Difference between revisions of "1996 OIM Problems/Problem 1"
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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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| + | ~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 | ||
Revision as of 09:58, 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...
~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.
