Equal sum?

True/False ?

In any set of ten 2 digit numbers, there always exist two non-empty disjoint sets A A and B B such that the sum of the numbers in A A is equal to sum of numbers in B B .

No,It is never true Yes,it is always true It is true for some sets only

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1 solution

Arjen Vreugdenhil
Jan 11, 2018

Given such a set S S , there exist 2 10 2 = 1022 2^{10} - 2 = 1022 different non-trivial subsets. The maximum sum in such a subset is 91 + + 99 = 855 91 + \cdots + 99 = 855 ; the minimum sum is 10 10 . Thus we have a mapping Σ : P ( S ) { , S } { 10 , , 945 } . \Sigma:\ \mathcal P(S) - \{\emptyset,S\} \to \{10, \cdots, 945\}. Since the codomain is strictly smaller than the domain, this function cannot be injective. There exist distinct A , B S A, B \subset S such that Σ ( A ) = Σ ( B ) \Sigma(A) = \Sigma(B) .

These sets are not necessarily disjoint. However, we may remove common elements: A = A ( A B ) A' = A - (A \cap B) and B = B ( A B ) B' = B - (A \cap B) . It is obvious that Σ ( A ) = Σ ( B ) \Sigma(A') = \Sigma(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 = { 16 , 17 , 24 , 32 , 34 , 64 , 68 } S = \{ 16,17,24,32,34,64,68 \} .

I don't think there are any sets of eight or nine 2-digit numbers for which this is true. However, 2 9 2 = 510 < 764 = 92 + + 100 2^9 - 2 = 510 < 764 = 92 + \cdots + 100 , so that the reasoning above won't work.

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