Imagine a prison consisting of \(64\) cells arranged like the squares of an \(8\times8\) chessboard. There are doors between all adjoining cells. A prisoner in one of the corner cells is told that he will be released, provided he can get into the diagonally opposite corner cell after passing through every other cell exactly once. Can the prisoner obtain his freedom?

This is not an original problem.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestImagine the cell floors painted like on a chess board, and suppose the starting cell floor is black. Then the cell floor of the opposite corner cell will be black as well.

Now every move by the prisoner results in a change in floor color. Further, each odd-numbered move leaves the prisoner in a white-floored cell. So after any path of \(63\) moves he will always be in a white-floored cell, which means he can never end up in the (black-floored) opposite corner cell after \(63\) moves, and hence there is no way he can complete the freedom-giving task. :(

Log in to reply

Poor prisoner :(

Log in to reply

Haha. Yes, so sad. :(

Log in to reply