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### Problem 110

Posted: Tue Sep 30, 2008 1:37 am
I'm not asking for a solution to 110. However, I have an theory which I think I can prove that would make solving 110 a lot easier, but there's something not quite right with it. Is there anyone to whom I could send a private message and he (or she) could respond either affirming my theory or pointing out my mistake? I hope this isn't against the rules... I just need someone to bounce my idea off of. By the by, if you are willing, PLEASE do not give away the answer in the PM. I very strongly want to come up with the solution myself.

Problem 110 (View Problem)

### Re: Looking for someone who has solved problem 110

Posted: Tue Sep 30, 2008 1:37 pm
I would look into your theory.

### Re: Looking for someone who has solved problem 110

Posted: Tue Sep 30, 2008 10:30 pm
Thanks!

### Re: Problem 110

Posted: Thu Nov 24, 2011 10:49 am
I apparently have the uncanny ability to work on a problem for days or weeks, and then find the bug minutes after posting here

### Re: Problem 110

Posted: Thu Nov 24, 2011 11:30 am
That is not so strange, actually.
Thinking about what to post may have triggered some circuits in your brain so that you see things sharper.
It is even a quite common experience that trying to explain your problem to someone else reveals (part of ) the solution to you.

### Problem 110

Posted: Sat Jan 27, 2018 5:00 pm
I checked the existing topics on the problem but none of them asserted the issue I want to expose.
I think the text of the problem is wrong or at least ambiguous because it should state that the (x,y) couples are not considered ordered, i.e. that it counts (x,y) and (y,x) as a single solution.
In problem 108 this is clarified with the example (even if stating it in the text wouldn't be bad even there), while in problem 110 the most reasonable assumption would be to consider the two permutation different solutions (since it is expressed as an equation), and this would make your solution wrong.

### Re: Problem 110

Posted: Sun Jan 28, 2018 7:27 pm
Please don't start a new topic for a problem if there already exists one.