MathLabs

Problem 1

Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum.
Step 4 of 4: Conclude the disjoint subsets have equal sum
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In plain words

Subtracting the shared items from both piles removes the same amount of weight from each side, so if the totals were equal before, they stay equal afterwards — and now the remaining piles no longer overlap.

sum(A′)=sum(A)−sum(A∩B)=sum(B)−sum(A∩B)=sum(B′)\text{sum}(A')=\text{sum}(A)-\text{sum}(A\cap B)=\text{sum}(B)-\text{sum}(A\cap B)=\text{sum}(B')
Detailed analysis

Because sum(A)=sum(A∩B)+sum(A′)\text{sum}(A)=\text{sum}(A\cap B)+\text{sum}(A') and sum(B)=sum(A∩B)+sum(B′)\text{sum}(B)=\text{sum}(A\cap B)+\text{sum}(B'), and sum(A)=sum(B)\text{sum}(A)=\text{sum}(B), subtracting the common term sum(A∩B)\text{sum}(A\cap B) from both sides gives sum(A′)=sum(B′)\text{sum}(A')=\text{sum}(B'). Thus A′A' and B′B' are two disjoint, nonempty subsets of the original ten two-digit numbers with the same sum, as required.