Let $(a_1, a_2, a_3, . . . , a_{2011})$ be a permutation (that is a rearrangement) of the numbers $1, 2, 3, . . . , 2011.$ Show that there exist two numbers $j,k$ such that $1 \le j < k \le 2011$ and $|a_j - j| = |a_k-k|$


closed as off-topic by Shaun, Vidyanshu Mishra, kingW3, José Carlos Santos, Marcus M Oct 5 '17 at 23:32

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  • $\begingroup$ As a suggestion: try it for numbers smaller than $2011$ first. $\endgroup$ – lulu Sep 25 '17 at 10:04
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    $\begingroup$ This looks like a job for the Pigeonhole Principle. $\endgroup$ – Shaun Sep 25 '17 at 10:09
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    $\begingroup$ I know the answer via the pigeonhole principle, but I am wondering if I can find a solution by induction. $\endgroup$ – астон вілла олоф мэллбэрг Sep 25 '17 at 10:10
  • $\begingroup$ @Shaun I was thinking that too, but I also know there's over 2 million possible selections of the indices. $\endgroup$ – user451844 Sep 25 '17 at 10:10
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    $\begingroup$ @LinuxGeek The downvotes are probably coming because your question is stated as is, and seemingly with no thought or effort from your side, although that is reflected in your comments. Also, the use of $2011$ reminds people of a contest problem, so they may think you have copied. BUT THAT ASIDE, if $|a_j - j| \neq |a_k - k|$ for all $j \neq k$, then as sets, $\{|a_j-j|\} = \{1,2,3,...,2010\}$, because there are only $2010 $ possible values of $|a_j-j|$. Summing over these, and using the summation formula for $\sum_{i=1}^n i = \frac{n(n+1)}{2}$ gives a contradiction. This sum will be odd. $\endgroup$ – астон вілла олоф мэллбэрг Sep 25 '17 at 10:21

Well, it wasn't hard to identify the problem for me. I have solved almost all past papers of RMO Delhi/Mumbai region. This is one of them.. See problem $2$.

  • $\begingroup$ Whatever reason considered legitimate, your link has more to like than the solutions. Therefore, +1/ Especially the sixth question. $\endgroup$ – астон вілла олоф мэллбэрг Sep 25 '17 at 10:51
  • $\begingroup$ This is where I came from sir. I couldn't get the solution at first, though now it seems trivial. I only wanted another solution like the ones discussed in the comment. I am preparing for RMO 2017 $\endgroup$ – MathDude3013 Sep 26 '17 at 16:55
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    $\begingroup$ @ LinuxGeek First of all I am not a sir (I maybe one or two year older than you), secondly best of luck for RMO $2017$ $\endgroup$ – Vidyanshu Mishra Sep 26 '17 at 17:35
  • $\begingroup$ @VidyanshuMishra I am 14 $\endgroup$ – MathDude3013 Oct 7 '17 at 17:54
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    $\begingroup$ Ha ha... I am $16$ and a half. @LinuxGeek $\endgroup$ – Vidyanshu Mishra Oct 7 '17 at 17:59

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