True/False ?
In any set of ten 2 digit numbers, there always exist two non-empty disjoint sets and such that the sum of the numbers in is equal to sum of numbers in .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Given such a set S , there exist 2 1 0 − 2 = 1 0 2 2 different non-trivial subsets. The maximum sum in such a subset is 9 1 + ⋯ + 9 9 = 8 5 5 ; the minimum sum is 1 0 . Thus we have a mapping Σ : P ( S ) − { ∅ , S } → { 1 0 , ⋯ , 9 4 5 } . Since the codomain is strictly smaller than the domain, this function cannot be injective. There exist distinct A , B ⊂ S such that Σ ( A ) = Σ ( B ) .
These sets are not necessarily disjoint. However, we may remove common elements: A ′ = A − ( A ∩ B ) and B ′ = B − ( A ∩ B ) . It is obvious that Σ ( A ′ ) = Σ ( B ′ ) , that they are not empty, and that they are disjoint.
Note that there exists a set of seven 2-digit numbers for which this property does not hold; for instance, S = { 1 6 , 1 7 , 2 4 , 3 2 , 3 4 , 6 4 , 6 8 } .
I don't think there are any sets of eight or nine 2-digit numbers for which this is true. However, 2 9 − 2 = 5 1 0 < 7 6 4 = 9 2 + ⋯ + 1 0 0 , so that the reasoning above won't work.