Riddle #481

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cleanstoryclever

In classic mythology, there is the story of the Sphinx, a monster with the body of a lion and the upper part of a woman. The Sphinx lay crouched on the top of a rock along the highroad to the city of Thebes, and stopped all travellers passing by, proposing to them a riddle. Those who failed to answer the riddle correctly were killed. This is the riddle the Sphinx asked the travellers: "What animal walks on four legs in the morning, two legs during the day, and three legs in the evening?"
This is part of the story of Oedipus, who replied to the Sphinx, "Man, who in childhood creeps on hands and knees, in manhood walks erect, and in old age with the aid of a staff." Morning, day and night are representative of the stages of life. The Sphinx was so mortified at the solving of her riddle that she cast herself down from the rock and perished.
89.00 %
535 votes
cleanlogicmathsimple

Create a number using only the digits 4,4,3,3,2,2,1 and 1. So I can only be eight digits. You have to make sure the ones are separated by one digit, the twos are separated by two digits the threes are separated with three digits and the fours are separated by four digits.
41312432.
85.76 %
59 votes
logicmathclever

You are standing in a pitch-dark room. A friend walks up and hands you a normal deck of 52 cards. He tells you that 13 of the 52 cards are face-up, the rest are face-down. These face-up cards are distributed randomly throughout the deck. Your task is to split up the deck into two piles, using all the cards, such that each pile has the same number of face-up cards. The room is pitch-dark, so you can't see the deck as you do this. How can you accomplish this seemingly impossible task?
Take the first 13 cards off the top of the deck and flip them over. This is the first pile. The second pile is just the remaining 39 cards as they started. This works because if there are N face-up cards in within the first 13 cards, then there will be (13 - N) face up cards in the remaining 39 cards. When you flip those first 13 cards, N of which are face-up, there will now be N cards face-down, and therefore (13 - N) cards face-up, which, as stated, is the same number of face-up cards in the second pile.
85.53 %
58 votes
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!
85.10 %
47 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.
84.83 %
64 votes
logiccleaninterview

While mixing sand, gravel, and cement for the foundation of a house, a worker noticed a small bird hopping along the top of the foundation wall. The bird misjudged a hop and fell down one of the holes between the blocks. The bird was down too far for anyone to reach it and the hole was too small for it to fly out of. Someone suggested using two sticks to reach down into the hole and pull the bird out, but this idea was rejected for fear it would injure the fragile bird. What would be the easiest way to get the bird out of the hole without injuring it?
Since they had plenty of sand available, they could pour a little at a time into the hole. The bird would constantly keep shifting its position so that it stood on the rising sand.
84.80 %
46 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.
84.78 %
55 votes