Is there a convex polyhedron with product of numbers of faces, vertices and edges equal to $3375$?
We are told that: $|E| \cdot |V| \cdot |F| = 3375$. From that we get $|E| = \frac{3375}{|V| \cdot |F|}$
Then I tried to use Eulers equation: $|F| + |V| = |E| + 2 \implies |F| + |V| = \frac{3375}{|V| \cdot |F|} + 2 \iff |V| \cdot |F|^2 + |V|^2 \cdot |F| = 3375 + 2|V| \cdot |F|$
And so I get:
$$|V| \cdot |F|^2 - 2|V| \cdot |F| + |V|^2 \cdot |F| = 3375$$
I know that from steinitz theorem, for convex polyhedron to exist it is necessary that:
- $|F| + |V| = |E| + 2$ (eulers equation)
- $3|V| \leq 2|E|$
- $3|F| \leq 2|E|$
So I tried to use those, but I don't know how / don't konw if that's the right direction.
*Also, the number $3375$ looks 'nice' in a way - it can be divided many times by $5$. Maybe there is some use in that.
Hint : Consider the parity of the numbers $|E|, |V|, |F|$ from $$|E| \cdot |V| \cdot |F| = 3375 \quad \text{and} \quad |F| + |V| = |E| + 2$$