Find the smallest positive integer $\mathrm{n}$ so that a cube with side $\mathrm{n}$ can be divided into 1996 cubes each with side a positive integer.
Answer: 13.

Divide all the cubes into unit cubes. Then the 1996 cubes must each contain at least one unit cube, so the large cube contains at least 1996 unit cubes. But $12^{3}=1728<1996<2197=13^{3}$, so it is certainly not possible for $\mathrm{n}<13$.

It can be achieved with 13 by $1.5^{3}+11.2^{3}+1984.1^{3}=13^{3}$ (actually packing the cubes together to form a $13 \times 13 \times 13$ cube is trivial since there are so many unit cubes).

## Problem 2

$M$ is the midpoint of the median $AD$ of the triangle $ABC$. The ray $BM$ meets $AC$ at $N$. Show that $AB$ is tangent to the circumcircle of $NBC$ iff $BM / BN = (BC / BN)^2$.