The biggest problem with the prisoner's proof is that his model is incomplete. One piece of that is that it merges two key variables - days on which the execution can happen, and days on which the ...

Assuming that the anchor will actually sink and reach the bottom of the very small lake, the water level of the lake will Reason: More intuitive reasoning:

There is no paradox. The teacher should get paid, one way or another. The key to understanding the situation is realizing there are multiple slightly different scenarios that are all being described ...

He should open the Explanation of the logic (assumes the inscriptions are intended to be helpful): I'm also assuming here that the dragon referred to by the silver chest's inscription is the same ...

The easiest explanation would be that in a 3x3 cube, only one cube is out of position, but in a 4x4 cube two cubes are out of position. In a 15 puzzle (the sliding puzzle where you try to put the ...

I found a different way to solve it. Note that this doesn't disprove @trolley813's proof - it only shows one part of the proof is incorrect:

This is a supplemental to Kendall's answer that proves its correctness: Because there are four bolts, each with two possible positions, there are 24=16 states. You might think we'd want to categorize ...

First of all, notice that by repeatedly applying $\bf o$, we can always reduce a number to 2. This should be readily apparent, but here's a non-rigorous proof of the fact: 2 has a binary ...

For simplicity, I'm going to refer to the associates as knight, knave, and joker. Also, I'm assuming that they won't reveal the location of the money. If they did, then obviously you would switch to ...

Let's lay down some groundwork to help us out: Let $req(n, d)$ be the number of digit $d$ required to write out all the numbers from 1 to $n$. Since each kit contains two of each digit, we have $2n$ ...

As @xnor said, we can show it is not possible by using a parity argument. Let's look at a few examples of smaller tables to illustrate the argument. Let's start with the obviously impossible 2x2: 0 1 ...

Here's the path (in green): My first step was to go through each one and mark (in orange) all the squares that could be seen by a zero. I marked the question mark and zero clues in red to help me ...

Given that your question is self referential and requests twice its own answer, we need something that is twice itself. Zero works - two times zero is still zero, so if the answer to your question is ...

There is no such number. Like Ross mentioned in his answer, this is related to Richard's paradox. At the heart of both this puzzle and Richard's paradox is (from the Wikipedia article) the ...

Easy: How does this guarantee that the message will not be decodable?

Here's one way to show the cube cannot be solved: Suppose the bottom face is white. Then the corner piece with yellow and white visible is in the right place, and is the fourth blue corner. That means ...

This might work: If Bob's boss is predictable, Bob can plan (or pretend) to This way, his boss will not know that only 50 pylons were used. It would be reasonable to assume that either 50 pylons ...

There is no paradox as long as you realize that, when you say "all adjectives can be categorized as autological or heterological", the "or" is not exclusive. To explain this a little bit better, let'...

How can you assure you get in the shortest amount of time? Unfortunately, if you know nothing about the design of the maze, then there is no optimal strategy. For example, it would be easy to design ...

I believe the twist here is that: Why do we need the twist? From there Now that we have some possibilities established for that person, let's move on to another one. We now know who is married to ...

The smallest number I could find is My process for finding it: An example of my process: UPDATE: Here's a number that is smaller that almost works: It doesn't quite work because the number ...

Variant 1: The oracle saying "not everyone has blue eyes" does not add any information here. Consider the case where there are only two people on the island, one of whom has blue eyes and the other ...

This is what was written on the paper: The quote is: How I figured this out:

Your "parallel induction" is already happening in the generally accepted solution When the solution says "A considers that B considers that ...", B, C, and so on are all arbitrary people. So ...

Mary will win 2 Euros if both of them play optimally. Frodoskywalker has given a strategy that Mary can use to accomplish this. Here's why this will happen: Mary will write down 41 M's, and Ursula ...

The numbers represent The next 5 numbers are

Like @dmg said, the culprit is I used a slightly different path to come to this conclusion, though: