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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.
1 6 11 15 21 25 29 31 35 39 45 49 54 59 2 3 9 13 19 23 27 33 37 41 47 51 57 58 3 11 13 18 23 27 28 32 33 37 42 47 49 57 5 6 11 15 21 25 29 31 35 39 45 49 54 55 6 7 9 17 19 27 29 31 33 41 43 51 53 54 7 8 17 18 19 28 29 31 32 41 42 43 52 53 8 11 13 18 23 27 28 32 33 37 42 47 49 52 9 12 17 19 22 23 27 33 37 38 41 43 48 51 11 15 16 21 24 25 26 34 35 36 39 44 45 49 12 13 19 22 23 27 29 31 33 37 38 41 47 48 13 21 22 23 27 28 29 31 32 33 37 38 39 47 16 17 19 21 26 27 29 31 33 34 39 41 43 44For size 8, only three solutions were found, whiche are:
2 17 25 32 37 38 47 52 53 58 67 68 73 80 88 103 11 25 26 29 31 41 44 46 59 61 64 74 76 79 80 94 17 23 25 32 37 38 47 52 53 58 67 68 73 80 82 88