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.

FEATURED

Soccer Ball

Almost everyone knows what 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.

$5 Bill

Bern tells Sandy, “This isn’t the $5 bill you left on the table. It is mine – I was keeping it between pages 15 and 16 of my textbook.”
Sandy retorts, “You are lying and I can prove it.” How does she know?

Pages 15 and 16 are on the same sheet of paper in the textbook (just like pages 1 and 2 are) and therefore the bill couldn’t be between them.

Water with Ice

You have a glass of water and an ice cube floating in it. When the ice cube melts, will the water level increase, decrease or remain the same?

It will remain the same. The amount of water that the ice cube displaces is equal to its mass. Since the mass does not change and the density of water is equal to 1, the extra water after melting will be the same amount as the displaced water before that.

FEATURED

Heads Up, Heads Down

You are blindfolded and on the table, in front of you, 50 coins are placed. You are told that X of them are heads up and the rest are heads down. Then you are asked to separate the coins into two groups and optionally flip some of them so that the number of heads in both groups becomes the same. How can you do this?

Separate the coins into one group of X coins and into another group of 50-X coins, then flip every coin in the first group. If in the first group there were Y heads up initially, then after flipping there would be X-Y – exactly the number of heads up in the second group.