When Manish was three years old he carved a nail into his favorite tree to mark his height. Six years later at age nine, Manish returned to see how much higher the nail was. If the tree grew by five centimeters each year, how much higher would the nail be.
The nail would be at the same height since trees grow at their tops.
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.
You've been placed on a course of expensive medication in which you are to take one tablet of Plusin and one tablet of Minusin daily. You must be careful that you take just one of each because taking more of either can have serious side effects. Taking Plusin without taking Minusin, or vice versa, can also be very serious, because they must be taken together in order to be effective. In summary, you must take exactly one of the Plusin pills and one of the Minusin pills at one time.
Therefore, you open up the Plusin bottle, and you tap one Plusin pill into your hand. You put that bottle aside and you open the Minusin bottle. You do the same, but by mistake, two Minusins fall into your hand with the Plusin pill.
Now, here's the problem. You weren't watching your hand as the pills fell into it, so you can't tell the Plusin pill apart from the two Minusin pills. The pills look identical. They are both the same size, same weight (10 micrograms), same color (Blue), same shape (perfect square), same everything, and they are not marked differently in any way.
What are you going to do?
You cannot tell which pill is which, and they cost $500 a piece, so you cannot afford to throw them away and start over again. How do you get your daily dose of exactly one Plusin and exactly one Minusin without wasting any of the pills?
Carefully cut each of the three pills in half, and carefully separate them into two piles, with half of each pill in each pile. You do not know which pill is which, but you are 100% sure that each of the two piles now contains two halves of Minusin and half of Plusin. Now go back into the Plusin bottle, take out a pill, cut it in half, and add one half to each stack. Now you have two stacks, each one containing two halves of Plusin and two halves of Minusin. Take one stack of pills today, and save the second stack for tomorrow.
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.
It was a Pink Island. There were 201 individuals (perfect logicians) lived in the island. Among them 100 people were blue eyed people, 100 were green eyed people and the leader was a black eyed one.
Except the leader, nobody knew how many individuals lived in the island. Neither have they known about the color of the eyes. The leader was a very strict person. Those people can never communicate with others. They even cannot make gestures to communicate. They can only talk and communicate with the leader. It was a prison for those 200 individuals.
However, the leader provided an opportunity to leave the island forever but on one condition. Every morning he questions the individuals about the color of the eyes! If any of the individuals say the right color, he would be released. Since they were unaware about the color of the eyes, all 200 individuals remained silent. When they say wrong color, they were eaten alive to death. Afraid of punishment, they remained silent.
One day, the leader announced that "at least 1 of you has green eyes! If you say you are the one, come and say, I will let you go if you are correct! But only one of you can come and tell me!"
How many green eyed individuals leave the island and in how many days?
All 100 green eyed individuals will leave on the 100th night.
Consider, there is only one green eyed individual lived in the island. He will look at all the remaining individuals who have blue eyes. So, he can get assured that he has green eyes!
Now consider 2 people with green eyes. Only reason the other green-eyed person wouldn't leave on the first night is because he sees another person with green eyes. Seeing no one else with green eyes, each of these two people realize it must be them. So both leaves on second night.
This is the same for any number. Five people with green eyes would leave on the fifth night and 100 on the 100th, all at once.
Search: Monty Hall problem
Why it's important for the solution that the leader said the new information "at least 1 of you has green eyes", when they must knew from the beginning, that there are no less than 99 green-eyed people on the island? Because they cannot depart the island without being certain, they cannot begin the process of leaving until the guru speaks, and common knowledge is attained.
Search: Common knowledge (logic)
Betty signals to the headwaiter in a restaurant, and says, "There is a fly in my tea."
The waiter says "No problem Madam. I will bring you a fresh cup of tea."
A few minutes later Betty shouts, "Get me the manager! This is the same cup of tea."
How did she know?
Hint: The tea is still hot.
Betty had already put sugar in her tea before sending it back. When the "new" cup came, it was already tasted sweet.
At a dinner party, many of the guests exchange greetings by shaking hands with each other while they wait for the host to finish cooking.
After all this handshaking, the host, who didn't take part in or see any of the handshaking, gets everybody's attention and says: "I know for a fact that at least two people at this party shook the same number of other people's hands."
How could the host know this? Note that nobody shakes his or her own hand.
Assume there are N people at the party.
Note that the least number of people that someone could shake hands with is 0, and the most someone could shake hands with is N-1 (which would mean that they shook hands with every other person).
Now, if everyone at the party really were to have shaken hands with a different number of people, then that means somone must have shaken hands with 0 people, someone must have shaken hands with 1 person, and so on, all the way up to someone who must have shaken hands with N-1 people. This is the only possible scenario, since there are N people at the party and N different numbers of possible people to shake hands with (all the numbers between 0 and N-1 inclusive).
But this situation isn't possible, because there can't be both a person who shook hands with 0 people (call him Person 0) and a person who shook hands with N-1 people (call him Person N-1). This is because Person 0 shook hands with nobody (and thus didn't shake hands with Person N-1), but Person N-1 shook hands with everybody (and thus did shake hands with Person 0). This is clearly a contradiction, and thus two of the people at the party must have shaken hands with the same number of people.
Pretend there were only 2 guests at the party. Then try 3, and 4, and so on. This should help you think about the problem.
Search: Pigeonhole principle
A grandfather's clock chimes the appropriate number of times to indicate the hour, as well as chiming once at each quarter hour. If you were in another room and hear the clock chime just once, what would be the longest period of time you would have to wait in order to be certain of the correct time?
You would have to wait 90 minutes between 12:15 and 1:45. Once you had heard seven single chimes, you would know that the next chime would be two chimes for 2 o'clock.
Using only and all the numbers 3, 3, 7, 7, along with the arithmetic operations +,-,*, and /, can you come up with a calculation that gives the number 24? No decimal points allowed.
[For example, to get the number 14, we could do 3 * (7 - (7 / 3))]
You have two lengths of rope. Each rope has the property that if you light it on fire at one end, it will take exactly 60 minutes to burn to the other end. Note that the ropes will not burn at a consistent speed the entire time (for example, it's possible that the first 90% of a rope will burn in 1 minute, and the last 10% will take the additional 59 minutes to burn).
Given these two ropes and a matchbook, can you find a way to measure out exactly 45 minutes?
The key observation here is that if you light a rope from both ends at the same time, it will burn in 1/2 the time it would have burned in if you had lit it on just one end.
Using this insight, you would light both ends of one rope, and one end of the other rope, all at the same time. The rope you lit at both ends will finish burning in 30 minutes. Once this happens, light the second end of the second rope. It will burn for another 15 minutes (since it would have burned for 30 more minutes without lighting the second end), completing the 45 minutes.
