Wine and Water

You have two 1 liter mugs – one of them halfway filled with water and the other one halfway filled with wine. You pour 300ml water from the first mug into the second one, stir it well, then pour 300ml of the mix from the second mug back to the first one. Now, do you have more water in the first mug than you have wine in the second one?

If you have X ml water in the first mug, then in the second mug you have 500-X ml water and X ml wine. Therefore you have exactly as much water in the first mug as you have wine in the second one.

Non-Negative

You have a rectangular grid and arbitrary real numbers in its cells. You are allowed repeatedly to multiply the elements in any row or any column by -1. Prove that you can make all row sums and all column sums non-negative simultaneously.

If there is any row or column in the grid with a negative sum, multiply it by -1. Since on every step the total sum of the numbers in the grid increases, we will be able to do this procedure only finitely many times. In the end, all row sums and column sums will be non-negative.

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.

Chessboard Infection

On a standard 8×8 chessboard there are 7 infected cells. Every minute each cell which has at least 2 infected neighbors gets infected as well. Is it possible for the entire chessboard to get infected eventually?

The total perimeter of the infected regions never increases. If there are 7 infected cells initially, their total perimeter is at most 28. The perimeter of an 8×8 square is 32. Therefore, it is impossible to infect the entire chessboard.

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.

Slowest Horse

Two friends made a bet whose horse is slower. After wondering for days what is the fastest and fairest way to figure out who wins the bet, they finally decided to ask a famous wise hermit for help. Upon giving them his advice, the two friends jumped on the horses and started racing back to the city as fast as they could. What did the hermit say?

He told them to switch their horses and whoever gets to the city first will win the bet.

A Maze Puzzle for the Day

Here’s a little maze puzzle I originally built a couple of years ago, that seems apropos to reprise now:

Can you make it from the A in the top left of this grid to the Z in the bottom right, always going either up one letter (for instance, A to B or G to H) or down one letter (for instance, N to M)? The alphabet wraps around, so you can go from Z up to A or A down to Z too. Try as hard as you can (and remember that you can always work backward if you get stuck forwards), and see where you get!

Remark: Solving the maze is not the same thing as solving the puzzle. Read those instructions carefully!

Notice this puzzle is published on April 1st. Actually, it doesn’t have a standard solution. If you connect every two consecutive letters which appear next to each other in the grid, you will get two disconnected components, one of which contains the START and the other contains the END. The first component has 5 dead-ends – at letters A, P, R, I, L, and the second component has 5 dead-ends – at letters F, O, O, L, S. These two spell out “April Fools”, which is the real solution of the maze.