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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.

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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?

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.

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A perfectly symmetrical square 4-legged table is standing in a room with a continuous but uneven floor. Is it always possible to position the table in such a way that it doesn’t wobble, i.e. all four legs are touching the floor?

The answer is yes. Let the feet of the table clockwise are labeled with 1, 2, 3, 4 clockwise. Place the table in the room such that 3 of its feet – say 1, 2, 3, touch the ground. If foot 4 is on the ground, then the problem is solved. Otherwise it is easy to see that we can not put it there if we keep legs 2 and 3 on the same places. Now start rotating the table clockwise, keeping feet 1, 2 and 3 on the ground at all times. If at some point foot 4 touches the ground as well, the problem is solved. Otherwise continue rotating until foot 1 goes to the place where foot 2 was and foot 2 goes to the place where foot 3 was. Foot 3 will be on the ground, but this contradicts the observation that initially we couldn’t place legs 2, 3 and 4 on the ground without replacing feet 2 and 3.

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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.

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You are living in a 100-floor apartment block. You know that there is one floor in the block, such that if you drop a light bulb from there or anywhere higher, it will crash upon hitting the ground. If you drop a light bulb from any floor underneath it however, the light bulb will remain intact. If you have two light bulbs at your disposal, how many drop attempts do you need such that you can surely find which the floor in question is?

The answer is 14 drops. You can do this by throwing the first bulb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 (notice that the difference decreases always by 1) until it crashes and then start throwing the second bulb from the floors in between. For example, if the first bulb crashes at floor 69, you start throwing the second bulb from floors 61, 62, 63, etc. This way the total number of throws would be always at most 14.

Proving that 14 is optimal is done using the same logic. In order to use at most 13 throws, the first throw should be made from floor 13 or lower. The second throw should be made from floor 13+12 or lower, the third throw should be made from floor 13+12+11 or lower, etc. Continuing with the same argument, we conclude that the 13th drop should be made from floor 13+12+…+2+1=91 or lower. However, if the first light bulb does not crash after the last throw, you will not be able to find out which number among 92-100 is X.

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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 area of all black parts 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 – in white. It is easy to see after doing this, the entire unit square will be colored in black (each of the 2015 squares covers it once completely). This implies that the total area of black pieces is no less than 1.