switch the colour until only one black square is left Consider a standard chess board (8 × 8 squares). In each move, you pick one row or one column and switch the colours of all 8 squares (from black to white or from white to black). Is it possible to do this in such a way that exactly one square is black? If possible, say how.
 A: At any stage $n$, let $A_n$ be the total number of white and $B_n$ the total number of black. At the beginning ($n=0$), $A_0-B_0$ is divisible by $4$. We show that any legal move changes $A_n-B_n$ by a multiple of  $4$. 
For let the number of white in the row/column we switch be say $a$, and the number of black be $b$. Note that $a+b=8$.
When we do the switching, the total number of white "increases" by $b-a$, and the number of black increases by $a-b$, so 
$$A_{n+1}-B_{n+1}=(A_n+b-a)-(B_n+a-b)=A_n-B_n+2(b-a).$$
But since $b+a$ is even, it follows that $b-a$ is even, so $2(b-a)$ is divisible by $4$, and therefore $A_{n+1}-B_{n+1}$ is divisible by $4$.
That shows we cannot end up with a single black, for then we would have $A-B=62$, and $62$ is not a multiple of $4$. 
Remark: Many "impossibility" arguments rely on finding an appropriate invariant. In our case, the remainder of $A-B$ on division by $4$ is unchanged by any legal move.
A: HINT: Show that after each switch you have an even number of white squares.
