10 Dots, 10 Coins

If you have 10 dots on a table, can you always cover them with 10 pennies without overlapping?

Remark: You can assume the dots are not too close to the edge of the table.

Assume the dots lie in a plane and the radius of a penny is 1. Make an infinite grid of circles with radii 1, as shown on the picture, and place it randomly in the plane.

If we choose any point in the plane, the probability that it will end up inside some circle of the grid is equal to S(C)/S(H), where S stands for “area”. A simple calculation shows that this ratio is bigger than 90%. Therefore the probability that some chosen point in the plane will not end up inside any circle is less than 10%. If we have 10 points, the probability that neither of them will end up inside a circle is less than 100%. Therefore we can place the grid in the plane so that all dots end up in some circles. Now just place the given coins where these circles are.

Crashing Light Bulbs

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.

Married Couples

In a small village there are 100 married couples living. Everyone in the village lives by the following two rules:

  1. If a husband cheats on his wife and she figures it out, the husband gets immediately killed.
  2. The wives gossip about all the infidelities in town, with the only exception that no woman is told whether her husband has cheated on her.

One day a traveller comes to the village and finds out that every man has cheated at least once on his wife. When he leaves, without being specific, he announces in front of everybody that at least one infidelity has occurred. What will happen in the next 100 days in the village?

Let us first see what will happen if there are N married couples in the village and K husbands have cheated, where K=1 or 2.

If K = 1, then on the first day the cheating husband would get killed and nobody else will die. If K = 2, then on the first day nobody will get killed. During the second day however, both women would think like this: “If my husband didn’t cheat on me, then the other woman would have immediately realized that she is being cheated on and would have killed her husband on the first day. This did not happen and therefore my husband has cheated on me.”. Then both men will get killed on the second day.

Now assume that if there are N couples on the island and K husbands have cheated, then all K cheaters will get killed on day K. Let us examine what will happen if there are N + 1 couples on the island and L husbands have cheated.

Every woman would think like this: “If I assume that my husband didn’t cheat on me, then the behavior of the remaining N couples will not be influenced by my family’s presence on the island.”. Therefore she has to wait and see when and how many men will get killed in the village. After L days pass however and nobody gets killed, every woman who has been cheated on will realize that her assumption is wrong and will kill her husband on the next day. Therefore if there are N + 1 couples on the island, again all L cheating husbands will get killed on day L.

Applying this inductive logic consecutively for 3 couples, 4 couples, 5 couples, etc., we see that when there are 100 married couples on the island, all men will get killed on day 100.

9 balls, 1 defective

You have 9 balls, 8 of which have the same weight. The remaining one is defective and heavier than the rest. You can use a balance scale to compare weights in order to find which is the defective ball. How many measurements do you need so that will be surely able to do it?

First we put 3 balls on the left side and 3 balls on the right side of the balance scale. If the scale tips to one side, then the defective ball is there. If not, the defective ball is among the remaining 3 balls. Once left with 3 balls only, we put one on each side of the scale. If the scale tips to one side, the defective ball is there. If not, the defective ball is the last remaining one. Clearly we can not find the defective ball with just one measurement, so the answer is 2.