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Specifically, if a set $S$ of integers that sums to $k$ is known to be able to be partitioned into $3$ subsets such that each subset sums to $\dfrac{k}{3}$, if $A$ is one such subset of $S$, is it always possible to partition the remaining integers of $S-A$ into two subsets that both sum to $\dfrac{k}{3}$?

For example, if $S = \{2, 3, 4, 6, 7, 8\}$ and sums to $30$, it can be divided into sets $\{2, 8\}$, $\{3, 7\}$, and $\{4, 6\}$ which each sum to $\dfrac{30}{3}=10$. This is a simple example, but if an arbitrary set $S$ that sums to $k$ is known to be able to be partitioned into 3 subsets that each sum to $\dfrac{k}{3}$, if one such subset $A$ is identified, is it guaranteed that the remaining integers in $S-A$ can also be partitioned into subsets that each sum to $\dfrac{k}{3}$?

My gut feeling is that the answer is yes, but I don't know how to prove it.

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I wonder if your feeling is correct if you require each subset to contain one third of the elements. – Ethan Bolker Jan 4 at 0:47
    
@EthanBolker My intuition is that it could still fail. Even so, "containing one third of the elements" wouldn't make sense if the total number of elements wasn't a multiple of $3$ to begin with. The same tactic I used to construct the counterexample below should probably still hold. Have precisely three odd elements and the correct partition has exactly one odd element each, and find a subset of correct size and total using all odd elements. – JMoravitz Jan 4 at 0:56
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@EthanBolker Playing with it a moment found $\{1,3,5,4,0,8,-2,12,-4\}=\{0,4,5\}\cup \{3,8,-2\}\cup \{1,12,-4\}$. partitioned into subsets of size three each of whose sum is equal to $9$. $\{1,3,5\}$ is also a subset of size three whose sum is $9$. All remaining elements are even and could therefore not be partitioned into sets adding to $9$. If the fact that some are negative bothers you, then add four to every number, or five or more. – JMoravitz Jan 4 at 1:00
    
@JMoravitz What if the set to partition is the interval $\{1, \ldots , 3n\}$? See arxiv.org/abs/1304.6756. published in Journal of Combinatorics (JOC) Vol. 7.1. – Ethan Bolker Jan 4 at 1:14
up vote 10 down vote accepted

$\{1,3,4,5,8,14,16\}$

can be partitioned into the sets

$\{1,16\},\{3,14\},\{4,5,8\}$, each of which add to $17$.

The subset $A=\{1,3,5,8\}$ is also a subset which adds to $17$, however the remaining integers $\{4,14,16\}$ are all even and therefore could not possibly be partitioned into subsets which add to an odd number.

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I'm not surprised by many upvotes for this answer, but given that I am surprised at so few for the question. – Ethan Bolker Jan 4 at 0:46

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