411. A parallelepiped is composed of identical cubes. Three faces of the parallelepiped, having a common vertex, were painted. It turned out that at least one face of half of all the cubes is painted. How many cubes have painted faces?
411. Answer: the number of painted cubes can take one of five values: $60, 72, 84, 90, 120$. Let the parallelepiped have dimensions $m \times n \times k(k \leqslant n \leqslant m)$. The total number of unpainted cubes will be $(m-1)(n-1)(k-1)$. According to the condition $m n k=$
$-2(m-1)(n-1)(k-1), \quad$ from which it follows that $2<k<5$. It remains to solve the equation in integers for $m$ and $n$ when $k=3$ and $k=4$.