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!
78.50 %
56 votes

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logicmathstory

A swan sits at the center of a perfectly circular lake. At an edge of the lake stands a ravenous monster waiting to devour the swan. The monster can not enter the water, but it will run around the circumference of the lake to try to catch the swan as soon as it reaches the shore. The monster moves at 4 times the speed of the swan, and it will always move in the direction along the shore that brings it closer to the swan the quickest. Both the swan and the the monster can change directions in an instant. The swan knows that if it can reach the lake's shore without the monster right on top of it, it can instantly escape into the surrounding forest. How can the swan succesfully escape?
Assume the radius of the lake is R feet. So the circumference of the lake is (2*pi*R). If the swan swims R/4 feet, (or, put another way, 0.25R feet) straight away from the center of the lake, and then begins swimming in a circle around the center, then it will be able to swim around this circle in the exact same amount of time as the monster will be able to run around the lake's shore (since this inner circle's circumference is 2*pi*(R/4), which is exactly 4 times shorter than the shore's circumference). From this point, the swan can move a millimeter inward toward the lake's center, and begin swimming around the center in a circle from this distance. It is now going around a very slightly smaller circle than it was a moment ago, and thus will be able to swim around this circle FASTER than the monster can run around the shore. The swan can keep swimming around this way, pulling further away each second, until finally it is on the opposite side of its inner circle from where the monster is on the shore. At this point, the swan aims directly toward the closest shore and begins swimming that way. At this point, the swan has to swim [0.75R feet + 1 millimeter] to get to shore. Meanwhile, the monster will have to run R*pi feet (half the circumference of the lake) to get to where the swan is headed. The monster runs four times as fast as the swan, but you can see that it has more than four times as far to run: [0.75R feet + 1 millimeter] * 4 < R*pi [This math could actually be incorrect if R were very very small, but in that case we could just say the swan swam inward even less than a millimeter, and make the math work out correctly.] Because the swan has less than a fourth of the distance to travel as the monster, it will reach the shore before the monster reaches where it is and successfully escape.
83.08 %
57 votes
logicmath

There are 1 million closed school lockers in a row, labeled 1 through 1,000,000. You first go through and flip every locker open. Then you go through and flip every other locker (locker 2, 4, 6, etc...). When you're done, all the even-numbered lockers are closed. You then go through and flip every third locker (3, 6, 9, etc...). "Flipping" mean you open it if it's closed, and close it if it's open. For example, as you go through this time, you close locker 3 (because it was still open after the previous run through), but you open locker 6, since you had closed it in the previous run through. Then you go through and flip every fourth locker (4, 8, 12, etc...), then every fifth locker (5, 10, 15, etc...), then every sixth locker (6, 12, 18, etc...) and so on. At the end, you're going through and flipping every 999,998th locker (which is just locker 999,998), then every 999,999th locker (which is just locker 999,999), and finally, every 1,000,000th locker (which is just locker 1,000,000). At the end of this, is locker 1,000,000 open or closed?
Locker 1,000,000 will be open. If you think about it, the number of times that each locker is flipped is equal to the number of factors it has. For example, locker 12 has factors 1, 2, 3, 4, 6, and 12, and will thus be flipped 6 times (it will end be flipped when you flip every one, every 2nd, every 3rd, every 4th, every 6th, and every 12th locker). It will end up closed, since flipping an even number of times will return it to its starting position. You can see that if a locker number has an even number of factors, it will end up closed. If it has an odd number of factors, it will end up open. As it turns out, the only types of numbers that have an odd number of factors are squares. This is because factors come in pairs, and for squares, one of those pairs is the square root, which is duplicated and thus doesn't count twice as a factor. For example, 12's factors are 1 x 12, 2 x 6, and 3 x 4 (6 total factors). On the other hand, 16's factors are 1 x 16, 2 x 8, and 4 x 4 (5 total factors). So lockers 1, 4, 9, 16, 25, etc... will all be open. Since 1,000,000 is a square number (1000 x 1000), it will be open as well.
82.08 %
68 votes
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
81.26 %
44 votes
logicmath

A women walks into a bank to cash out her check. By mistake the bank teller gives her dollar amount in change, and her cent amount in dollars. On the way home she spends 5 cents, and then suddenly she notices that she has twice the amount of her check. How much was her check amount?
The check was for dollars 31.63. The bank teller gave her dollars 63.31 She spent .05, and then she had dollars 63.26, which is twice the check. Let x be the dolars of the check, and y be the cent. The check was for 100x + y cent He was given 100y + x cent Also 100y + x - 5 = 2(100x + y) Expanding this out and rearranging, we find: 98y = 199x + 5 Which doesn't look like enough information to solve the problem except that x and y must be whole numbers, so: 199x ≡ -5 (mod 98) 98*2*x + 3x ≡ -5 (mod 98) 3x ≡ -5 ≡ 93 (mod 98) This quickly leads to x = 31 and then y = 63 Alternative solution by substitution: 98y = 199x + 5 y = (199x + 5)/98 = 2x + (3x + 5)/98 Since x and y are whole numbers, so must be (3x + 5)/98. Call it z = (3x+5)/98 so 98z = 3x + 5, or 3x = 98z - 5 or x = (98z - 5)/3 or x = 32z-1 + (2z-2)/3. Since everything is a whole number, so must be (2z-2)/3. Call it w = (2z-2)/3, so 3w = 2z-2 so z = (3w+2)/2 or z = w + 1 + w/2. So w/2 must be whole, or w must be even. So try w = 2. Then z = 4. Then x = 129. Then y = 262. if you decrease y by 199 and x by 98, the answer is the same: y = 63 and x = 31.
81.23 %
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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.
80.65 %
82 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.
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Frank and some of the boys were exchanging old war stories. James offered one about how his grandfather (Captain Smith) led a battalion against a German division during World War I. Through brilliant maneuvers he defeated them and captured valuable territory. Within a few months after the battle he was presented with a sword bearing the inscription: "To Captain Smith for Bravery, Daring and Leadership, World War One, from the Men of Battalion 8." Frank looked at James and said, "You really don't expect anyone to believe that yarn, do you?" What is wrong with the story?
It wasn't called World War One until much later. It was called the Great War at first, because they did not know during that war and immediately afterward that there would be a second World War (WW II).
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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
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