TITLE: If a set of integers can be partitioned into 3 subsets with equal sums, if one such subset is identified, can the remaining be partitioned as well?
QUESTION [7 upvotes]: 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.

REPLY [11 votes]: $\{1,3,4,5,8,14,16\}$
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.