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An Arbitrary sum?


Best Answer vinay.singh84, 09 April 2013 - 10:05 PM

Spoiler for you can see it in the tables.
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#1 BMAD

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Posted 09 April 2013 - 05:43 PM

The first 2n positive integers are arbitrarily divided into two groups of n numbers each. The numbers in the first group are sorted in ascending order: a1 < a2 < ... < an; the numbers in the second group are sorted in descending order: b1 > b2 > ... > bn.

Find, with proof, the value of the sum |a1 − b1| + |a2 − b2| + ... + |an − bn|.

 

The table below is generated at random, in two stages. Firstly, a value of n in the range 2..20 is chosen at random. Then, n numbers in the range 1..2n are chosen at random. These numbers are sorted to form the sequence ai; the remaining numbers are sorted to form the sequence bi. For each i, the value |ai − bi| is calculated, and the sum is given.

 

Here are some generated random table . Use the tables to formulate a conjecture about each pair, (ai, bi.) Prove your conjecture, and hence find the value of the sum.


Edited by BMAD, 09 April 2013 - 05:45 PM.

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#2 BMAD

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Posted 09 April 2013 - 05:49 PM

It wouldn't let me paste my random tables so here you go...

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#3 vinay.singh84

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Posted 09 April 2013 - 10:05 PM   Best Answer

Spoiler for you can see it in the tables.

Edited by vinay.singh84, 09 April 2013 - 10:06 PM.

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#4 witzar

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Posted 10 April 2013 - 09:06 AM

Spoiler for


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