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

(Created page with "== Problem == Let <math>n</math> be a natural number. A cube with edge <math>n</math> can be divided into 1996 cubes whose edges are also natural numbers. Determine the smalle...")
 
 
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== Solution ==
 
== Solution ==
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A cube with edge <math>n</math> can be divided at the most into <math>n^3</math> cubes with side 1.  Since <math>12^3 < 1996 < 13^3</math> then smallest <math>n</math> 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.
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Since <math>13^3-1996=201</math>, 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, <math>k^3</math> cubes into a cube of edge <math>k</math> and so on...
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We can express the volume of the cube as:
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<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>
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Starting with <math>k_1=13^3=2197</math> with 201 cubes left we find that <math>201=(25)(8^3)+1</math>.  So, we can make 25 cubes of edge 2.  If we set <math>k_2</math>=25 we need to subtract 200 from <math>k_1</math> and our numbers are now at: 
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<math>k_1=1997</math>, <math>k_2=25</math>, <math>\sum_{i}^{}k_i=2022</math>.  So we still have 26 cubes more than we need.
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So we can take 27 cubes of edge 1 (subtract 27 from <math>k_1</math>) and combine them into one cube of edge three and our new numbers are:
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<math>k_1=1970</math>, <math>k_2=25</math>, <math>k_3=1</math>, <math>\sum_{i}^{}k_i=1970+25+1=1996</math>.  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.
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Since smallest <math>n</math> is greater than 12, and <math>n=13</math> is possible then smallest possible <math>n</math> is 13.
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~Tomas Diaz. ~orders@tomasdiaz.com
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{{Alternate solutions}}
  
 
== See also ==
 
== See also ==
 
https://www.oma.org.ar/enunciados/ibe11.htm
 
https://www.oma.org.ar/enunciados/ibe11.htm

Latest revision as of 09:29, 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 $\sum_{i}^{}k_i=1996$

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.

See also

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

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