Difference between revisions of "1996 OIM Problems/Problem 1"

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We can express the volume of the cube as:
 
We can express the volume of the cube as:
  
<math>V=\sum_{i}^{}2^ik_i=13^3</math> where <math>k_i</math> is the quantity of cubes of edge <math>i</math>, Hence, we want to find solve this equation for <math>k_i</math> with <math>V=\sum_{i}^{}k_i=1996</math>
+
<math>V=\sum_{i}^{}2^ik_i=13^3</math> where <math>k_i</math> is the quantity of cubes of edge <math>i</math>, Hence, we want to find solve this equation for <math>k_i</math> with <math>\sum_{i}^{}k_i=1996</math>
  
 
~Tomas Diaz. ~orders@tomasdiaz.com
 
~Tomas Diaz. ~orders@tomasdiaz.com

Revision as of 10:11, 23 December 2023

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

~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.

See also

https://www.oma.org.ar/enunciados/ibe11.htm