ISL 2001 C1

by Wolstenholme, Aug 1, 2014, 9:22 PM

Let $ A = (a_1, a_2, \ldots, a_{2001}) $ be a sequence of positive integers. Let $ m $ be the number of $ 3 $-element subsequences $ (a_i,a_j,a_k) $ with $ 1 \leq i < j < k \leq 2001 $, such that $ a_j = a_i + 1 $ and $ a_k = a_j + 1 $. Considering all such sequences $ A $, find the greatest value of $ m $.

Solution:

I claim that the answer is $ 667^3 $ which can be obtained by letting $ a_i = 1, a_j = 2, a_k = 3 $ for all $ i \equiv 1 \pmod{3}, j \equiv 2 \pmod{3}, k \equiv 0 \pmod{3} $.

Now I shall show that $ 667^3 $ is the highest possible number of triples. Let $ r, s, t $ be the number of $ a_i $ that are congruent to $ 0, 1, 2 \pmod 3 $ respectively. It is clear that since each relevant $ 3 $-element subset forms a complete residue system modulo $ 3 $ so the maximum number of such subsets is $ rst $.

Now by AM-GM we have that $ rst \leq \frac{(r + s + t)^3}{27} = 667^3 $ as desired, so we are done.

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