My Math Forum Diophantine Problem

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 March 14th, 2012, 02:16 AM #1 Newbie   Joined: Nov 2011 Posts: 2 Thanks: 0 Diophantine Problem I have absolute no idea on how to do this, i am finding it hard to interpret. 1. Find four numbers, the sum of every arrangement three at a time being given; say 22, 24, 27 and 20. -Could you set up a list of equations and solve them simultaneously?
March 14th, 2012, 03:58 AM   #2
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Re: Diophantine Problem

Hello, johnsy123!

Quote:
 1. Find four numbers, the sum of every arrangement three at a time being given; say 22, 24, 27 and 20.

$\text{Let the four numbers be }a,\,b,\,c,\,d.$

$\text{W\!e have: }\:\begin{Bmatrix}a\,+\,b\,+\,c &=& 22 && [1] \\ \\ \\
a\,+\,b\,+\,d &=& 24 && [2] \\ \\ \\
a\,+\,c\,+\,d &=& 27 && [3] \\ \\ \\
b\,+\,c\,+\,d &=& 20 && [4] \end{Bmatrix}$

$\text{Add the equations: }\:3a\,+\,3b\,+\,3c\,+\,3d \:=\:93 \;\;\;\Rightarrow\;\;\;a\,+\,b\,+\,c\,+\,d \:=\:31 \;\;[5]$

$\text{Subtract [1] - [2]: }\:d\,-\,c\:=\:2 \;\;\;\Rightarrow\;\;\;d \:=\:c\,+\,2\;\;[6]$

$\text{Subtract [2] - [3]: }\:c\,-\,b\:=\:3 \;\;\;\Rightarrow\;\;\;c \:=\:b\,+\,3\;\;[7]$

$\text{Subtract [3] - [4]: }\:a\,-\,b \:=\:7 \;\;\;\Rightarrow\;\;\;a \:=\:b\,+\,7\;\;[8]$

$\text{Substitute [7] into [6]: }\:d \:=\:(b\,+\,3)\,+\,2 \;\;\;\Rightarrow\;\;\;d \:=\:b\,+\,5\;\;[9]$

$\text{Substitute [8], [7], [9] into [5]: }\;(b\,+\,7)\,+\,b\,+\,(b\,+\,3)\,+\,(b\,+\,5) \;=\;31$

[color=beige]. . . [/color]$4b\,+\,15 \:=\;31 \;\;\;\Rightarrow\;\;\;4b \:=\:16 \;\;\;\Rightarrow\;\;\;b\:=\:4$

$\text{Therefore: }\:\begin{Bmatrix}a=&11 \\ \\ \\ b=&4 \\ \\ \\ c=&7 \\ \\ \\ d=&9 \end{Bmatrix}=$

 March 14th, 2012, 05:10 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Diophantine Problem Code: matsolve([1,1,1,0;1,1,0,1;1,0,1,1;0,1,1,1],[22,24,27,20]~)

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