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.
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Gabriel Wong
·
4 years ago

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@Gabriel Wong
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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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Raja Metronetizen
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4 years ago

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@Raja Metronetizen
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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.

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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Raja Metronetizen
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4 years ago

@The Destroyer
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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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Raja Metronetizen
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4 years ago

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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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Aditya Parson
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4 years ago

@Riya Gupta
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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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Aditya Parson
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4 years ago

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@Aditya Parson
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ok thankssss but genuinely i didn't got the question what it says....
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Riya Gupta
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4 years ago

@Riya Gupta
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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.
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Aditya Parson
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4 years ago

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@Aditya Parson
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ohkkk.......now i got it........thank u so much....:)

then i support u for the answer u have given.....
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Riya Gupta
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4 years ago

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Comment deleted
11 months ago

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@Raja Metronetizen
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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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Aditya Parson
·
4 years ago

## Comments

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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. – Gabriel Wong · 4 years ago

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– Raja Metronetizen · 4 years ago

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

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. – Gabriel Wong · 4 years ago

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– Raja Metronetizen · 4 years ago

ok. got to it.....thanx a lot....Log in to reply

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...... – Raja Metronetizen · 4 years ago

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– Riya Gupta · 4 years ago

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

– The Destroyer · 4 years ago

wow-_-...Log in to reply

– Raja Metronetizen · 4 years ago

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.....:(Log in to reply

when i take n as 1 the answer is even but it comes odd when i take n as 2. – Aditya Parson · 4 years ago

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– Riya Gupta · 4 years ago

how?Log in to reply

– Aditya Parson · 4 years ago

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.Log in to reply

– Riya Gupta · 4 years ago

ok thankssss but genuinely i didn't got the question what it says....Log in to reply

– Aditya Parson · 4 years ago

I can explain, provided you are online for a while?Log in to reply

– Riya Gupta · 4 years ago

yeah i am plz do xplain it.....i am really thankful to u....:)Log in to reply

– Aditya Parson · 4 years ago

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.Log in to reply

then i support u for the answer u have given..... – Riya Gupta · 4 years ago

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– Aditya Parson · 4 years ago

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.Log in to reply