Difficulty: Hard

100 coins are placed on a rectangular table, such that no more coins can be added without overlapping. Show that you can cover the entire table with 400 coins (overlapping allowed).

Since we can not place any more coins on the table, each point of it is at distance at most 2r from the center of some coin, where r is the radius of the coin. Now shrink the entire table twice in width and length, then replace every shrunk coin with a full sized one. This way the small table will be completely covered because every point of it will be at distance at most r from the center of some coin. Add three more of these smaller tables, covered with coins, to create a covering of the big table.

There are 5 people who possess a box. You are allowed to secure the box with as many different locks as you like and distribute any combination of keys for these locks to any people among the 5. Find the least number of locks needed, so that no 2 people can open the box, but any cannot people can open it.

For every subset of 2 people you pick among the 5, there should be a lock which none of the 2 can unlock, and each of the remaining 3 people can unlock. Clearly, the lock in question cannot be the same for any two different subsets of 2 people you choose. Therefore the number of locks you need is at least the number of different 2-element subsets of a 5-element set, which is 5!/(2!3!)=10. This number is sufficient as well – just give keys to a different group of 3 people for every lock.

Can you relabel two 6-sided dices, so that every face has a positive number of dots, and also their sum has the same probability distribution?

Yes, you can do this. The easiest way is to use generating functions. Using simple polynomial algebra, you can see that

(x + x₂ + x₃ + x₄ + x₅ + x₆)² = (x + 2x₂ + 2x₃ + x₄)(x + x₃ + x₄ + x₅ + x₆ + x₈).

Therefore, if you take a dice with spots {1, 2, 2, 3, 3, 4}, and a dice with spots {1, 3, 4, 5, 6, 8}, their sum will have the same probability distribution.

The lion plays a deadly game against a group of 100 zebras that takes place in the steppe (an infinite plane). The lion starts in the origin with coordinates (0,0), while the 100 zebras may arbitrarily pick their 100 starting positions. The lion and the group of zebras move alternately:

• In a lion move, the lion moves from its current position to a position at most 100 meters away.
• In a zebra move, one of the 100 zebras moves from its current position to a position at most 100 meters away.
• The lion wins the game as soon as he manages to catch one of the zebras.

Will the lion always win the game after a finite number of moves? Or is there a strategy for the zebras that lets them survive forever?

The zebras can survive forever. They choose 100 parallel strips with width 300m each, then start on points on their mid-lines. If the lion lands on some zebra’s strip, the zebra simply jumps 100m away from the lion, along its mid-line.

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 in 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.