The positive integers are partitioned into 2 sequences $a_1<a_2<\dots$ and $b_1<b_2<\dots$ such that $b_n=a_n+n$ for every positive integer $n$. Show that $a_n+b_n=a_{b_n}$.
Problem
Source: 2023 Serbia TST Problem 3
Tags: Serbia, TST, algebra, broken
22.05.2023 16:13
Here's my goofy solution from contest, there are other cooler solutions that I'll let someone else post.
22.05.2023 16:20
@gvole Can you please post the rest of the problems from Serbia TST (except for the ISL ones, if there are such)? Sorry for posting something not related to the problem.
22.05.2023 16:21
I am checking if I am allowed to do so. They haven't disclosed anything about the origin of the problems so far.
22.05.2023 16:30
and the conclusion is now straightforward.
24.05.2023 23:22
This problem is easy once you know IMO 1993 P5.
24.05.2023 23:54
math90 wrote: This problem is easy once you know IMO 1993 P5. How exactly? I knew that problem and didn't see any obvious connections?
25.05.2023 00:56
In this case $a_n=f(n)$ and $b_n=f(f(n))$.
25.05.2023 01:14
Oh, true. That's cool. This year's problems won't be remembered for their originality, that's for sure!
25.05.2023 01:41
about specific types of partitions, so it can also give a clue.