olymmath

ISL 2009 C2

  • Answer

For any integer n≥2n \geq 2, let N(n)N(n) be the maximal number of triples (ai,bi,ci),i=1,…,N(n)\left(a_{i}, b_{i}, c_{i}\right), i=1, \ldots, N(n), consisting of nonnegative integers ai,bia_{i}, b_{i} and cic_{i} such that the following two conditions are satisfied: (1) ai+bi+ci=na_{i}+b_{i}+c_{i}=n for all i=1,…,N(n)i=1, \ldots, N(n), (2) If i≠ji \neq j, then ai≠aj,bi≠bja_{i} \neq a_{j}, b_{i} \neq b_{j} and ci≠cjc_{i} \neq c_{j}. Determine N(n)N(n) for all n≥2n \geq 2. Comment. The original problem was formulated for mm-tuples instead for triples. The numbers N(m,n)N(m, n) are then defined similarly to N(n)N(n) in the case m=3m=3. The numbers N(3,n)N(3, n) and N(n,n)N(n, n) should be determined. The case m=3m=3 is the same as in the present problem. The upper bound for N(n,n)N(n, n) can be proved by a simple generalization. The construction of a set of triples attaining the bound can be easily done by induction from nn to n+2n+2.

Dataset text NuminaMath olympiads. Quoted with source.

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N(n)=\left\lfloor\frac{2 n}{3}\right\rfloor+1

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