Here's a small puzzle with an ingenious solution:

Consider the set of numbers {-4, -3, -2, -1, 0, 1, 2, 3, 4}. Two players alternately choose one number at a time from the set (without replacement).

The first player who obtains any three out of his or her selected numbers (this may happen after (s)he has chosen 3, 4 or even 5 numbers) that sum to zero wins the game.

Now, the question is that, does either player have a forced win? That is to say, can either player always choose in a way such that a win is guaranteed?

note:
this problem is not original. I found the game on this blog .

http://recreational-math.blogspot.in

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:

TopNewestYour game is analogous to Tic Tac Toe. So no, there is no forced win.

Log in to reply