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|?

Submitted by Federico Kereki

Rating: 3.6667 (3 votes)

Solution: (Hide)

n² -- and here's is an explanation.
A sneaky way of solving this: if there's an unique answer, it must be same if we take A={1,2...n} and B={n+1,n+2...2n}. In that case the sum is (2n-1)+(2n-3)+...+1, which can be seen to be n².

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
answer	K Sengupta	2010-05-29 04:10:32
re: Proof (spoiler)	Richard	2005-07-13 02:18:32
Proof (spoiler)	Tristan	2005-07-12 23:05:07
re(2): thoughts -HTML issues	Josh70679	2005-07-12 17:57:07
re: thoughts -oops (spoiler)	Josh70679	2005-07-12 17:35:37
By inspection... no proof !	pcbouhid	2005-07-12 17:24:15
re: thoughts	Richard	2005-07-12 17:15:22
thoughts	Josh70679	2005-07-12 16:51:52

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