Behind 1 door is a coffer overflowing with jewels and gold, along with an exit. Behind the other door is an enormous, hungry lion that will pounce on anyone opening the door.
You do not know which door leads to the treasure and exit, and which door leads to the lion.
In the room you are in are 2 individuals. The first is a knight, who always tells the truth, and a knave, who always lies.
Both of these individuals know what is behind each door. You do not know which individual is the knight, or which one is the knave.
You may ask one of the individuals exactly 1 question.
What should you ask in order to be certain that you will open the door with the coffer behind it, instead of the hungry lion?
Hint: This will require logical thinking.
You ask one of the individuals what the other one would say if you asked him which door is holding back the hungry lion and then open this door.
You have a pile of 24 coins. Twenty-three of these coins have the same weight, and one is heavier.
Your task is to determine which coin is heavier and do so in the minimum number of weighings.
You are given a beam balance (scale), which will compare the weight of any two sets of coins out of the total set of 24 coins.
How many weighings are required to identify the heavier coin?
Hint: Split the coins into groups of 8 for the first weighing.
Answer: It can be done in three weighings.
Weighing 1: Break the coins into three piles of eight. Weigh one group of eight against another group of eight. If the scale balances, then the group that hasn't been weighed has the heavier coin. If the scale tips, then that group contains the heavier coin.
Weighing 2: Break the group of eight that has the heavier coin into three groups (three coins, three coins, and two coins). Weigh one set of three against the other set of three. If it balances, the group of two has the heavier coin. If the scale tips, then that group has the heavier coin.
Weighing 3: If the heavier coin is in the group of two, then just weigh one coin against the other to determine the heaviest coin. If the heavier coin is in a group of three, then take two of those coins and weigh them against each other. If the scale balances, the coin that hasn't been weighed is the heavier coin. If the scale tips, then that is the heavier coin.
You can imagine an arrow in flight, toward a target. For the arrow to reach the target, the arrow must first travel half of the overall distance from the starting point to the target. Next, the arrow must travel half of the remaining distance.
For example, if the starting distance was 10m, the arrow first travels 5m, then 2.5m.
If you extend this concept further, you can imagine the resulting distances getting smaller and smaller. Will the arrow ever reach the target?
Hint: This puzzle needs some very careful thinking.
Since the arrow does indeed hit the target, it must be true that 1/2 + 1/4 + 1/8 + ... = 1.
This is because the sum of an infinite series can be a finite number.
Imagine a prisoner in a prison. He is sentenced to death and has been told that he will be killed on one day of the following week. He has been assured that the day will be a surprise to him, so he will not be anticipating the hangman on a particular day, so keeping his stress levels in check.
The prisoner starts to think to himself, if I am still alive on Thursday, then clearly I shall be hanged on Friday, this would mean that I then know the day of my death, therefore I cannot be hanged on Friday. Now then, if I am still alive on Wednesday, then clearly I shall be hanged on Thursday, since I have already ruled out Friday. The prisoner works back with this logic, finally concluding that he cannot after all be hanged, without already knowing which day it was.
Casually, resting on his laurels, sitting in his prison cell on Tuesday, the warden arrives to take him to be hanged, the prisoner was obviously surprised!
Hint: Be careful not to let your brain melt on this one.
Answer: This puzzle is a classic paradox. You are led through a sequence of seemingly valid arguments which lead to a conclusion, which quite clearly cannot be true.
Mr Smith has lots of pound coins, ten boxes in all. Each box contains 100 pound coins, but one box contains coins which are all counterfeit and are slightly lighter, 1/16 of an ounce lighter to be exact.
The problem lies in the fact that they all look identical, the only way to tell them apart is to weigh them.
Mr Smith knows the correct weight for a box, but how many weighings are required to determine which box contains the counterfeit ones?
Hint: The answer is a lot less than 100 weighings.
Answer: One weighing is enough.
Take one coin from the first box, two from the second and so on.
When the coins are weighed, the number of 1/16ths light will tell us which box contains the counterfeits.
For example if it was box 5, the weighing would be 5/16 too light.
Three people check into a hotel.
They pay £30 to the manager and go to their room.
The manager suddenly remembers that the room rate is £25 and gives £5 to the porter to return to the people.
On the way to the room the porter reasons that £5 would be difficult to share among three people so they keep £2 and give £1 to each person.
Now each person paid £10 and got back £1.
So they paid £9 each, totalling £27. The porter has £2, totalling £29.
Where is the missing £1?
Hint: Be careful of what you're adding.
We have to be careful what we are adding together.
Originally, they paid £30, they each received back £1, they now have only paid £27.
Of this £27, £25 went to the manager for the room and £2 went to the porter.
Think of words ending in -GRY. Angry and hungry are two of them. There are only three words in the English language. What is the third word? The word is something that everyone uses every day. If you have listened carefully, I have already told you what it is.
Hint: Is this a trick question?
Answer: There are only two: angry and hungry. The rec.puzzles archive offers a large collection of words that end in -GRY, but none of them could be considered even remotely common. There are many generally unsatisfying "trick" answers to the problem, which depend on a specific wording of the question or that the question be spoken instead of written. There seems to be no agreement among puzzle historians about which form is the original, or even the age of the problem. In any event, it is apparent that the frequent mutations of the puzzle statement over the years have erased whatever answer was intended by the original author. The usual trick is to play on the expression "the English Language", you are then asked for the third word - which is of course Language!
There are three houses, and three utilities: water, gas and electricity.
Your task is to connect each house to all three utilities.
Therefore each house will have three lines and each utility will also have three lines.
However, you cannot cross lines! You cannot pass lines through houses or utilities. You cannot share lines.
Can you draw the 9 lines required?
This puzzle is a classic one which has no solution in 2D.
However, if you place the items on a doughnut shape in 3D you can solve it.
In the picture below, the electricity is linked to House 3 by going over the top and re-entering through the hole in the middle.
There is a concert that starts in just 17 minutes and all of the band members must all cross a bridge to get there. The four members begin on the same side of the bridge and you must help them to get across to the other side.
Due the age of the bridge, a maximum of two people can cross at one time. To make matters worse, it is night-time and there is only one torch. The torch is always required when crossing the bridge and the torch must be walked back and forth, it cannot be thrown, etc. Each band member walks at a different speed and a pair must walk together at the rate of the slower man:
Alan takes 1 minute to cross
Bill takes 2 minutes to cross
Carl takes 5 minutes to cross
Dave takes 10 minutes to cross
For example, if Alan and Dave walk across first, it takes them 10 minutes to cross. If Alan then returns with the torch, a total of 11 minutes will have passed. There is no trick behind this, it is the simple movement of resources in the appropriate order.
Hint: Always using the quickest person doesn't necessarily help.
Alan and Bill cross - 2 minutes
Alan returns - 1 minute
Carl and Dave cross - 10 minutes
Bill returns - 2 minutes
Alan and Bill cross - 2 minutes
Total - 17 minutes