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 $\sum_{i}^{}k_i=1996$

Starting with $k_1=13^3=2197$ with 201 cubes left we find that $201=(25)(8^3)+1$ . So, we can make 25 cubes of edge 2. If we set $k_2$ =25 we need to subtract 200 from $k_1$ and our numbers are now at:

$k_1=1997$ , $k_2=25$ , $\sum_{i}^{}k_i=2022$ . So we still have 26 cubes more than we need.

So we can take 27 cubes of edge 1 (subtract 27 from $k_1$ ) and combine them into one cube of edge three and our new numbers are:

$k_1=1970$ , $k_2=25$ , $k_3=1$ , $\sum_{i}^{}k_i=1970+25+1=1996$ . And this proves that we can divide a cube of edge 13 into 1970 cubes of edge 1, plus 25 cubes of edge 2, plus one cube of edge 3.

Since smallest $n$ is greater than 12, and $n=13$ is possible then smallest possible $n$ is 13.

~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.

Page

Toolbox

Search

1996 OIM Problems/Problem 1

Problem

Solution

See also