1996 OIM Problems/Problem 1

Problem

Let $n$ be a natural number. A cube with edge $n$ can be divided into 1996 cubes whose edges are also natural numbers. Determine the smallest possible value of $n$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

A cube with edge $n$ can be divided at the most into $n^3$ cubes with side 1. Since $12^3 < 1996 < 13^3$ then smallest $n$ cannot be less or equal to 12. Now we need to find out if it is possible to divide a cube of edge 13 into 1996 cubes.

Since $13^3-1996=201$ , this means that I have 201 extra cubes of side 1. Now we need to find out if we can if I can combine groups of these into other cubes of larger edge sides until my total of cubes is 1996. That is, I can combine 8 cubes into a cube of edge 2, 27 cubes into a cube of edge 3, $k^3$ cubes into a cube of edge $k$ and so on...

We can express the volume of the cube as:

$V=\sum_{i}^{}2^ik_i=13^3$ where $k_i$ is the quantity of cubes of edge $i$ , Hence, we want to find solve this equation for $k_i$ with $V=\sum_{i}^{}k_i=1996$

~Tomas Diaz. ~orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1996 OIM Problems/Problem 1

Problem

Solution

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