Permalink Submitted by Anonymous on August 31, 2011

Instead of using the Klein group, I used the symbols 1 and 0, and added mod 2. Assuming this is legitimate, I could derive the same result about possible positions when one marble is left. For example, if the center hole has a 1 then there will be an odd number of 1's in the remaining holes, meaning that the sum is odd. All holes labeled 0 are eliminated as possibilities. Using the same type of symmetry arguments, there are only 2 types of position that do not have 0.

## Using mod 2 arithmetic

Instead of using the Klein group, I used the symbols 1 and 0, and added mod 2. Assuming this is legitimate, I could derive the same result about possible positions when one marble is left. For example, if the center hole has a 1 then there will be an odd number of 1's in the remaining holes, meaning that the sum is odd. All holes labeled 0 are eliminated as possibilities. Using the same type of symmetry arguments, there are only 2 types of position that do not have 0.