Did you solve it? Are you smarter than a maths jammer?

Alex works out the mutilated chessboard answer Photograph: Guardian

Earlier today I set you two mutilated chessboard puzzles.

1) The classic:

Imagine you have a chessboard and 32 dominoes. Each domino is exactly the size of two adjacent squares on the board, which means there is a way of placing the 32 dominoes so that they cover all 64 of the chessboard squares.

Now mutilate the chessboard by cutting off two squares at diagonally opposite corners of the board, and lose one of the dominoes. Is it possible to place the 31 dominoes on the board so that all the remaining 62 squares are covered? Show how it can be done, or prove it impossible.

2) The other problem:

Mutilate the chessboard by cutting out any two squares, one of each colour. Is it possible to place the 31 dominoes on the board so that all the remaining 62 squares are covered? Show how it can be done, or prove it impossible.

Watch the solution in video form:

Mutilated chessboard solutions

The answers

1) The board missing two opposite corners cannot be covered with 31 dominoes.

Each domino will always cover two adjacent squares of the chessboard. Since adjacent squares have different colours, each domino placed on the board must therefore cover two different colours.

Imagine it is possible to cover the mutilated board. You place down the dominoes and you will get to the stage when there are only two free squares remaining. The insight here is to realise that these two remaining squares must be of the same colour. (This is because the two opposite corners that were cut out the board were of the same colour, so the mutilated board no longer has equal numbers of squares in each colour.) If the two remaining squares are of the same colour, they are not adjacent and cannot be covered by the last domino.

2) The board missing any two squares, one of each colour, can be covered with 31 dominoes.

First, consider a path through the board that visits every square just once.

Here’s a path through every square. There are many of them, and they are all valid.

Now remove one square of each colour.

I chose these two squares, but the argument is valid with any two of different colour.

What you have done is to cut this path into two pieces. Each path must be of even length, and can therefore be covered. The argument follows for all paths, and all choices of two differently-coloured squares.

If you found those problems too easy, then Colin Wright has set the following extra challenges:

1. Can you cover the board with dominoes if you cut out two squares of each colour?
   
2. Can you cover the board with dominoes if you cut out three squares of each colour?
   

In both cases the remaining squares must be connected, so no single or group of squares is cut off from the rest.

The maths is interesting and Colin has kindly provided the answers and a discussion on his website.

I post a puzzle here on a Monday every two weeks. If you like this sort of thing check out my other Guardian blog Adventures in Numberland. You can also check me out on Twitter, Facebook, Google+ and my personal website.

If you like gazing at mathematical images and colouring them in, my latest book, Snowflake Seashell Star: Colouring Adventures in Numberland, is out now.

And if know of any great puzzles that you would like me to set here, get in touch.