Waste less time on Facebook — follow Brilliant.
×

6th Day: Solve this interesting problem

Thanks to all who join this discussion.

Note by Raja Metronetizen
3 years, 8 months ago

No vote yet
3 votes

Comments

Sort by:

Top Newest

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 · 3 years, 8 months ago

Log in to reply

@Gabriel Wong ..thanx a lot....would you please elaborate a little more about the idea about the parity of the sum changes and the conclusion.....regards.... Raja Metronetizen · 3 years, 8 months ago

Log in to reply

@Raja Metronetizen 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 · 3 years, 8 months ago

Log in to reply

@Gabriel Wong ok. got to it.....thanx a lot.... Raja Metronetizen · 3 years, 8 months ago

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 · 3 years, 8 months ago

Log in to reply

@Raja Metronetizen awesum yar....then also y u think always bad as i saw on post "1+1"? Riya Gupta · 3 years, 8 months ago

Log in to reply

@Raja Metronetizen wow-_-... The Destroyer · 3 years, 8 months ago

Log in to reply

@The Destroyer 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.....:( Raja Metronetizen · 3 years, 8 months ago

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 · 3 years, 8 months ago

Log in to reply

@Aditya Parson how? Riya Gupta · 3 years, 8 months ago

Log in to reply

@Riya Gupta 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 · 3 years, 8 months ago

Log in to reply

@Aditya Parson ok thankssss but genuinely i didn't got the question what it says.... Riya Gupta · 3 years, 8 months ago

Log in to reply

@Riya Gupta I can explain, provided you are online for a while? Aditya Parson · 3 years, 8 months ago

Log in to reply

@Aditya Parson yeah i am plz do xplain it.....i am really thankful to u....:) Riya Gupta · 3 years, 8 months ago

Log in to reply

@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. Aditya Parson · 3 years, 8 months ago

Log in to reply

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

then i support u for the answer u have given..... Riya Gupta · 3 years, 8 months ago

Log in to reply

Comment deleted 7 months ago

Log in to reply

@Raja Metronetizen 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. Aditya Parson · 3 years, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...