Let $ n$ be a natural number. A cube of edge $ n$ may be divided in 1996 cubes whose edges length are also natural numbers. Find the minimum possible value for $ n$.
First notice that $n^3\geq 1996$ which implies that $n\geq 13$.
Now we just find a configuration for $n=3$.
If we use $1984$ cubes of edge $1$, $11$ of edge $2$ and $1$ of edge $5$ then the we have $1996$ cubes which form a cube of edge $13$.
Claim 1
$ n \ge 13$
Proof:
If we want to divide a cube into the maximum number of cubes, we want the edge to be $1$. So each cube with edge $n$ can be divided into at most $n^3$ cubes.
If $n = 12$ can be divided into at most $12^3 = 1728$ cubes. So $n \ge 13$
Claim 2
$n = 13$.
Proof:
Let $e$ be the length of edge of the cube.
We divide the main cube with $n=13$ into $2197$ cubes. Then we have $201$ cubes more than we want. Hence we make $25$ cubes with $e=2$ of $200$ cubes with $e=1$. Then we have $200-25 = 175$ cubes less, hence $26$ cubes more than we want. So we make a cube with $e=3$ of $27$ cubes with $e=1$: So we have $27- 1= 26$ cubes less. So we have $1996$ cubes.