Get from START to FINISH in this “Bad Idea” maze, created by David Modray.
The solution is shown below.
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. Everyday 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 already have been in the living room. If this assertion is false, all 100 prisoners are executed. If it is correct, all prisoners are 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 they will be set free?
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 199 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!”
Source: Puzzling StackExchange
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.
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.
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.