perplexus.info :: Numbers : Any order will do

Any order will do (Posted on 2005-07-12)

The numbers from 1 to 2n are separated into two lists of n numbers each, A and B; elements are ordered so a₁<a₂<...<a_n and b₁>b₂>...>b_n.

How much is |a₁-b₁|+|a₂-b₂|+...+|a_n-b_n|?

See The Solution Submitted by Federico Kereki

Rating: 3.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

re: Proof (spoiler)

| Comment 7 of 8 |

(In reply to Proof (spoiler) by Tristan)

Very nice approach with the X, Y, A, and B.

One notes that n+1 + n+2 + ... + 2n - 1 -2 - ... - n =

= (n+1 - 1) + (n+2 - 2) + ... + (2n - n) =

= n + n + ... + n = n².

Posted by Richard on 2005-07-13 02:18:32

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