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

SOLUTION

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.

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?

SOLUTION

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.

Three friends, A, B and C, want to find out what their average salary is without disclosing their own salaries to the others. How can they do it using only verbal communication?

SOLUTION

A tells B some number, then B adds his salary to it and tells the result to C, then C adds his salary and tells the result to A. Now A subtracts the number he told B in the beginning, adds his own salary and divides by 3. Repeat the same procedure with B and C starting first.

Because it is impossible for the covers to drop inside the holes, no matter how you position them on top. Also, when they have a circular shape, they are easier to roll.

At 12 PM a scientist put a bacteria in a container. The bacteria reproduce by splitting every minute into two. If at exactly 1 PM the container was full, at what time was it half-full?

SOLUTION

Since every minute the bacteria double their quantity, the container was half-full at 12:59 PM.

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

The oldest pirate proposes how to split the money.

Everybody votes, including the proposer.

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?

SOLUTION

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.

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?

SOLUTION

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

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!

SOLUTION

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.

Please note:
This action will also remove this member from your connections and send a report to the site admin.
Please allow a few minutes for this process to complete.