A man walks into a bar and asks the bartender for a glass of water. The bartender reaches under the bar and brings out a gun and aims it at the man. The man says thank you and leaves. What happened?
The man had the hiccups and the water helped him stop it, and the gun scared him which also help stop his hiccups as well.
Four cars come to a four way stop, all coming from a different direction. They can't decide who got there first, so they all entered the intersection at the same time. They do not crash into each other. How is this possible?
You have twelve balls, identical in every way except that one of them weighs slightly less or more than the balls.
You have a balance scale, and are allowed to do 3 weighings to determine which ball has the different weight, and whether the ball weighs more or less than the other balls.
What process would you use to weigh the balls in order to figure out which ball weighs a different amount, and whether it weighs more or less than the other balls?
Take eight balls, and put four on one side of the scale, and four on the other.
If the scale is balanced, that means the odd ball out is in the other 4 balls.
Let's call these 4 balls O1, O2, O3, and O4.
Take O1, O2, and O3 and put them on one side of the scale, and take 3 balls from the 8 "normal" balls that you originally weighed, and put them on the other side of the scale.
If the O1, O2, and O3 balls are heavier, that means the odd ball out is among these, and is heavier. Weigh O1 and O2 against each other. If one of them is heavier than the other, this is the odd ball out, and it is heavier. Otherwise, O3 is the odd ball out, and it is heavier.
If the O1, O2, and O3 balls are lighter, that means the odd ball out is among these, and is lighter. Weigh O1 and O2 against each other. If one of them is lighter than the other, this is the odd ball out, and it is lighter. Otherwise, O3 is the odd ball out, and it is lighter.
If these two sets of 3 balls weigh the same amount, then O4 is the odd ball out. Weight it against one of the "normal" balls from the first weighing. If O4 is heavier, then it is heavier, if it's lighter, then it's lighter.
If the scale isn't balanced, then the odd ball out is among these 8 balls.
Let's call the four balls on the side of the scale that was heavier H1, H2, H3, and H4 ("H" for "maybe heavier").
Let's call the four balls on the side of the scale that was lighter L1, L2, L3, and L4 ("L" for "maybe lighter").
Let's also call each ball from the 4 in the original weighing that we know aren't the odd balls out "Normal" balls.
So now weigh [H1, H2, L1] against [H3, L2, Normal].
-If the [H1, H2, L1] side is heavier (and thus the [H3, L2, Normal] side is lighter), then this means that either H1 or H2 is the odd ball out and is heavier, or L2 is the odd ball out and is lighter.
-So measure [H1, L2] against 2 of the "Normal" balls.
-If [H1, L2] are heavier, then H1 is the odd ball out, and is heavier.
-If [H1, L2] are lighter, then L2 is the odd ball out, and is lighter.
-If the scale is balanced, then H2 is the odd ball out, and is heavier.
-If the [H1, H2, L1] side is lighter (and thus the [H3, L2, Normal] side is heavier), then this means that either L1 is the odd ball out, and is lighter, or H3 is the odd ball out, and is heavier.
-So measure L1 and H3 against two "normal" balls.
-If the [L1, H3] side is lighter, then L1 is the odd ball out, and is lighter.
-Otherwise, if the [L1, H3] side is heavier, then H3 is the odd ball out, and is heavier.
If the [H1, H2, L1] side and the [H3, L2, Normal] side weigh the same, then we know that either H4 is the odd ball out, and is heavier, or one of L3 or L4 is the odd ball out, and is lighter.
So weight [H4, L3] against two of the "Normal" balls.
If the [H4, L3] side is heavier, then H4 is the odd ball out, and is heavier.
If the [H4, L3] side is lighter, then L3 is the odd ball out, and is lighter.
If the [H4, L3] side weighs the same as the [Normal, Normal] side, then L4 is the odd ball out, and is lighter.
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.
You have two jugs, one that holds exactly 3 gallons, and one that holds exactly 5 gallons. Using just these two jugs and a fire hose, how can you measure out exactly 4 gallons of water?
Fill the 5-gallon jug to the top, and then pour it into the 3-gallon jug until the 3-gallon jug is full. You now have 2 gallons remaining in the 5-gallon jug. Pour out the 3-gallon jug, and then pour the 2 gallons from the 5-gallon jug into the 3-gallon jug. Finally, fill the 5-gallon jug to the top and pour it into the 3-gallon jug until it's full. Since there was only space left for 1 more gallon in the 3-gallon jug, you now have exactly 4 gallons in the 5-gallon jug.
How can you divide a pizza into 8 equal slices using only 3 straight cuts?
Cut 1: Cut the pizza straight down the middle into two halves.
Cut 2: Keeping the two halves in the place, cut the pizza straight down the middle at right angles to the first cut (you will be left with 4 equal quarters)
Cut 3: Pile the 4 quarters on top of each other and cut through the middle of the pile. You will be left with 8 equal slices.
An egg has to fall 100 feet, but it can't break upon landing (or in the air). Its fall can't be slowed down, nor can its landing be cushioned in any way. How is it done?
Drop it from more than 100 feet high. It won't break for the first 100 feet.
