Best difficult riddles

logicmath

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.
81.23 %
51 votes
funnycrazy

By what process could you make a 'Tea-Table' into food?
If you take away the T, it would be eatable.
81.11 %
71 votes
cleanlogicstory

100 men are in a room, each wearing either a white or black hat. Nobody knows the color of his own hat, although everyone can see everyone else's hat. The men are not allowed to communicate with each other at all (and thus nobody will ever be able to figure out the color of his own hat). The men need to line up against the wall such that all the men with black hats are next to each other, and all the men with white hats are next to each other. How can they do this without communicating? You can assume they came up with a shared strategy before coming into the room.
The men go to stand agains the wall one at a time. If a man goes to stand against the wall and all of the men already against the wall have the same color hat, then he just goes and stands at either end of the line. However, if a man goes to stand against the wall and there are men with both black and white hats already against the wall, he goes and stands between the two men with different colored hats. This will maintain the state that the line contains men with one colored hats on one side, and men with the other colored hats on the other side, and when the last man goes and stands against the wall, we'll still have the desired outcome.
81.02 %
64 votes
logiccleansimple

Hussey has been caught stealing goats, and is brought into court for justice. The judge is his ex-wife Amy Hussey, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Hussey from close range. To make things a little better for Hussey, Amy Hussey tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot. If Hussey is still alive, she will then either take another shot, or spin the chamber again before shooting. Hussey is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower. He steels himself as Amy Hussey loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Amy Hussey asks, 'Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?' What should Hussey choose?
Hussey should have Amy Hussey pull the trigger again without spinning. We know that the first chamber Amy Hussey fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Hussey has Amy Hussey pull the trigger again, the probability that a bullet will be fired is 1/4. If Amy Hussey spins the chamber again, the probability that she shoots Hussey would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
81.02 %
64 votes
logicmathstorycleaninterview

You are somewhere on Earth. You walk due south 1 mile, then due east 1 mile, then due north 1 mile. When you finish this 3-mile walk, you are back exactly where you started. It turns out there are an infinite number of different points on earth where you might be. Can you describe them all? It's important to note that this set of points should contain both an infinite number of different latitudes, and an infinite number of different longitudes (though the same latitudes and longitudes can be repeated multiple times); if it doesn't, you haven't thought of all the points.
One of the points is the North Pole. If you go south one mile, and then east one mile, you're still exactly one mile south of the North Pole, so you'll be back where you started when you go north one mile. To think of the next set of points, imagine the latitude slighty north of the South Pole, where the length of the longitudinal line around the Earth is exactly one mile (put another way, imagine the latitude slightly north of the South Pole where if you were to walk due east one mile, you would end up exactly where you started). Any point exactly one mile north of this latitude is another one of the points you could be at, because you would walk south one mile, then walk east a mile around and end up where you started the eastward walk, and then walk back north one mile to your starting point. So this adds an infinite number of other points we could be at. However, we have not yet met the requirement that our set of points has an infinite number of different latitudes. To meet this requirement and see the rest of the points you might be at, we just generalize the previous set of points. Imagine the latitude slightly north of the South Pole that is 1/2 mile in distance. Also imagine the latitudes in this area that are 1/3 miles in distance, 1/4 miles in distance, 1/5 miles, 1/6 miles, and so on. If you are at any of these latitudes and you walk exactly one mile east, you will end up exactly where you started. Thus, any point that is one mile north of ANY of these latitudes is another one of the points you might have started at, since you'll walk one mile south, then one mile east and end up where you started your eastward walk, and finally, one mile north back to where you started.
81.02 %
64 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.
80.94 %
57 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.
80.94 %
57 votes