All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Not so trivial (Posted on 2016-03-18) Difficulty: 4 of 5
In a certain chess position, each row and each column contains an odd number of pieces.

Prove that the total number of pieces on black squares is an even number.

Source: Proizvolov, “Problems Teaching Us How to Think".

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Parity Solution | Comment 1 of 6
There must be an even number of pieces (the sum of eight odd numbers is even).  Furthermore, each of the four stripe parities must each be even (sum of four odd numbers is even): odd rows, even rows, odd columns, and even columns.

Let X be the number of pieces on odd row white tiles.  Let Y be the number of pieces on the odd row black tiles.  X+Y is even, which means that X and Y are the same parity.

Let V be the number of pieces on odd column white tiles.  Let W be the number of pieces on the odd column black tiles.  V+W is even, which means that V and W are the same parity.

The odd row white tiles and the odd column white tiles are the same tiles, so X=V.  Then all four X, Y, V, and W are the same parity.

All the black tiles are the union of the odd row black tiles and the odd column black tiles.  Then the number of pieces on black tiles is Y+W.  Since Y and W are the same parity their sum must be even!

Edited on March 24, 2016, 11:40 am
  Posted by Brian Smith on 2016-03-18 23:47:02

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information