The chart-problem; problem solving In how many ways can we construct a $6\times 6$ chart with only $1$ and $-1$ such that in every row and column, the product is always positive?
 A: There are $\color{red}{2^{25}}$ ways of filling a $6\times 6$ grid with $0$ or $1$ in such a way that along every row or column there is an even number of $1$s. We may first fill a $5\times 5$ subgrid in any way we like, then complete those rows and columns in the only way they fit our constraints. The only square we still have to fill has to be filled with the parity of the number of $1s$ in the chosen $5\times 5$ subgrid:
$$\begin{pmatrix} 0 & 0 & 1 & 0 & 0 & ? \\
                  0 & 1 & 1 & 1 & 1 & ? \\
                  1 & 1 & 0 & 0 & 1 & ? \\
                  1 & 0 & 1 & 1 & 0 & ? \\
                  0 & 1 & 1 & 0 & 1 & ? \\
                  ? & ? & ? & ? & ? & ?  \end{pmatrix}\mapsto
\begin{pmatrix} 0 & 0 & 1 & 0 & 0 & \color{red}{1} \\
                  0 & 1 & 1 & 1 & 1 & \color{red}{0} \\
                  1 & 1 & 0 & 0 & 1 & \color{red}{1} \\
                  1 & 0 & 1 & 1 & 0 & \color{red}{1} \\
                  0 & 1 & 1 & 0 & 1 & \color{red}{1} \\
                  \color{red}{0} & \color{red}{1} & \color{red}{0} & \color{red}{0} & \color{red}{1} & \color{blue}{0}  \end{pmatrix}.$$
A: There are $2^{25} $ combinations. Like Jack's answer fill the $5 \times 5$ subgrid. I will prove that there is a unique $6 \times 6$ grid with the desired property.
First we add the correct $1$'s or $0$'s in the first five places of the last row and column. We only have to determine the last digit in the lower left corner.
We can chose this number so that there is an even amount of 1's in the last column. I will prove that in this case the last row also has an even amount of 1's.
Consider the 6x6 grid as constructud above and call it $A$.
I do all algebraic calculations in $\mathbb{Z}_2$. So $1+1=0$. 
Denote $e_i = \begin{pmatrix}
  0 \\
  \vdots \\
1 \\
\vdots \\
  0
 \end{pmatrix}$ where $1$ is placed in the $i$-th row. 
Because in each column we have an even amount of $1$'s:
$$\begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix} A .e_i= 0 , \forall i $$
Because in the first 5 rows we have an even amount of $1$'s:
$$ e_j .A \begin{pmatrix}
  1 \\
  \vdots \\
  1
 \end{pmatrix} = 0 , \forall j \neq 6 $$
If we can prove that the last one also holdsfor $j=6$ we are done.
We have: $$e_6 = \begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix} - e_1 -e_2- \cdots - e_5$$
We have:
 $$  \: \:  \: \: e_6 .A \begin{pmatrix}
  1 \\
  \vdots \\
  1
 \end{pmatrix} =\left( \begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix} - e_1 -e_2- \cdots - e_5 \right)A \begin{pmatrix}
  1 \\
  \vdots \\
  1
 \end{pmatrix} \: \: \: \: \: \:  \: \: (1)$$
We calculate
$$\begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix}A\begin{pmatrix}
  1 \\
  \vdots \\
  1
 \end{pmatrix} = \begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix}A(e_1+e_2+\cdots+e_6) = 0+ 0+\cdots+0 = 0$$
If we plug this in in $(1)$ and use the property of the columns the result follows. Thus starting with the $5 \times 5$ subgrid and constructing the $6 \times 6$ grid always gives you a correct grid. 
