Warning: this puzzle involves mature themes that are inappropriate for younger audiences. If you are not an adult, please skip this puzzle.
SHOW PUZZLE
3 men must have sex with 1 woman, but they have only 2 condoms. Each of the 4 people has some unique STD which they don’t want to transfer to the rest. What can they do?
SOLUTION
They can start by putting the two condoms on top of each other and letting the first man use them. After that, the second man can take the inner condom out and use just the outer condom. Finally, the third man can take the removed inner condom, turn it inside out, and place it back inside the outer condom. Then he, he can use the two condoms simultaneously.
On the ground there is a stick and 10 ants standing on top of it. All ants have the same constant speed and each of them can travel along the entire stick in exactly 1 minute (if it is left alone). The ants start moving simultaneously straightforward, either towards the left or the right end of the stick. When two ants collide with each other, they both turn around and continue moving in the opposite directions. How much time at most would it take until all ants fall off the stick?
SOLUTION
Imagine the ants are just dots moving along the stick. Now it looks looks like all dots keep moving in their initially chosen directions and just occasionally pass by each other. Therefore it will take no more than a minute until they fall off the stick. If any of them starts at one end of the stick and moves towards the other end, then the time it will take for it to fall off will be exactly 1 minute.
There are 100 prisoners in solitary cells. There is a central living room with one light bulb in it, which can be either on or off initially. No prisoner can see the light bulb from his or her own cell. Every day, the warden picks a prisoner at random and that prisoner visits the living room. While there, the prisoner can toggle the light bulb if he wishes to do so. Also, at any time, every prisoner has the option of asserting that all 100 prisoners have already been in the living room. If this assertion is false, all 100 prisoners will be executed. If it is correct, all prisoners will be set free.
The prisoners are allowed to get together one night in the courtyard and come up with a plan. What plan should they agree on, so that eventually someone will make a correct assertion and will set everyone free, assuming the warden will bring each of them an infinite number of times to the central living room?
SOLUTION
First, the prisoners should elect one of them to be a leader and the rest – followers. The first two times a follower visits the living room and sees that the light bulb is turned off, he should turn it on; after that he shouldn’t touch it anymore. Every time the leader visits the living room and sees that the light bulb is turned on, he should turn it off. After the leader turns off the lightbulb 198 times, this will mean that all followers have already visited the room. Then he can make the assertion and set everyone free.
Two adventurers, Merlin and Hermes, approached a large iron door built into a cliff face.”Well…”, said Hermes, “What do we do now?”. Merlin produced an old, large piece of crumpled paper from his pocket. “Hrm…”, Merlin mumbled. “It says here that we must speak the six letter keyword to open the door and enter the secret chamber, but I don’t remember seeing any signs as to what that keyword might be…”
After a bit of searching, Hermes notices something etched into the ground. “Come over here!”, he yelled, pointing frantically. And sure enough, barely visible and obscured by dust, was a series of lines of different colors etched into the ground:
“Ah”, Merlin said, “So that is the keyword.” Hermes was lost and confused. After staring at it for another thirty seconds, he grumbled “What keyword!? All I see is a bunch of lines!”. Merlin simply responded, “You’re just looking at it the wrong way. It’s obvious!”
Isn’t it?
SOLUTION
The signs are engraved letters on the ground and Merlin and Hermes are looking at them from above (the italic “looking at it the wrong way” is a hint). The darker a part from some sign is, the farther from the ground it is. The only letters which could correspond to this description are U – N – L – I – N – K. Therefore the keyword is “UNLINK”.
A boy draws 2015 unit squares on a piece of paper, all oriented the same way, possibly overlapping each other. Then the colors the resulting picture in black and white chess-wise, such that any area belonging to an even number of squares is painted white and any area belonging to an odd number of squares is painted black.
Prove that the total black area is at least one.
SOLUTION
Draw a grid in the plane which is parallel to the sides of the squares. Then, take the content of each cell of the grid and translate it (move it) to some chosen unit square. The points in that unit square which are covered by odd number of black pieces color in black, the rest color in white. It is easy to see that after doing this, the entire unit square will be colored in black (each of the 2015 squares cover it once completely). This implies that the total black area is no less than 1.
In the Padurea forest there are 100 rest stops. There are 1000 trails, each connecting a pair of rest stops. Each trail has some particular level of difficulty with no two trails having the same difficulty. An intrepid hiker, Sendeirismo has decided to spend a vacation by taking a hike consisting of 20 trails of ever increasing difficulty. Can he be sure that it can be done?
He is free to choose the starting rest stop and the 20 trails from a sequence where the start of one trail is the end of a previous one.
SOLUTION
Place one hiker in each of the rest stops. Now, go through the trails in the forest one by one, in increasing difficulty, and every time you pick a trail, let the two hikers in its ends change places. This way the 100 hikers would traverse 2000 trails in total, and therefore one of them would traverse at least 20 trails.
A battleship starts moving at 12 PM from an integer point on the real line with constant speed, landing on every hour again on an integer point. Every day at midnight you can shoot at an arbitrary point on the real plane, trying to destroy the battleship. Can you find a strategy with which you will eventually succeed to do this?
SOLUTION
If we know the starting point of the battleship and its speed, then we can determine its position at any time after 12 PM.
There are countably many combinations (X, Y) of starting point and speed. We can order them in the following way:
(0, 0) – starting point 0, speed 0; (0, 1) – starting point 0, speed +1; (1, 0) – starting point 1, speed 0; (0, -1) – starting point 0, speed -1; (1, 1) – starting point 1, speed +1; (-1, 0) – starting point -1, speed 0; (0, 2) – starting point 0, speed +2; (1, -1) – starting point 1, speed -1; (-1, 1) – starting point -1, speed 1; (2, 0)- starting point 2, speed 0, and so on. Of course, we can choose the ordering in many different ways.
Now we can start exhausting all possibilities one after another. First we assume the combination is (0, 0), calculate where the battleship would be at midnight during the first day and shoot there. Then we assume the combination is (0, 1), calculate where the battleship would be at midnight during the second day and shoot there. If we continue like this, eventually we will hit the battleship.
In which direction is this bus moving – towards the left or towards the right?
SOLUTION
Since we can not see the passenger doors, they should be on the other side of the bus, and therefore the bus should be on the other side of the street. Here we assume the bus is from a right-hand traffic country.
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