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Proof: Assume that S-{i_{1}, i_{2}} cannot be partitioned, while S can be partioned, than in all partitions of S i_{1} and i_{2} must be in different sets. There must be a partitioning P_{1}, P_{2}, .. ,P_{n} such that P_{1} contains i_{1} and P_{2} contains i_{2}. Let {i_{1}, i_{3}, .. ,i_{k}} (with k>3) the set P_{1}, than i_{3}, .. ,i_{k} sum up to m-i_{1}, which by assumption is equal to i_{2}. It is now possible to construct a partitioning P'_{1}, P'_{2}, .. ,P_{n} where P'_{2} equal P_{2} with i_{2} replaced by i_{3}, .. ,i_{k} and P'_{1} equal to {i_{1}, i_{2}}. It is clear that P'_{2}, .. ,P_{n} is a partitioning of S-{i_{1}, i_{2}} 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:
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