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> | + | <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. | ||
~Tomas Diaz. ~orders@tomasdiaz.com | ~Tomas Diaz. ~orders@tomasdiaz.com | ||
Latest revision as of 09:29, 23 December 2023
Problem
Let 
be a natural number. A cube with edge can be divided into 1996 cubes whose edges are also natural numbers. Determine the smallest possible value of .
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
A cube with edge can be divided at the most into 
cubes with side 1. Since 
then smallest 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 
, 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, 
cubes into a cube of edge 
and so on...
We can express the volume of the cube as:
where 
is the quantity of cubes of edge 
, Hence, we want to find solve this equation for with 
Starting with 
with 201 cubes left we find that 
. So, we can make 25 cubes of edge 2. If we set 
=25 we need to subtract 200 from 
and our numbers are now at:
, 
, 
. So we still have 26 cubes more than we need.
So we can take 27 cubes of edge 1 (subtract 27 from ) and combine them into one cube of edge three and our new numbers are:
, , 
, 
. 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 is greater than 12, and 
is possible then smallest possible 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.
