logicEmperor Akbar once ruled over India. He was a wise and intelligent ruler; and he had in his court the Nine Gems, his nine advisors, who were each known for a particular skill. One of these Gems was Birbal, known for his wit and wisdom. The story below is one of the examples of his wit. Do you have it in you to find the answer? One day the Emperor Akbar stumbled on a small rock in the royal gardens and momentarily went off balance. He was in a bad mood that day and the incident only served to make him more angry. Finding a target for his mood of the day, he ordered the gardener's arrest and execution. Birbal heard of this and visited the gardener in the cell where he was being held awaiting execution. Birbal had known the gardener for many years and also knew of the gardener's immense respect and sense of loyalty for the king. He decided to help the gardener escape the death sentence and explained his plan to the gardener, who reluctantly agreed to go along. The next day the gardener was asked what his last wish was before he was hanged, as was custom. The gardener requested an audience with the emperor. This wish was granted, but when the man neared the throne he tried to attack the emperor. The emperor was shocked and demanded an explanation. The gardener looked at Birbal, who stepped forward and explained why the gardener had attacked the emperor. The emperor immediately realised how unjust he had been and ordered the release of the gardener. How did Birbal manage this?

"Your Majesty," said Birbal, "there is probably no person more loyal to you than this unfortunate gardener. Fearing that people would say you hanged him for a silly reason and question your sense of justice, he went out of his way to give you a genuine reason for hanging him."

## Similar riddles

See also best riddles or new riddles.

logicYou 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.

logicOn the game show et´s Make a Deal, Monty Hall shows you three doors. Behind one of the doors is a new car, the other two hide goats. You choose one door, perhaps #1. Now Monty shows you what´s behind door #2 and it´s a goat.He gives you the chance to stay with original pick or select door #3. What do you do?

You should always abandon your original choice in favor of the remaining door (#3). When you make your first choice the chance of winning is 1 in 3 or 33%. When you switch doors, you turn a 2 in 3 chance of losing in the first round into a 2 in 3 chance of winning in the second round.

logicThere are 4 big houses in my home town. They are made from these materials: red marbles, green marbles, white marbles and blue marbles.
Mrs Jennifer's house is somewhere to the left of the green marbles one and the third one along is white marbles.
Mrs Sharon owns a red marbles house and Mr Cruz does not live at either end, but lives somewhere to the right of the blue marbles house.
Mr Danny lives in the fourth house, while the first house is not made from red marbles.
Who lives where, and what is their house made from ?

From, left to right:
#1 Mrs Jennifer - blue marbles
#2 Mrs Sharon - red marbles
#3 Mr Cruz - white marbles
#4 Mr Danny - green marbles
If we separate and label the clues, and label the houses #1, #2, #3, #4 from left to right we can see that:
a. Mrs Jennifer's house is somewhere to the left of the green marbles one.
b. The third one along is white marbles.
c. Mrs Sharon owns a red marbles house
d. Mr Cruz does not live at either end.
e. Mr Cruz lives somewhere to the right of the blue marbles house.
f. Mr Danny lives in the fourth house
g. The first house is not made from red marbles.
By (g) #1 isn't made from red marbles, and by (b) nor is #3. By (f) Mr Danny lives in #4 therefore by (c) #2 must be red marbles, and Mrs Sharon lives there.
Therefore by (d) Mr Cruz must live in #3, which, by (b) is the white marbles house. By (a) #4 must be green marbles (otherwise Mrs Jennifer couldn't be to its left) and by (f) Mr Danny lives there.
Which leaves Mrs Jennifer, living in #1, the blue marbles house.

logicYou die and the devil says he'll let you go to heaven if you beat him in a game. The devil sits you down at a perfectly round table. He gives himself and you an infinite pile of quarters. He says, "OK, we'll take turns putting one quarter down, no overlapping allowed, and the quarters must rest flat on the table surface. The first guy who can't put a quarter down loses." You guys are about to start playing, and the devil says that he'll go first. However, at this point you immediately interject, and ask if you can go first instead. You make this interjection because you are very smart and can place quarters perfectly, and you know that if you go first, you can guarantee victory. Explain how you can guarantee victory.

You place a quarter right in the center of the table. After that, whenever the devil places a quarter on the table, mimic his placement on the opposite side of the table.. If he has a place to place a quarter, so will you. The devil will run out of places to put a quarter before you do.

logicThis 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?
HINT: 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.

logicMad Ade's Uncle, Phil Space, who doesn't like what passes for art these days, ran into the National Gallery and caused millions of pounds of damage to several masterpieces. Later that day, Uncle Phil was invited to meet the manager and was warmly thanked for his actions. How come?

Uncle Phil is a fireman The water from his house damaged the paintings as he put out a fire in the Gallery, but in the process rescuing hundreds of millions of pounds worth.

logicA 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.

logicYou 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!

logicA deliveryman comes to a house to drop off a package. He asks the woman who lives there how many children she has.
"Three," she says. "And I bet you can't guess their ages."
"Ok, give me a hint," the deliveryman says.
"Well, if you multiply their ages together, you get 36," she says. "And if you add their ages together, the sum is equal to our house number."
The deliveryman looks at the house number nailed to the front of her house. "I need another hint," he says.
The woman thinks for a moment. "My youngest son will have a lot to learn from his older brothers," she says.
The deliveryman's eyes light up and he tells her the ages of her three children. What are their ages?

Their ages are 1, 6, and 6. We can figure this out as follows:
Given that their ages multiply out to 36, the possible ages for the children are:
1, 1, 36 (sum = 38)
1, 2, 18 (sum = 21)
1, 3, 12 (sum = 16)
1, 4, 9 (sum = 14)
1, 6, 6 (sum = 13)
2, 2, 9 (sum = 13)
2, 3, 6 (sum = 11)
3, 3, 4 (sum = 10)
When the woman tells the deliveryman that the children's ages add up to her street number, he still doesn't know their ages. The only way this could happen is that there is more than one possible way for the children's ages to add up to the number on the house (or else he would have known their ages when he looked at the house number). Looking back at the possible values for the children's ages, you can see that there is only one situation in which there are multiple possible values for the children's ages that add up to the same sum, and that is if their ages are either 1, 6, and 6 (sums up to 13), or 2, 2, and 9 (also sums up to 13). So these are now the only possible values for their ages.
When the woman then tells him that her youngest son has two older brothers (who we can tell are clearly a number of years older), the only possible situation is that their ages are 1, 6, and 6.

logicshortThere were two women, standing and facing the opposite ways. The first lady was facing south and the second lady was facing north. But, they could see each other. How is that possible?

They were holding the mirror.