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Canadian Mathematical Olympiad

Combinatorics Difficulty 6.3 National Olympiad Prove it Canada

Problem:

Let SS be a set of n≥3n \geq 3 positive real numbers. Show that the largest possible number of distinct integer powers of three that can be written as the sum of three distinct elements of SS is n−2n-2.

Solution

Solution:

We will show by induction that for all n≥3n \geq 3, it holds that at most n−2n-2 powers of three are sums of three distinct elements of SS for any set SS of positive real numbers with ∣S∣=n|S|=n. This is trivially true when n=3n=3.

Let n≥4n \geq 4 and consider the largest element x∈Sx \in S. The sum of xx and any two other elements of SS is strictly between xx and 3x3x. Therefore xx can be used as a summand for at most one power of three. By the induction hypothesis, at most n−3n-3 powers of three are sums of three distinct elements of S∖{x}S \setminus \{x\}. This completes the induction.

Even if it was not asked to prove, we will now show that the optimal answer n−2n-2 is reached. Observe that the set S={1,2,32−3,33−3,…,3n−3}S=\{1,2,3^{2}-3,3^{3}-3, \ldots, 3^{n}-3\} is such that 32,33,…,3n3^{2}, 3^{3}, \ldots, 3^{n} can be expressed as sums of three distinct elements of SS. This makes use of the fact that each term of the form 3k−33^{k}-3 can be used in exactly one sum of three terms equal to 3k3^{k}.

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