The problem statement:
Start with the positive integers 1,...,4n-1. In one move you may replace any two integers by their difference. Prove that an even integer will be left after 4n-2 steps.
My question:
Is it fair to assume that the sequence progresses as 1,2,...,4n-1? if not, is there anything that I can infer about the original sequence from the information I am given?
I tried getting the sum(s) = 1+...+4n-1 and then observing how it behaves if it is decreased by2min(x,y), which I got by doing x+y-(x-y)=2y = 2min(x,y). But I can't proceed any further because I don't know how to represent the sum. Had I known anything about the sum and its nature (odd or even) I would've been further down the proof.
Any help/suggestion would be much appreciated.
These "moves" do not change the parity of the sum (weather it is odd or even). This is because the sum of 2 numbers and the difference of those numbers have the same parity.
You can see that it starts even because adding numbers on "opposite ends" gives even numbers and the middle term is even. That is: $(1+4n-1)+(2+4n-2)+...+(2n-1+2n+1)+2n$ is even. (This approach generalizes to the fact that the sum of the first k integers is $\frac{k*(k+1)}{2}$)
Since the sum starts even and every move preserves parity, it will be even after every move.