Theorem: Let S be a set of natural numbers that sum up to n.m
and that can be partitioned into n sets such that each set sums up
to m, than for any pair i1 and i2
in S such that i1+i2=m, it holds
that S-{i1, i2} can always be
partitioned into n-1 sets that each sum up to m.
Proof: Assume that S-{i1, i2}
cannot be partitioned, while S can be partioned, than in all partitions of
S i1 and i2 must be in different sets.
There must be a partitioning P1, P2, .. ,Pn
such that P1 contains i1 and P2
contains i2. Let {i1, i3, ..
,ik} (with k>3) the set P1, than
i3, .. ,ik sum up to
m-i1, which by assumption is equal to
i2. It is now possible to construct a partitioning
P'1, P'2, .. ,Pn where P'2
equal P2 with i2 replaced by i3,
.. ,ik and P'1 equal to {i1,
i2}. It is clear that P'2, .. ,Pn
is a partitioning of S-{i1, i2} such
that each set sums up to m.
After some puzzling, I discovered that there is no similar theorem for each
combination of