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. Suppose the positive integer n is odd. First Al writes the numbers $1, 2,..., 2n$ on the blackboard. Then he picks any two numbers a, b, erases them, and writes, instead, $|a − b|$. Prove that an odd number will remain at the end.

I have proved it in this way, please check if it's correct:-
Clearly, there will be n odds and n even numbers
possibilities of picking a and b integers:-
$2$ odds -> $|a-b|$= even number Hence, sum of the series ->$(n-2)$odds $+$ evens $=$ odd integer
when a & b are both even numbers -> sum = $n$ odds $+$ $(n-2)$ even $+$ even $=$ an odd integer
Without losing generality,
when a is odd and b is even -> sum = $(n-1)$odd $+$ $n$ even $+$ odd $=$ odd integer
Hence, the last integer left shall also be an odd int. Is my proof correct, please check, if not then kindly give me a hint, not a full solution...since I am preparing for math olympiad
Thank you

Death Champion
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    This is hard to read. here is a tutorial on typesetting for this site. Also, I recommend avoiding crude shorthand like "no.s" and "int". As it looks like you are working case by case, I suggest writing out each case. There are not very many. Just waving your hands and saying that they all work out is not convincing. – lulu Mar 14 '22 at 10:58
  • @lulu Is there any way of doing this without case by case, then please post it, I want to see that answer as well so that I can get another new approach – Death Champion Mar 14 '22 at 11:37
  • Of course there is another approach, but edit this one first. There's nothing wrong with the case by case method, it's just that if you are going to argue that way you have to be systematic about it. The dangers, of course, are that you'll either forget a case or that you'll make some minor error in one of them that ends up invalidating the entire argument. – lulu Mar 14 '22 at 11:42
  • Should stress: if you are preparing for an olympiad, it is critical that you write clearly and completely. What you've written here....well, if anyone took the time to supply the missing pieces, I'd credit them with the proof. – lulu Mar 14 '22 at 11:43
  • What more do I have to add then...I don't know... please help – Death Champion Mar 14 '22 at 12:55
  • This is still hard to read. Please edit to remove crude slang like "no.s" and "int". Personally, I'd never read anything written like that. And write out your cases separately, and clearly. Where, for instance, do you use the fact that $n$ is odd? – lulu Mar 14 '22 at 12:58
  • Just to say, a short proof of the desired result is: The sum is odd initially (it is $n(2n+1)$ and $n$ is odd). The parity of the sum is conserved by the operation since, taking $a≥b$ without loss of generality, $|a-b|=a-b$ and $a+b\equiv a-b \pmod 2$. – lulu Mar 14 '22 at 13:00
  • @lulu it is already given that n is odd – Death Champion Mar 14 '22 at 13:24
  • Yes, I asked where you used that assumption. In my argument, I use it to show that the initial sum is odd. – lulu Mar 14 '22 at 13:37
  • @lulu I have used that in all the three cases like sum of (n-2) odds + evens= odd integer because n is odd – Death Champion Mar 14 '22 at 13:52
  • In that case your proof is incomplete, as it is only true that there are $n$ odd and $n$ even terms initially. Thus, all you did was to prove that the sum is odd after the first use of the operation. The point, of course, is that the operation preserves the parity of any sum. As your sum starts out odd initially it must stay odd all the way down the line. – lulu Mar 14 '22 at 14:55
  • Oh alright I got your point, so, yeah I have to explain a lot of things including conservation of parity then... thankyou very much – Death Champion Mar 14 '22 at 16:17

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