For a fixed $k$ with $4 \le k \le 9$ consider the set of all positive integers with $k$ decimal digits such that each of the digits from $1$ to $k$ occurs exactly once.
Show that it is possible to partition this set into two disjoint subsets such that the sum of the cubes of the numbers in the first set is equal to the sum of the cubes in the second set.
One set contains all integers for which the decimal digits are an even pemutation of $\{1,2,\cdots,k\}$. There other set contains all odd permutations.
Interestingly, this partition is not unique, at least for $k = 4$. My code found $5$ partitions: https://gist.github.com/Sky-Nik/dc75241e2df6a4a1a5ab4f2d2f459b1b
NikitaSkybytskyi wrote:
Interestingly, this partition is not unique, at least for $k = 4$. My code found $5$ partitions: https://gist.github.com/Sky-Nik/dc75241e2df6a4a1a5ab4f2d2f459b1b
For k=4, since the even/odd permutation partition holds for up to 5th powers, it is not surprising that if only the cube is required, the partition is not unique.
But what about only requiring the fourth or fifth power? Is the partition unique?