Observe that the parity of the sum changes every move.As the number of moves is 2n-1 which is odd, the parity of the remaining integer is opposite that of the original parity of sum.

The sum is initially (2n)(2n-1)/2 which is n(2n-1). Thus if n is even, the remaining integer is odd; and if n is odd the remaining integer is even.

Let the sum of the elements be S. Observe that when x and y are removed, S becomes S-x-y+abs(x-y)+1.

Observe that abs(x-y) - x - y is always even. Thus S and S-x-y+abs(x-y)+1 are of opposite parity.

There are 2n-1 steps to remove 2n-1 elements, leaving 1 final integer. As the parity of the sum changes with every step, it changes 2n-1 times and thus ends up opposite the original.

sorry not to see anyone to join this interesting question....did you think that i copied from a book?.....NOPE!!.....i just created this problem myself ...... respond to this teasing problem of number theory......

why....r u delighted....?......to see the question or to see my comments to urge you to participate.....?.....is this problem a child's play to you.....!?!....i could not understand you.....:(

let n be 1 that means we have 1,2.
So if you solve the two we get 2 according to question.
now when n=2 we have 1,2,3,4. which if you evaluate comes to 1.

@Riya Gupta
–
the problem states that let there be integers 1 to 2n.
now you can at random pick any two integers between 1 and 2n and then apply the algorithm. The result of the algorithm replaces the two integers which were previously there. You continue to do this until there is only a single integer left.

well actually when n is odd the answer is even and vice versa. You can show that through mathematical induction. Starting with n=1 for proving that the answer will be even, when n is odd.
And for even numbers n the sum is odd.

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TopNewestObserve that the parity of the sum changes every move.As the number of moves is 2n-1 which is odd, the parity of the remaining integer is opposite that of the original parity of sum.

The sum is initially (2n)(2n-1)/2 which is n(2n-1). Thus if n is even, the remaining integer is odd; and if n is odd the remaining integer is even.

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..thanx a lot....would you please elaborate a little more about the idea about the parity of the sum changes and the conclusion.....regards....

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Let the sum of the elements be S. Observe that when x and y are removed, S becomes S-x-y+abs(x-y)+1.

Observe that abs(x-y) - x - y is always even. Thus S and S-x-y+abs(x-y)+1 are of opposite parity.

There are 2n-1 steps to remove 2n-1 elements, leaving 1 final integer. As the parity of the sum changes with every step, it changes 2n-1 times and thus ends up opposite the original.

The rest is simple.

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sorry not to see anyone to join this interesting question....did you think that i copied from a book?.....NOPE!!.....i just created this problem myself ...... respond to this teasing problem of number theory......

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awesum yar....then also y u think always bad as i saw on post "1+1"?

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wow-_-...

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why....r u delighted....?......to see the question or to see my comments to urge you to participate.....?.....is this problem a child's play to you.....!?!....i could not understand you.....:(

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when i take n as 1 the answer is even but it comes odd when i take n as 2.

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how?

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let n be 1 that means we have 1,2. So if you solve the two we get 2 according to question. now when n=2 we have 1,2,3,4. which if you evaluate comes to 1.

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then i support u for the answer u have given.....

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Comment deleted Jun 12, 2016

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well actually when n is odd the answer is even and vice versa. You can show that through mathematical induction. Starting with n=1 for proving that the answer will be even, when n is odd. And for even numbers n the sum is odd.

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