Yes, that is what I am thinking. Nothing I tried looks promising.An interesting problem, but I wonder what "sum of possible a" mean? Perhaps the sum of all possible a?
I did find two solutions but my method involved brute force and intense algebra. Since the question has 2020, it's probably a competition problem. It's likely to have a prettier solution than mine.I am quite interested in this problem.
Here is what i have done so far.
I supposed p(r) =1 and solved for r. Then I set r=2020 + p(2020).
I tried other things but nothing need is coming out.
Any hints?
I'll share it with you through a private message. I don't want to hijack the thread.Post your solution and maybe someone can improve on it.
I feel that this problem is difficult enough where you can post your solution especially since your solution is not the way it needs to be done during a competition. Of course if you are uncomfortable with posting the solution here then please send me a private message.I'll share it with you through a private message. I don't want to hijack the thread.
An interesting observation.Does anyone have a better way of solving this problem or maybe some other ideas?
To atone for my mistake, it appears that [imath]p(x+p(x))=p(x)p(x+1)[/imath]Actually that should be p([x) + 2x +a + 1]
I guess that you get a two minute penalty in the penalty box (the corner)