7.50 Given that $n$ planes divide a cube into 300 parts, find the minimum value of $n$.


［Solution］First, note that it is easy to prove by mathematical induction that $n$ lines can divide the plane into at most
$$
p_{n}=\frac{1}{2} n(n+1)+1
$$

parts, and this is achieved if and only if the $n$ lines are in general position, i.e., no two lines are parallel and no three lines intersect at the same point. In this case, the $n$ lines exactly divide the plane into $p_{n}$ parts.
Second, we use mathematical induction to prove that $n$ planes can divide the space into at most
$$
q_{n}=\frac{1}{6}\left(n^{3}+5 n+6\right)
$$

parts, and this is achieved if and only if the $n$ planes are in general position, i.e., the intersection lines of any two planes with the third plane are not parallel, and no four planes intersect at the same point. In this case, the $n$ planes exactly divide the space into $q_{n}$ parts.

When $n=0$, $q(0)=1$, i.e., the proposition holds for $n=0$. Assume the proposition holds for $n=k$. When $n=k+1$, by the induction hypothesis, the first $k$ planes can divide the space into at most $q_{k}$ parts, and exactly into $q_{k}$ parts if the $k$ planes are in general position. Now, add one more plane, which intersects the first $k$ planes in $k$ lines. These $k$ lines can divide the $(k+1)$-th plane into at most $p_{k}$ parts, and exactly into $p_{k}$ parts if the $k$ lines are in general position. Each of these at most $p_{k}$ planar regions divides its corresponding spatial region into two, thus increasing the number of spatial regions by at most $p_{k}$. Therefore, we have
$$
\begin{aligned}
q_{k+1} & \leqslant q_{k}+p_{k}=\frac{1}{6}\left(k^{3}+5 k+6\right)+\frac{1}{2} k(k+1)+1 \\
& =\frac{1}{6}\left[(k+1)^{3}+5(k+1)+6\right] .
\end{aligned}
$$

The equality holds if and only if the $k+1$ planes are in general position, completing the induction proof.
From (2), we have
$$
q_{12}=299<300<378=q_{13},
$$

Thus, to divide a cube into 300 parts, 12 planes are insufficient, and at least 13 planes are required.

Using 12 planes in general position to divide the space into 299 parts, then take a sufficiently large cube that contains all the $C_{12}^{3}$ intersection points of 3 planes within the cube, and ensure that the vertex $A$ of the cube is not on any of the 12 planes. Thus, the 12 planes divide the cube into 299 parts. Let the shortest of the 3 edges emanating from vertex $A$ be $\alpha$. On the 3 edges emanating from vertex $A$, take points $P, Q, R$ such that $A P=A Q=A R=\frac{\alpha}{2}$, then the plane $P Q R$ divides the polyhedral part containing point $A$ into two parts without intersecting any other parts. Thus, these 13 planes exactly divide the cube into 300 parts.
In summary, the minimum number of planes $n$ required is 13.