Pennies on a checkerboard. Here is a question on pennies on checkerboard. It isnt a homework question. I saw it in a book. 
Pennies are placed on an 8 × 8 checkerboard in an alternating pattern of heads and
tails.

i. You are allowed to make moves where in each move you turn over exactly two pennies
   that lie next to each other in the same row or column. Can you take a sequence of
   moves that leaves just one penny face up?

ii. You are allowed to make moves where in each move you turn over exactly three
    neighboring pennies that lie in the same row or column of the checkerboard. Can you
    take a sequence of moves that leaves just one penny face up?

I am able to do the first part using parity. The answer to the first part is "NO". But I cant figure out how to go about the second part. Please explain.
Thank you.
 A: Consider the following coloring, where a, b, and c are colors, and lower-case means tails, while upper case means heads:
aBcAbCaB 
BcAbCaBc 
cAbCaBcA 
AbCaBcAb 
bCaBcAbC 
CaBcAbCa 
aBcAbCaB 
BcAbCaBc 

Note that any legal move switches the status of exactly one coin of each color. This means that parity conservation of number of heads holds for pairs of colors. For instance, "the number of heads of color A + the number of heads of color B" always has the same parity. But in the original configuration all three colors have an even number of heads. Therefore, you can never reach a configuration in which only one color has an odd number of heads, which would be the case if there were only one head on the board.
Interestingly, the same argument doesn't work for tails, unless you take the mirror image of the coloring. The original configuration has an odd number of tails for both a and c, and an even number for b, so one could (theoretically) have a situation in which tails(a) ~= tails(c) ~= 0 (mod 2) and tails(b) ~= 1 (mod 2).
