I have a short parity problem on Numberplay (the NYTimes blog) this week! Thanks to Gary Antonick for the great image and the opportunity to interact with his fantastic readers.I have some variations in mind, to keep making the problem harder once people finish off this easy beginning. I’m curious what variations you can think of to add to the collection!

Feel free to comment here or there or both.

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OK, I’ll play… 😉

For the original game, player 1 will win iff the sum of the eight numbers chosen is odd. Since +1=-1 mod 2, there is nothing player 2 can do to affect the overall parity.

If x is allowed, player 2 can win. If all 8 numbers player 1 chooses are odd, player 2 wins by choosing 6 +’s and an x. Otherwise player 2 wins by choosing 7 x’s.

oops — small mistake in my analysis of the game with x’s. If all 8 numbers player 1 chooses are odd, player 2 wins by choosing 7 +’s. Otherwise, player 2 wins by choosing 7 x’s.

I think what I had in mind there was that player 2 is allowed just one x sign — so there are two questions, what if they must use exactly one x sign, and what if they may use one or none?

Then there are the obvious extensions and variations of that.

Ah… Sorry about the misinterpretation. In that case player 2 can with either by playing 7 +’s, or, if that doesn’t work, 6 +’s and a x where the x is between an odd and an even. This position is guaranteed to exist (or else 7 +’s would have won) and is also guaranteed to change the parity of the result.

Cool puzzle, Josh — I am curious to see where it ends up going.😉

Nice job, but there’s still one of the two questions from my previous comment left to defeat: what if player 2 must use exactly one x sign?

Sorry — my reading comprehension has gone to the dogs. 😉

For no x’s : Player 1 wins by choosing 7 odds and an even.

For 0 or 1 x’s: Player 2 wins by choosing no x’s (if that works) or by placing an x between an odd and an even.

For exactly 1 x: Player 1 wins by choosing 8 odds. (this also works if Player 2 must use an odd number of x’s.

Great. Now let’s keep it going!

What about exactly 2 x’s?

What about exactly one x, and one set of parentheses?

2 x’s: Player 1 wins by choosing 7 odds and an even. No matter where player 2 places the x’s, the result will be a sum of 5 odds and an even.

1 x, 1 () : Gotta think this one out more.

BTW — I really like how this problem is evolving, Josh. Thanks for bringing it to my attention!

Great, glad you’re having fun!

The next place to go is to restrict player 1’s choice of numbers. For instance, what if they have to choose from 1 through 10, without replacement? In other words, they can have at most 5 odds. Can they still win? A lot of your strategies so far won’t work any more, but maybe there are other ways to win.

There’s some good discussion over on the Numberplay blog’s comments too.

Reblogged this on Metroplex Math Circle.