Distributing numbers
Given the distribution of numbers shown above, it is requested:
- Distribute the numbers in two groups A and B of 8 numbers each.
- The sum of all the numbers in group A must equal the sum of all the numbers in group B.
- The sum of all the squares of the numbers in group A must be equal to the sum of all the squares in group B.
Is it possible to satisfy the requested conditions for the distribution of numbers?
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2 solutions
Yes, { 1 , 4 , 6 , 7 , 1 0 , 1 1 , 1 3 , 1 6 } . Pretty amazing.
I guess the natural question is: for any positive integer k , is there a number N such that { 1 , … , N } is the union of two disjoint subsets such that, for all nonnegative integers d ≤ k , the sum of the d th powers of the elements of each subset is the same? Your observation shows that if k = 3 then N = 1 6 works.
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If k=4, then N=32 works and if k=5, then N=64 works.
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So...given a positive integer k , can we always split { 1 , 2 , … , 2 k + 1 } into two subsets S 1 and S 2 such that, for every integer d such that 0 ≤ d ≤ k , x ∈ S 1 ∑ x d = y ∈ S 2 ∑ y d ?
It turns out there are seven ways to do this. Here are all seven possibilities for the set that contains 1 (found by a computer search): { 1 , 2 , 7 , 8 , 1 1 , 1 2 , 1 3 , 1 4 } { 1 , 4 , 6 , 7 , 9 , 1 2 , 1 4 , 1 5 } { 1 , 4 , 6 , 7 , 1 0 , 1 1 , 1 3 , 1 6 } { 1 , 4 , 5 , 8 , 1 0 , 1 1 , 1 4 , 1 5 } { 1 , 3 , 6 , 8 , 1 0 , 1 2 , 1 3 , 1 5 } { 1 , 3 , 6 , 9 , 1 0 , 1 1 , 1 2 , 1 6 } { 1 , 5 , 6 , 7 , 8 , 1 1 , 1 4 , 1 6 }
In fact, there is even a solution in which the sums of the cubes are equal as well.