I think this one is the shortest and simplest so far: If you want it to work with negative values of $x$, just P.S. I just noticed this is very similar to @Joe K's formula, but avoids using ...

Since it seems the question was mis-read, and it wasn't my intention to frustrate anyone, I provide the correct answer as intended: Or, more visually:

@Deusovi's answer is totally correct, but I want to add here the general approach for solving such problems as well. No need to upvote, since I did not invent the technique, and you can see it ...

First produce 48 by putting the traces up/down/up/down/up in the freezer. Then use 18 of the ice cubes and put them in the corners of the botton 6 trays, produce 66 more ice cubes. Do this 2 more ...

It is Consider the remainder of the number when divided by $13$. Initially we start with remainder $3$ ($81\equiv 3 \mod 13$), we have to get remainder $4$ ($82\equiv 4 \mod 13$). Applying step 1, ...

@Gamow already offered a nice solution, but I decided to give an alternative one. The good thing about this one is that even if the prisoners are not told when they will start getting steaks for ...

OK, I barely made it with 10 moves, but it seems pretty hard to prove this is optimal (and I suspect the answer may be lower). The last move you make is Rh8, which will be the mate.

EDITED - thanks to @Meelo for pointing out my mistake. We can do it in We place the C.Coli's at a4, a8, b2, c1, e2, g1, h3, h6. Now we prove that this is optimal. Every day there are either 1 new ...

We will prove the problem for 8 friends instead of 5 and see that this is the optimal bound. Proof. Let us imagine that each of the professor's friends has a twin who visits him on the days when the ...

Rigorous proof and exact bounds: If the starting amount of honey in each pot is less or equal to $17/7$ kilograms, then Pooh can always do this. If the starting amount of honey in each pot can be ...

I encountered this problem in Peter Winkler's book "Mathematical Puzzles: A Connoisseur's Collection". I didn't check the solution there, so it might be more elegant than my proof below, but anyway: ...

We consider the directed graph G=(V,E) of all Twitter users, such that V={1,2,...,300} and $(i,j)\in E$ if and only if user $i$ follows user $j$. Since every node has outgoing degree of $1$, the graph ...

This is a system of equivalences. You may want to check the Chinese Remainder Theorem (which says that there is just one solution modulo 420): https://en.wikipedia.org/wiki/Chinese_remainder_theorem ...

Since this problem is possibly a bit too mathematical for Puzzling StackExchange, I decided to post the solution. I believe it contains few interesting ideas, so hope you like it. It is easy to check ...

For completeness, decided to add the optimality proof. Assume the opposite - i.e. you can cover the board with tetraminos, such that there is not a pond of size larger than 3. Now consider the 4 ...

OPTIMALITY SOLUTION GUIDELINES As several people pointed out already, it is not hard to give an example with 9 tetraminos. I also believe this is the correct answer and even though proving it ...

I made up the puzzle late last night and even though it seemed good back then, I feel it may be too misleading now. In order to save you the hassles, decided to post the solution as well. This may be ...

@M Oehm already gave a construction, but for completeness I will add the analysis which yield all solutions. Remark: Never mind, I just noticed @Rand Al'Thor already posted something similar. ...

The answer is The question is equivalent to analyzing the intersection of a cube and a sphere which share a common center. Thus the question gets reduced to figuring out whether such intersection, ...

Yes, there exists a unidirectional unit square in the grid. Take the shape with smallest area in the grid which has unidirectional boundary. If it is not a unit square, then it contains an inner edge....