## A Short, Brutal Riddle

Left alone, I’m a word with five letters.
I’m honest and fair, I’ll admit.
Rearranged, I’m of no use to trains.
Again, and I’m an overt place, warm and well lit.

What am I?

Source: Puzzling StackExchange

The answer is LIAR. After rearranging the letters, you can get RAIL – important for trains, or LAIR – a dark, hidden place. Since the riddler is a liar, the resulting words are exactly the opposite of his descriptions.

## David Copperfield

David Copperfield and his assistant perform the following magic trick. The assistant offers to a person from the audience to pick 5 arbitrary cards from a regular deck and then hand them back to him. After the assistant sees the cards, he returns one of them to the audience member and gives the rest one by one to David Copperfield. After the magician receives the fourth card, he correctly guesses what card the audience member holds in his hand. How did they perform the trick?

Out of the five cards there will be (at least) two of the same suit, assume they are clubs. Now imagine all clubs are arranged on a circle in a cyclic manner – A, 2, 3, … J, Q, K (clock-wise), and locate the two chosen ones on it. There are two arks on the circle which are connecting them and exactly one of them will contain X cards, with X between 0 and 5. Now the assistant will pass to David Copperfield first the clubs card which is located on the left end of this ark, will return to the audience member the clubs card which is located on the right end of it and with the remaining three cards will encode the number X. In order to do this, he will arrange the three extra cards in increasing order – first clubs A-K, then diamonds A-K, then hearts A-K and finally spades A-K. Let us call the smallest card under this ordering “1”, the middle one “2” and the largest one “3”. Now depending on the value of X, the assistant will pass the cards “1”, “2” and “3” in the following order:

X=0 -> 1, 2, 3
X=1 -> 1, 3, 2
X=2 -> 2, 1, 3
X=3 -> 2, 3, 1
X=4 -> 3, 1, 2
X=5 -> 3, 2, 1

In this way David Copperfield will know the suit of the audience member’s card and also with what number he should increase the card he received first in order to get value as well. Therefore he will be able to guess correctly.

## Merlin and Hermes: Mysterious Lines

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?

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”.

## Black and White

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.