# A number theory problem [Permutation] [closed]

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 MOct 5 '17 at 23:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Vidyanshu Mishra, kingW3, José Carlos Santos, Marcus M
If this question can be reworded to fit the rules in the help center, please edit the question.

• As a suggestion: try it for numbers smaller than $2011$ first. – lulu Sep 25 '17 at 10:04
• This looks like a job for the Pigeonhole Principle. – Shaun Sep 25 '17 at 10:09
• I know the answer via the pigeonhole principle, but I am wondering if I can find a solution by induction. – астон вілла олоф мэллбэрг Sep 25 '17 at 10:10
• @Shaun I was thinking that too, but I also know there's over 2 million possible selections of the indices. – user451844 Sep 25 '17 at 10:10
• @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. – 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 $2$.
• @ LinuxGeek First of all I am not a sir (I maybe one or two year older than you), secondly best of luck for RMO $2017$ – Vidyanshu Mishra Sep 26 '17 at 17:35
• Ha ha... I am $16$ and a half. @LinuxGeek – Vidyanshu Mishra Oct 7 '17 at 17:59