IMO 1983 Problem 5

I knew this construction, so it's not by me.

Is it possible to choose $1983$ distinct positive integers, all less than or equal to $10^5$, no three of which are consecutive terms of an arithmetic progression?

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Yes. Consider all natural numbers no more than $11$ digits in base $3$ representation without digits $2$. The largest is $$1+3+\dots+3^{10}<10^5,$$and there are $2^{11}>1983$ of them.