## One potato chip to the other

What did one potato chip say to the other?

Shall we go for a dip?

What did one potato chip say to the other?

Shall we go for a dip?

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).

How far can you run into the woods?

Half way. If you go in any further, you’d be running OUT of the woods.

Why do Chinese men eat more rice than Japanese men do?

There are more Chinese men than Japanese men.

What do squirrels give each other for Valentine’s Day?

Forget-me-nuts.

Two men working at a construction site were up for a challenge, and they were pretty mad at each other.
Finally, at lunch break, they confronted one another.
One man, obviously stronger, said "See that wheelbarrow? I'm willin' to bet $100 (that's all I have in my wallet here) that you can't wheel something to that cone and back that I can't do twice as far. Do you have a bet?"
The other man, too dignified to decline, shook his hand, but he had a plan formulating.
He looked at the objects lying around: a pile of 400 bricks, a steel beam, the 10 men that had gathered around to watch, his pickup truck, a stack of ten bags of concrete mix, and then he finalized his plan.
"All right," he said, and revealed his object.
That night, the strong man went home thoroughly teased and $100 poorer.
What did the other man choose?

He looked the man right in the eye and said "get in."

Can a man legally marry his widow's sister in the state of California?

NO since she is a 'widow', the guy's dead!

The Miller next took the company aside and showed them nine sacks of flour that were standing as depicted in the sketch.
"Now, hearken, all and some," said he, "while that I do set ye the riddle of the nine sacks of flour.
And mark ye, my lords and masters, that there be single sacks on the outside, pairs next unto them, and three together in the middle thereof.
By Saint Benedict, it doth so happen that if we do but multiply the pair, 28, by the single one, 7, the answer is 196, which is of a truth the number shown by the sacks in the middle.
Yet it be not true that the other pair, 34, when so multiplied by its neighbour, 5, will also make 196.
Wherefore I do beg you, gentle sirs, so to place anew the nine sacks with as little trouble as possible that each pair when thus multiplied by its single neighbour shall make the number in the middle."
As the Miller has stipulated in effect that as few bags as possible shall be moved, there is only one answer to this puzzle, which everybody should be able to solve.

The way to arrange the sacks of flour is as follows: 2, 78, 156, 39, 4. Here each pair when multiplied by its single neighbour makes the number in the middle, and only five of the sacks need be moved.
There are just three other ways in which they might have been arranged (4, 39, 156, 78, 2; or 3, 58, 174, 29, 6; or 6, 29, 174, 58, 3), but they all require the moving of seven sacks.

There are 5 pirates in a ship. Pirates have hierarchy C1, C2, C3, C4 and C5. C1 designation is the highest and C5 is the lowest.
These pirates have three characteristics:
a. Every pirate is so greedy that he can even take lives to make more money.
b. Every pirate desperately wants to stay alive.
c. They are all very intelligent.
There are total 100 gold coins on the ship. The person with the highest designation on the deck is expected to make the distribution. If the majority on the deck does not agree to the distribution proposed, the highest designation pirate will be thrown out of the ship (or simply killed). The first priority of the pirates is to stay alive and second to maximize the gold they get. Pirate 5 devises a plan which he knows will be accepted for sure and will maximize his gold. What is his plan?

To understand the answer,we need to reduce this problem to only 2 pirates. So what happens if there are only 2 pirates. Pirate 2 can easily propose that he gets all the 100 gold coins. Since he constitutes 50% of the pirates, the proposal has to be accepted leaving Pirate 1 with nothing.
Now let's look at 3 pirates situation, Pirate 3 knows that if his proposal does not get accepted, then pirate 2 will get all the gold and pirate 1 will get nothing. So he decides to bribe pirate 1 with one gold coin. Pirate 1 knows that one gold coin is better than nothing so he has to back pirate 3. Pirate 3 proposes {pirate 1, pirate 2, pirate 3} {1, 0, 99}. Since pirate 1 and 3 will vote for it, it will be accepted.
If there are 4 pirates, pirate 4 needs to get one more pirate to vote for his proposal. Pirate 4 realizes that if he dies, pirate 2 will get nothing (according to the proposal with 3 pirates) so he can easily bribe pirate 2 with one gold coin to get his vote. So the distribution will be {0, 1, 0, 99}.
Smart right?
Now can you figure out the distribution with 5 pirates? Let's see. Pirate 5 needs 2 votes and he knows that if he dies, pirate 1 and 3 will get nothing. He can easily bribe pirates 1 and 3 with one gold coin each to get their vote. In the end, he proposes {1, 0, 1, 0, 98}. This proposal will get accepted and provide the maximum amount of gold to pirate 5.

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.