6th Day: Solve this interesting problem

Thanks to all who join this discussion.

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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......

Sayan Chaudhuri - 8 years, 1 month ago

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

A Former Brilliant Member - 8 years, 1 month ago

wow-_-...

the Destroyer - 8 years, 1 month 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.....:(

Sayan Chaudhuri - 8 years, 1 month ago

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.

Gabriel Wong - 8 years, 1 month 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....

Sayan Chaudhuri - 8 years, 1 month ago

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.

Gabriel Wong - 8 years, 1 month ago

@Gabriel Wong – ok. got to it.....thanx a lot....

Sayan Chaudhuri - 8 years, 1 month ago

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

Aditya Parson - 8 years, 1 month ago

how?

A Former Brilliant Member - 8 years, 1 month 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.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – ok thankssss but genuinely i didn't got the question what it says....

A Former Brilliant Member - 8 years, 1 month ago

@A Former Brilliant Member – I can explain, provided you are online for a while?

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – yeah i am plz do xplain it.....i am really thankful to u....:)

A Former Brilliant Member - 8 years, 1 month ago

@A Former Brilliant Member – 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.

Aditya Parson - 8 years, 1 month ago

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

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

A Former Brilliant Member - 8 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$