Three strictly increasing sequences \[a_1, a_2, a_3, \ldots,\qquad b_1, b_2, b_3, \ldots,\qquad c_1, c_2, c_3, \ldots\] of positive integers are given. Every positive integer belongs to exactly one of the three sequences. For every positive integer $n$, the following conditions hold: (a) $c_{a_n}=b_n+1$; (b) $a_{n+1}>b_n$; (c) the number $c_{n+1}c_{n}-(n+1)c_{n+1}-nc_n$ is even. Find $a_{2010}$, $b_{2010}$ and $c_{2010}$. (4th Middle European Mathematical Olympiad, Team Competition, Problem 1)
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Tags: induction, algebra solved, algebra
Matematika
13.09.2010 10:14
I think that this problem is perfect for team competition because if you write first 16 numbers and conclude where they belong and from that you can conclude general rule it is easy to prove that the solutions are $ a_{n}=n^{2}\ \ b_{n}=(n+1)^{2}-2 $ and $ c_{i^2}=(i+1)^2-1$ for all natural i $ c_{i^2+j}=(i+1)^2+j$ for all natural i and j for which $(i+1)^2>j>i^2$ so $a_{2010}=2010^{2}$ $b_{2010}=2011^{2}-2$ $c_{2010}=45^2+74=2099$ Hope I didn't mees up with calculations
Swistak
13.09.2010 14:54
Matematika wrote: I think that this problem is perfect for team competition because if you write first 16 numbers and conclude where they belong and from that you can conclude general rule Yes, if you write first 16 numbers you can guess general rule, but this is why this problem is totally idiotical not perfect! We should do million supposes and get from this million contradictions...
Matematika
14.09.2010 20:15
I respect your opinion but the thing for all such problems is that you can simply use induction when you notice the rule and prove it using that. As well if on the competition you have 6 people as a team somebody can always spend some time to find rules in this type of problems. I would type the induction proof but I'm too lazy for that and it isn't anything smart(for that part I agree).