Riddle #773

logicstorymath

You are standing in a house in the middle of the countryside. There is a small hole in one of the interior walls of the house, through which 100 identical wires are protruding. From this hole, the wires run underground all the way to a small shed exactly 1 mile away from the house, and are protruding from one of the shed's walls so that they are accessible from inside the shed. The ends of the wires coming out of the house wall each have a small tag on them, labeled with each number from 1 to 100 (so one of the wires is labeled "1", one is labeled "2", and so on, all the way through "100"). Your task is to label the ends of the wires protruding from the shed wall with the same number as the other end of the wire from the house (so, for example, the wire with its end labeled "47" in the house should have its other end in the shed labeled "47" as well). To help you label the ends of the wires in the shed, there are an unlimited supply of batteries in the house, and a single lightbulb in the shed. The way it works is that in the house, you can take any two wires and attach them to a single battery. If you then go to the shed and touch those two wires to the lightbulb, it will light up. The lightbulb will only light up if you touch it to two wires that are attached to the same battery. You can use as many of the batteries as you want, but you cannot attach any given wire to more than one battery at a time. Also, you cannot attach more than two wires to a given battery at one time. (Basically, each battery you use will have exactly two wires attached to it). Note that you don't have to attach all of the wires to batteries if you don't want to. Your goal, starting in the house, is to travel as little distance as possible in order to label all of the wires in the shed. You tell a few friends about the task at hand. "That will require you to travel 15 miles!" of of them exclaims. "Pish posh," yells another. "You'll only have to travel 5 miles!" "That's nonsense," a third replies. "You can do it in 3 miles!" Which of your friends is correct? And what strategy would you use to travel that number of miles to label all of the wires in the shed?
Believe it or not, you can do it travelling only 3 miles! The answer is rather elegant. Starting from the house, don't attach wires 1 and 2 to any batteries, but for the remaining wires, attach them in consecutive pairs to batteries (so attach wires 3 and 4 to the same battery, attach wires 5 and 6 to the same battery, and so on all the way through wires 99 and 100). Now travel 1 mile to the shed, and using the lightbulb, find all pairs of wires that light it up. Put a rubberband around each pair or wires that light up the lightbulb. The two wires that don't light up any lightbulbs are wires 1 and 2 (though you don't know yet which one of them is wire 1 and which is wire 2). Put a rubberband around this pair of wires as well, but mark it so you remember that they are wires 1 and 2. Now go 1 mile back to the house, and attach odd-numbered wires to batteries in the following pairs: (1 and 3), (5 and 7), (9 and 11), and so on, all the way through (97 and 99). Similarly, attach even-numbered wires to batteries in the following pairs: (4 and 6), (8 and 10), (12 and 14), and so on, all the way through (96 and 98). Note that in this round, we didn't attach wire 2 or wire 100 to any batteries. Finally, travel 1 mile back to the shed. You're now in a position to label all of the wires here. First, remember we know the pair of wires that are, collectively, wires 1 and 2. So test wires 1 and 2 with all the other wires to see what pair lights up the lightbulb. The wire from wires 1 and 2 that doesn't light up the bulb is wire 2 (which, remember, we didn't connect to a battery), and the other is wire 1, so we can label these as such. Furthermore, the wire that, with wire 1, lights up a lightbulb, is wire 3 (remember how we connected the wires this round). Now, the other wire in the rubber band with wire 3 is wire 4 (we know this from the first round), and the wire that, with wire 4, lights up the lightbulb, is wire 6 (again, because of how we connected the wires to batteries this round). We can continue labeling batteries this way (next we'll label wire 7, which is rubber-banded to wire 6, and then we'll label wire 9, which lights up the lightbulb with wire 7, and so on). At the end, we'll label wire 97, and then wire 99 (which lights up the lightbulb with wire 97), and finally wire 100 (which isn't connected to a battery this round, but is rubber-banded to wire 99). And we're done, having travelled only 3 miles!
76.32 %
61 votes

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logicmath

Can you arrange four 9's and use of at most 2 math symbols, make the total be 100?
99 / .99 or 99 + 9/9
78.55 %
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You have a basket of infinite size (meaning it can hold an infinite number of objects). You also have an infinite number of balls, each with a different number on it, starting at 1 and going up (1, 2, 3, etc...). A genie suddenly appears and proposes a game that will take exactly one minute. The game is as follows: The genie will start timing 1 minute on his stopwatch. Where there is 1/2 a minute remaining in the game, he'll put balls 1, 2, and 3 into the basket. At the exact same moment, you will grab a ball out of the basket (which could be one of the balls he just put in, or any ball that is already in the basket) and throw it away. Then when 3/4 of the minute has passed, he'll put in balls 4, 5, and 6, and again, you'll take a ball out and throw it away. Similarly, at 7/8 of a minute, he'll put in balls 7, 8, and 9, and you'll take out and throw away one ball. Similarly, at 15/16 of a minute, he'll put in balls 10, 11, and 12, and you'll take out and throw away one ball. And so on....After the minute is up, the genie will have put in an infinite number of balls, and you'll have thrown away an infinite number of balls. Assume that you pull out a ball at the exact same time the genie puts in 3 balls, and that the amount of time this takes is infinitesimally small. You are allowed to choose each ball that you pull out as the game progresses (for example, you could choose to always pull out the ball that is divisible by 3, which would be 3, then 6, then 9, and so on...). You play the game, and after the minute is up, you note that there are an infinite number of balls in the basket. The next day you tell your friend about the game you played with the genie. "That's weird," your friend says. "I played the exact same game with the genie yesterday, except that at the end of my game there were 0 balls left in the basket." How is it possible that you could end up with these two different results?
Your strategy for choosing which ball to throw away could have been one of many. One such strategy that would leave an infinite number of balls in the basket at the end of the game is to always choose the ball that is divisible by 3 (so 3, then 6, then 9, and so on...). Thus, at the end of the game, any ball of the format 3n+1 (i.e. 1, 4, 7, etc...), or of the format 3n+2 (i.e. 2, 5, 8, etc...) would still be in the basket. Since there will be an infinite number of such balls that the genie has put in, there will be an infinite number of balls in the basket. Your friend could have had a number of strategies for leaving 0 balls in the basket. Any strategy that guarantees that every ball n will be removed after an infinite number of removals will result in 0 balls in the basket. One such strategy is to always choose the lowest-numbered ball in the basket. So first 1, then 2, then 3, and so on. This will result in an empty basket at the game's end. To see this, assume that there is some ball in the basket at the end of the game. This ball must have some number n. But we know this ball was thrown out after the n-th round of throwing balls away, so it couldn't be in there. This contradiction shows that there couldn't be any balls left in the basket at the end of the game. An interesting aside is that your friend could have also used the strategy of choosing a ball at random to throw away, and this would have resulted in an empty basket at the end of the game. This is because after an infinite number of balls being thrown away, the probability of any given ball being thrown away reaches 100% when they are chosen at random.
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logicmath

If, Fernando + Alonso + McLaren = 6 Fernando x Alonso = 2 Alonso x McLaren = 6 Then, McLaren x Fernando = ?
3 or 0.75 Explanation: Rewriting the last 2 equations in terms of Alonso, Fernando = 2/Alonso McLaren = 6/Alonso Replacing above values in equation "Fernando + Alonso + McLaren = 6" 2/Alonso + Alonso + 6/Alonso =6 (2 + Alonso^2 + 6)/Alonso = 6 8 + Alonso^2 = 6Alonso Alonso^2 - 6Alonso + 8 = 0 (Alonso - 4) (Alonso - 2) = 0 Therefore; Alonso = 4 or 2 Let's take value of Alonso as 2 Fernando = 2/2 = 1 McLaren = 6/2 = 3 Therefore; McLaren x Fernando = 3 x 1 = 3 Let's take value of Alonso as 4 Fernando = 2/4 = 0.5 McLaren = 6/4 = 1.5 Therefore; McLaren x Fernando = 1.5 x 0.5 = 0.75
77.88 %
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interviewlogicmath

Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?
It is 17 mins. 1 and 2 go first, then 1 comes back. Then 7 and 10 go and 2 comes back. Then 1 and 2 go again, it makes a total of 17 minutes.
77.76 %
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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)
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interviewlogicmath

A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine. Fortunately (or say unfortunately) the bad king's guards catch the servant after he has only poisoned one bottle. Alas, the guards don't know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect. The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners – believing he can fob off such a low death rate – and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time. Explain what is in mind of the king, how will he be able to do so?
Think in terms of binary numbers. (now don’t read the solution, give a try). Number the bottles 1 to 1000 and write the number in binary format. bottle 1 = 0000000001 (10 digit binary) bottle 2 = 0000000010 bottle 500 = 0111110100 bottle 1000 = 1111101000 Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc. prisoner = 10 9 8 7 6 5 4 3 2 1 bottle 924 = 1 1 1 0 0 1 1 1 0 0 For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die. After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned. 1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.
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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.
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logicmathsimplecleanclever

There are 100 ants on a board that is 1 meter long, each facing either left or right and walking at a pace of 1 meter per minute. The board is so narrow that the ants cannot pass each other; when two ants walk into each other, they each instantly turn around and continue walking in the opposite direction. When an ant reaches the end of the board, it falls off the edge. From the moment the ants start walking, what is the longest amount of time that could pass before all the ants have fallen off the plank? You can assume that each ant has infinitely small length.
The longest amount of time that could pass would be 1 minute. If you were looking at the board from the side and could only see the silhouettes of the board and the ants, then when two ants walked into each other and turned around, it would look to you as if the ants had walked right by each other. In fact, the effect of two ants walking into each other and then turning around is essentially the same as two ants walking past one another: we just have two ants at that point walking in opposite directions. So we can treat the board as if the ants are walking past each other. In this case, the longest any ant can be on the board is 1 minute (since the board is 1 meter long and the ants walk at 1 meter per minute). Thus, after 1 minute, all the ants will be off the board.
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