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.

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.

In the Padurea Forest

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.

Battleship

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?

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.

Soccer Ball

Almost everyone knows how a soccer ball looks like – there are several black regular pentagons on it and around each of them – five white regular hexagons. Around each hexagon there are three pentagons and three more hexagons. However, can you figure out how many pentagons and hexagons are there in total on the surface of the ball?

Let the number of pentagons is equal to P and the number of hexagons is equal to H. Then the number of edges is equal to (5P + 6H)/2 – that’s because every pentagon has five edges, every hexagon has 6 edges and every edge belongs to 2 sides. Also, the number of vertices is equal to (5P + 6H)/3 – that’s because every pentagon has five vertices, every hexagon has 6 vertices and every vertex belongs to 3 sides. Now using Euler’s Theorem we get P + H + (5P + 6H)/3 – (5P + 6H)/2 = 2, or equivalently P/6=2 and therefore P = 12. Since around every pentagon there are exactly 5 hexagons and around every hexagon there are exactly 3 pentagons, we get H = 5P/3 = 20. Therefore there are 12 pentagons and 20 hexagons on a soccer ball.