Binary polyhedron I would like to propose this problem. I think that Euler's characteristic could be useful.
A polyhedron with more than 8 vertices is called binary if we can assign to each vertex a number from the set $\left\{-1,1\right\}$, in such way that the product of numbers in any face is $-1$.


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*a) Show that the number of vertices of binary polyhedron is divisible by $8$.

*b) Prove that a polyhedron with $2000$ vertices is binary.

 A: (a) is false. There are binary polyhedra with any number of vertices $\ge4$, since there are polyhedra with any number of vertices $\ge4$ all of whose faces have an odd number of edges. (When the number of vertices is $2k$, take a cone over a $(2k-1)$-gon; when the number of vertices is $2k+1$, take two such cones and glue them together by their $(2k-1)$-gons.) Then simply assign $-1$ to all vertices.
A: I think that polyhedra must be changed to prism. Understanding by prism two regular n-gon(up and down) and the vertices joined by edges. In this case a) and b) are true.
a) Taking the product of all numbers on each face we have
$(-1)^{n+2}=\prod_L\times(-1)\times(-1)=1$, hence $n$ is even. 
Now consider blocks of three consecutive faces, we have $6$ possible cases,
$111,113,311,331,133,333$, where each number count the $-1's$ in each face. So each block could have $2,4$ or $6$ $-1's$. If $n=4k+2$, then we obtain $k$ blocks and $1$ face remaining(with $1$ or $3$ $-1's$). Hence the number of $-1's$($V_{-1}$) is odd. But at the same time $1=\prod_{up}\dot\prod_{down}=\prod_v=(-1)^{V_{-1}}$, so $V_{-1}$ is even, contradiction. Finally, $V=2n=8k$.
b) We make a cut on an edge and the result is a strip formed by $1000$ faces. Consider the first block of $3$ faces, and the distribution 


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*(111-1)

*(1-111)


Repeat this pattern on each block, with exception of last one, in which we write


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*(11-1-1)

*(1-1-11)


and end with $11$, the final edge, which matches with first one.
