Partition of twelve integers with a special property.

Question

Twelve numbers are selected from positive integers $1\le n\le27$ and partitioned into the following two sets $A$ and $B$ such that the power sums of $A $and $B$ are the same up to the fifth power.

$$A=\{1,5,10,18,23,27\}$$

$$B=\{2,3,13,15,25,26\}$$

We notice that we can also select twelve positive integers $1\le n\le23 $ and form sets $C$ and $D$ with the same property.

$$C=\{1,6,7,17,18,23\}$$

$$D=\{2,3,11,13,21,22\}$$

My question is what is the minimum $k$ for which this selection of $12$ integers $1\le n\le k$ is possible.

Answer 1

for the very symmetric way you constructed those, you have already found the best possible. For each, you can subtract the midpoint to get sextuples symmetric around zero:

? g=(x^2 - 25) * ( x^2 -36) * ( x^2 - 121)
%9 = x^6 - 182*x^4 + 8281*x^2 - 108900
? h=(x^2 - 1) * ( x^2 -81) * ( x^2 - 100)
%10 = x^6 - 182*x^4 + 8281*x^2 - 8100
? 
? i=(x^2 - 1) * ( x^2 -121) * ( x^2 - 144)
%11 = x^6 - 266*x^4 + 17689*x^2 - 17424
? j=(x^2 - 16) * ( x^2 -81) * ( x^2 - 169)
%12 = x^6 - 266*x^4 + 17689*x^2 - 219024
? j-i
%13 = -201600
? 

Here are your two and some worse examples:

jagy@phobeusjunior:~$ ./mse
 11 6 5     10 9 1
 13 9 4     12 11 1
 17 9 8     16 13 3
 18 11 7     17 14 3
 18 13 5     17 15 2
 19 11 8     17 16 1
 20 11 9     19 15 4

It seems possible that some less symmetric examples might do slightly better, but that is a much more elaborate and slow program