Hourglasses

You have a 7-minute hourglass and an 11-minute hourglass. How can you measure exactly 15 minutes with them?

Turn both hourglasses upside down. When the time of the 7-minute hourglass runs out, turn it upside down. When the time of the 11-minute hourglass runs out, turn the 7-minute hourglass upside down again. Finally, when the time of the 7-minute hourglass runs out, exactly 15 minutes will have been passed.

Pirate’s Treasure

Five pirates steal a treasure which contains 100 gold coins. The rules for splitting the treasure among the pirates are the following:

  1. The oldest pirate proposes how to split the money.
  2. Everybody votes, including the proposer.
  3. If there are more than 50% negative votes, the proposer gets thrown in the water and the procedure repeats.

Given that the pirates are very smart and bloodthirsty (if they can kill another without losing money, they will do it), how should the oldest pirate suggest to split the money among the five of the in order to maximize his profit?

Solve the problem backward. Let the pirates be called A, B, C, D, E, where A is older than B, B is older than C, C is older than D and D is older than E.
If there are only two pirates left – D and E, then the D will keep all the treasure for himself.
If there are three pirates left – C, D, and E, C can propose to give just 1 coin to E and keep the rest for himself. Pirate E will agree because otherwise, he will get nothing.
If there are four pirates left – B, C, D, and E, then B can propose to give just 1 coin to D and keep the rest for himself. Pirate D will agree because otherwise, he will get nothing.
Now if there are five pirates – A, B, C, D and E, A should give coins to at least two other pirates, because otherwise at least three of them will vote negative. Clearly, B will always vote negative, unless he gets offered 100 coins and D will also vote negative, unless he gets 2 coins or more. Pirate A can offer to give one 1 coin to C, 1 coin to E and keep the rest for himself and this is the only optimal proposal – 98:0:1:0:1.

Missionaries, Cannibals

Three missionaries and three cannibals must cross a river with a boat which can carry at most two people at a time. However, if on one of the two banks of the river the missionaries get outnumbered by the cannibals, they will get eaten. How can all 6 men cross the river without anybody gets eaten?

Remark: The boat cannot cross the river with no people on board.

Label the missionaries M1, M2, M3 and the cannibals C1, C2, C3. Then:

1. M1 and C1 cross the river, M1 comes back.
2. C2 and C3 cross the river, C2 comes back.
3. M1 and M2 cross the river, M1 and C1 come back.
4. M1 and M3 cross the river, C3 comes back.
5. C1 and C2 cross the river, C1 comes back.
6. C1 and C3 cross the river.

Now, everyone is on the other side.

Friends and Enemies

Show that in each group of 6 people, there are either 3 who know each other, or 3 who do not know each other.

Let’s call the people A, B, C, D, E, F. Person A either knows at least 3 among B, C, D, E, F, or does not know at least 3 among B, C, D, E, F.

Assume the first possibility – A knows B, C, D. If B and C know each other, C and D know each other, or B and D know each other, then we find a group of 3 people who know each other. Otherwise, B, C, and D form a group in which no-one knows the others.

If A doesn’t know at least 3 among B, C, D, E, F, the arguments are the same.