Baltic Way 1997 · Problem 2

Algebra

Given a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers in which every positive integer occurs exactly once. Prove that there exist integers $\ell$ and $m$ , $1<\ell<m$ , such that $a_{1}+a_{m}=2 a_{\ell}$ .

Change pool

When you’re ready

Review material becomes available with the next Daily.

Review

Topics

Sequences and recurrences · Equations and inequalities

Solutions

Solution

Solution:

Let $\ell$ be the least index such that $a_{\ell}>a_{1}$ . Since $2 a_{\ell}-a_{1}$ is a positive integer larger than $a_{1}$ , it occurs in the given sequence beyond $a_{\ell}$ . In other words, there exists an index $m>\ell$ such that $a_{m}=2 a_{\ell}-a_{1}$ . This completes the proof.

Contest context

Results from Baltic Way 1997

11 teams

Mean score: 3.6 / 5
Scores of 4 or 5: 8 / 11
Estonia: 5 / 5

Score distribution

All team scores

Team	Score
Poland	5 / 5
Germany	5 / 5
Estonia	5 / 5
Sweden	5 / 5
Denmark	5 / 5
Latvia	5 / 5
Finland	5 / 5
Norway	5 / 5
St. Petersburg	0 / 5
Iceland	0 / 5
Lithuania	0 / 5

Review

Topics

Solutions

Contest context

Elsewhere