccleverstory

A monk leaves at sunrise and walks on a path from the front door of his monastery to the top of a nearby mountain. He arrives at the mountain summit exactly at sundown. The next day, he rises again at sunrise and descends down to his monastery, following the same path that he took up the mountain. Assuming sunrise and sunset occured at the same time on each of the two days, prove that the monk must have been at some spot on the path at the same exact time on both days.
Imagine that instead of the same monk walking down the mountain on the second day, that it was actually a different monk. Let's call the monk who walked up the mountain monk A, and the monk who walked down the mountain monk B. Now pretend that instead of walking down the mountain on the second day, monk B actually walked down the mountain on the first day (the same day monk A walks up the mountain). Monk A and monk B will walk past each other at some point on their walks. This moment when they cross paths is the time of day at which the actual monk was at the same point on both days. Because in the new scenario monk A and monk B MUST cross paths, this moment must exist.
72.33 %
69 votes
logicstoryclean

A man comes to a small hotel where he wishes to stay for 7 nights. He reaches into his pockets and realizes that he has no money, and the only item he has to offer is a gold chain, which consists of 7 rings connected in a row (not in a loop). The hotel proprietor tells the man that it will cost 1 ring per night, which will add up to all 7 rings for the 7 nights. "Ok," the man says. "I'll give you all 7 rings right now to pre-pay for my stay." "No," the proprietor says. "I don't like to be in other people's debt, so I cannot accept all the rings up front." "Alright," the man responds. "I'll wait until after the seventh night, and then give you all of the rings." "No," the proprietor says again. "I don't like to ever be owed anything. You'll need to make sure you've paid me the exact correct amount after each night." The man thinks for a minute, and then says "I'll just cut each of my rings off of the chain, and then give you one each night." "I do not want cut rings," the proprietor says. "However, I'm willing to let you cut one of the rings if you must." The man thinks for a few minutes and then figures out a way to abide by the proprietor's rules and stay the 7 nights in the hotel. What is his plan?
The man cuts the ring that is third away from the end of the chain. This leaves him with 3 smaller chains of length 1, 2, and 4. Then, he gives rings to the proprietor as follows: After night 1, give the proprietor the single ring After night 2, take the single ring back and give the proprietor the 2-ring chain After night 3, give the proprietor the single ring, totalling 3 rings with the proprietor After night 4, take back the single ring and the 2-ring chain, and give the proprietor the 4-ring chain After night 5, give the proprietor the single ring, totalling 5 rings with the proprietor After night 6, take back the single ring and give the proprietor the 2-ring chain, totalling 6 rings with the proprietor After night 7, give the proprietor the single ring, totalling 7 rings with the proprietor
72.32 %
86 votes
logicmathstory

The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels. What is the most bananas you can bring over to your destination?
First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A. So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again. Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between. When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward. In the first part, P1, to shift the bananas by 1Km, the Camel will have to Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana Note: After point 5 the Camel does not need to return to point A again. So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas. After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas. Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km. Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Note: After point 3 the Camel does not need to return to the starting point of P2. So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas. After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas. The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example. Hence the length of part P2 is 333 Km. Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas. He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.
72.17 %
98 votes
logiccleanstory

One day a really rich old man with two sons died. In his will he said that he would give one of his sons all of his fortune. He gave each of his sons a horse and said they would compete in a horse race from Los Angeles to Sacramento, but the son whose horse came in second would get the money. So one day they started the race. After one whole day they had only ridden one mile. At night they decided they should stop at a hotel. While they were booking in they told their problem to the wise old clerk, who made a suggestion. The next day the two brothers rode as fast as they could. What did the clerk suggest that they do?
The clerk told them to swap horses. The father said that whoever's horse crossed the finish line second would get the money. He didn't say that the owner of the horse had to be on it.
72.17 %
98 votes
logicstoryclever

You have just purchased a small company called Company X. Company X has N employees, and everyone is either an engineer or a manager. You know for sure that there are more engineers than managers at the company. Everyone at Company X knows everyone else's position, and you are able to ask any employee about the position of any other employee. For example, you could approach employee A and ask "Is employee B an engineer or a manager?" You can only direct your question to one employee at a time, and can only ask about one other employee at a time. You're allowed to ask the same employee multiple questions if you want. Your goal is to find at least one engineer to solve a huge problem that has just hit the company's factory. The problem is so urgent that you only have time to ask N-1 total questions. The major problem with questioning the employees, however, is that while the engineers will always tell you the truth about other employees' roles, the managers may lie to you if they like. You can assume that the managers will do their best to confuse you. How can you find at least one engineer by asking at most N-1 questions?
You can find at least one engineer using the following process: Put all of the employees in a conference room. If there happen to be an even number of employees, pick one at random and send him home for the day so that we start with an odd number of employees. Note that there will still be more engineers than managers after we send this employee home. Then call them out one at a time in any order. You will be forming them into a line as follows: If there is nobody currently in the line, put the employee you just called out in the line. Otherwise, if there is anybody in the line, then we do the following. Let's call the employee currently at the front of the line Employee_Front, and call the employee who we just called out of the conference room Employee_Next. So ask Employee_Front if Employee_Next is a manager or an engineer. If Employee_Front says "manager", then send both Employee_Front and Employee_Next home for the day. However, if Employee_Front says "engineer", then put Employee_Next at the front of the line. Keep doing this until you've called everyone out of the conference room. Notice that at this point, you'll have asked N-1 or less questions (you asked at most one question each time you called an employee out except for the first employee, when you didn't ask a question, so that's at most N-1 questions). When you're done calling everyone out of the conference room, the person at the front of the line is an engineer. So you've found your engineer! But the real question: how does this work? We can prove this works by showing a few things. First, let's show that if there are any engineers in the line, then they must be in front of any managers. We'll show this with a proof by contradiction. Assume that there is a manager in front of an engineer somewhere in the line. Then it must have been the case that at some point, that engineer was Employee_Front and that manager was Employee_Next. But then Employee_Front would have said "manager" (since he is an engineer and always tells the truth), and we would have sent them both home. This contradicts their being in the line at all, and thus we know that there can never be a manager in front of an engineer in the line. So now we know that after the process is done, if there are any engineers in the line, then they will be at the front of the line. That means that all we have to prove now is that there will be at least one engineer in the line at the end of the process, and we'll know that there will be an engineer at the front. So let's show that there will be at least one engineer in the line. To see why, consider what happens when we ask Employee_Front about Employee_Next, and Employee_Front says "manager". We know for sure that in this case, Employee_Front and Employee_Next are not both engineers, because if this were the case, then Employee_Front would have definitely says "engineer". Put another way, at least one of Employee_Front and Employee_Next is a manager. So by sending them both home, we know we are sending home at least one manager, and thus, we are keeping the balance in the remaining employees that there are more engineers than managers. Thus, once the process is over, there will be more engineers than managers in the line (this is also sufficient to show that there will be at least one person in the line once the process is over). And so, there must be at least one engineer in the line. Put altogether, we proved that at the end of the process, there will be at least one engineer in the line and that any engineers in the line must be in front of any managers, and so we know that the person at the front of the line will be an engineer.
72.14 %
77 votes
logiccleansimplestory

A guard is stationed at the entrance to a bridge. He is tasked to shoot anyone who tries to cross to the other side of the bridge, and to turn away anyone who comes in from the opposite side of the bridge. You are on his side of the bridge and want to escape to the other side. Because the bridge is old and rickety, anyone who tries to cross it does so at a constant speed, and it always takes exactly 10 minutes to cross. The guard comes out of his post every 6 minutes and looks down the bridge for any people trying to leave, and at all other times he sits in his post and snoozes. You know you can sneak past him when he's sleeping, but the problem is that you won't be able to make it all the way to the other side of the bridge before he sees you (since he comes out every 6 minutes, but it takes 10 minutes to cross). One day a brilliant idea comes to you, and soon you've successfully crossed to the other side of the bridge without being shot. How did you do it?
Right after the guard goes back to his post after checking the bridge, you sneak by and make your way down the bridge. After a little bit less than 6 minutes, you turn around and start walking back toward the guard. He will come out and see you, and assume that you are a visitor coming from the other side of the bridge, since you're only about 4 minutes from the end of the other side of the bridge. He will go back into his post since he doesn't plan to turn you away until you reach him, and then you turn back around and make your way the rest of the way to the other side of the bridge.
72.10 %
106 votes
logicstoryclever

It was a Pink Island. There were 201 individuals (perfect logicians) lived in the island. Among them 100 people were blue eyed people, 100 were green eyed people and the leader was a black eyed one. Except the leader, nobody knew how many individuals lived in the island. Neither have they known about the color of the eyes. The leader was a very strict person. Those people can never communicate with others. They even cannot make gestures to communicate. They can only talk and communicate with the leader. It was a prison for those 200 individuals. However, the leader provided an opportunity to leave the island forever but on one condition. Every morning he questions the individuals about the color of the eyes! If any of the individuals say the right color, he would be released. Since they were unaware about the color of the eyes, all 200 individuals remained silent. When they say wrong color, they were eaten alive to death. Afraid of punishment, they remained silent. One day, the leader announced that "at least 1 of you has green eyes! If you say you are the one, come and say, I will let you go if you are correct! But only one of you can come and tell me!" How many green eyed individuals leave the island and in how many days?
All 100 green eyed individuals will leave on the 100th night. Consider, there is only one green eyed individual lived in the island. He will look at all the remaining individuals who have blue eyes. So, he can get assured that he has green eyes! Now consider 2 people with green eyes. Only reason the other green-eyed person wouldn't leave on the first night is because he sees another person with green eyes. Seeing no one else with green eyes, each of these two people realize it must be them. So both leaves on second night. This is the same for any number. Five people with green eyes would leave on the fifth night and 100 on the 100th, all at once. Search: Monty Hall problem Why it's important for the solution that the leader said the new information "at least 1 of you has green eyes", when they must knew from the beginning, that there are no less than 99 green-eyed people on the island? Because they cannot depart the island without being certain, they cannot begin the process of leaving until the guru speaks, and common knowledge is attained. Search: Common knowledge (logic)
71.95 %
68 votes
logicsimplecleanstory

Jack and Joe were on vacation and driving along a deserted country road from the town of Kaysville to the town of Lynnsville. They came to a multiple fork in the road. The sign post had been knocked down and they were faced with choosing one of five different directions. Since they had left their map at the last gas station and there was no one around to ask, how could Jack and Joe find their way to Lynnsville?
They need to stand the signpost up so that the arm reading Kaysville points in the direction of Kaysville, the town they had just come from. With one arm pointing the correct way, the other arms will also point in the right directions.
71.93 %
93 votes
logicstory

This teaser is based on a weird but true story from a few years ago. A complaint was received by the president of a major car company: "This is the fourth time I have written you, and I don't blame you for not answering me because I must sound crazy, but it is a fact that we have a tradition in our family of having ice cream for dessert after dinner each night. Every night after we've eaten, the family votes on which flavor of ice cream we should have and I drive down to the store to get it. I recently purchased a new Pantsmobile from your company and since then my trips to the store have created a problem. You see, every time I buy vanilla ice cream my car won't start. If I get any other kind of ice cream the car starts just fine. I want you to know I'm serious about this question, no matter how silly it sounds: 'What is there about a Pantsmobile that makes it not start when I get vanilla ice cream, and easy to start whenever I get any other kind?'" The Pantsmobile company President was understandably skeptical about the letter, but he sent an engineer to check it out anyway. He had arranged to meet the man just after dinner time, so the two hopped into the car and drove to the grocery store. The man bought vanilla ice cream that night and, sure enough, after they came back to the car it wouldn't start for several minutes. The engineer returned for three more nights. The first night, the man got chocolate. The car started right away. The second night, he got strawberry and again the car started right up. The third night he bought vanilla and the car failed to start. There was a logical reason why the man's car wouldn't start when he bought vanilla ice cream. What was it? The man lived in an extremely hot city, and this took place during the summer. Also, the layout of the grocery store was such that it took the man less time to buy vanilla ice cream.
Vanilla ice cream was the most popular flavor and was on display in a little case near the express check out, while the other flavors were in the back of the store and took more time to select and check out. This mattered because the man's car was experiencing vapor lock, which is excess heat boiling the fuel in the fuel line and the resulting air bubbles blocking the flow of fuel until the car has enough time to cool.. When the car was running there was enough pressure to move the bubbles along, but not when the car was trying to start.
71.88 %
55 votes