Logic riddles

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
logicsimplecleanstory

A king has 100 identical servants, each with a different rank between 1 and 100. At the end of each day, each servant comes into the king's quarters, one-by-one, in a random order, and announces his rank to let the king know that he is done working for the day. For example, servant 14 comes in and says "Servant 14, reporting in." One day, the king's aide comes in and tells the king that one of the servants is missing, though he isn't sure which one. Before the other servants begin reporting in for the night, the king asks for a piece of paper to write on to help him figure out which servant is missing. Unfortunately, all that's available is a very small piece that can only hold one number at a time. The king is free to erase what he writes and write something new as many times as he likes, but he can only have one number written down at a time. The king's memory is bad and he won't be able to remember all the exact numbers as the servants report in, so he must use the paper to help him. How can he use the paper such that once the final servant has reported in, he'll know exactly which servant is missing?
When the first servant comes in, the king should write down his number. For each other servant that reports in, the king should add that servant's number to the current number written on the paper, and then write this new number on the paper. Once the final servant has reported in, the number on the paper should equal (1 + 2 + 3 + ... + 99 + 100) - MissingServantsNumber Since (1 + 2 + 3 + ... + 99 + 100) = 5050, we can rephrase this to say that the number on the paper should equal 5050 - MissingServantsNumber So to figure out the missing servant's number, the king simply needs to subtract the number written on his paper from 5050: MissingServantsNumber = 5050 - NumberWrittenOnThePaper
76.32 %
61 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.
76.32 %
61 votes
logictricky

Fred went to a hardware store in Boston with Alex, Ben, and George. He noted that a hammer cost ten times as much as a screwdriver and a power saw cost ten times as much as a hammer. The storekeeper said that Ben could buy a power saw, George could buy a screwdriver and Alex could buy a hammer. Based on this what would the storekeeper let Fred buy? Alex's full name is Alexander and Ben's full name is Benjamin. George was Alex's boss and good friend.
Fred could buy all three (the power saw, hammer and screw driver) since he had $111 with him (a $1 bill - George Washington, a $10 Alexander Hamilton, and a $100 bill - Ben Franklin). Boston is in the USA and therefore uses the US currency I just described.
76.26 %
66 votes
logicmath

You have been given the task of transporting 3,000 apples 1,000 miles from Appleland to Bananaville. Your truck can carry 1,000 apples at a time. Every time you travel a mile towards Bananaville you must pay a tax of 1 apple but you pay nothing when going in the other direction (towards Appleland). What is highest number of apples you can get to Bananaville?
833 apples. Step one: First you want to make 3 trips of 1,000 apples 333 miles. You will be left with 2,001 apples and 667 miles to go. Step two: Next you want to take 2 trips of 1,000 apples 500 miles. You will be left with 1,000 apples and 167 miles to go (you have to leave an apple behind). Step three: Finally, you travel the last 167 miles with one load of 1,000 apples and are left with 833 apples in Bananaville.
76.22 %
71 votes
logicsimple

There are 3 switches outside of a room, all in the 'off' setting. One of them controls a lightbulb inside the room, the other two do nothing. You cannot see into the room, and once you open the door to the room, you cannot flip any of the switches any more. Before going into the room, how would you flip the switches in order to be able to tell which switch controls the light bulb?
Flip the first switch and keep it flipped for five minutes. Then unflip it, and flip the second switch. Go into the room. If the lightbulb is off but warm, the first switch controls it. If the light is on, the second switch controls it. If the light is off and cool, the third switch controls it.
76.22 %
71 votes
cleanlogiccleversimple

One company had two factories, in different parts of the country, that were making the same style of shoes. In both factories, workers were stealing shoes. How, without using any security, could that company stop the stealing?
Make one factory make the left shoe, and the other make the right shoe.
76.22 %
71 votes
logicmath

You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races. You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in. What is the least number of races you can conduct to figure out which 3 horses are fastest?
You need to conduct 7 races. First, separate the horses into 5 groups of 5 horses each, and race the horses in each of these groups. Let's call these groups A, B, C, D and E, and within each group let's label them in the order they finished. So for example, in group A, A1 finished 1st, A2 finished 2nd, A3 finished 3rd, and so on. We can rule out the bottom two finishers in each race (A4 and A5, B4 and B5, C4 and C5, D4 and D5, and E4 and E5), since we know of at least 3 horses that are faster than them (specifically, the horses that beat them in their respective races). This table shows our remaining horses: A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3 For our 6th race, let's race the top finishers in each group: A1, B1, C1, D1 and E1. Let's assume that the order of finishers is: A1, B1, C1, D1, E1 (so A1 finished first, E1 finished last). We now know that horse D1 cannot be in the top 3, because it is slower than C1, B1 and A1 (it lost to them in the 6th race). Thus, D2 and D3 can also not be in the to 3 (since they are slower than D1). Similarly, E1, E2 and E3 cannot be in the top 3 because they are all slower than D1 (which we already know isn't in the top 3). Let's look at our updated table, having removed these horses that can't be in the top 3: A1 B1 C1 A2 B2 C2 A3 B3 C3 We can actually rule out a few more horses. C2 and C3 cannot be in the top 3 because they are both slower than C1 (and thus are also slower than B1 and A1). And B3 also can't be in the top 3 because it is slower than B2 and B1 (and thus is also slower than A1). So let's further update our table: A1 B1 C1 A2 B2 A3 We actually already know that A1 is our fastest horse (since it directly or indirectly beat all the remaining horses). So now we just need to find the other two fastest horses out of A2, A3, B1, B2 and C1. So for our 7th race, we simply race these 5 horses, and the top two finishers, plus A1, are our 3 fastest horses.
76.22 %
71 votes