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 January 17th, 2013, 08:35 AM #1 Newbie   Joined: Jan 2013 Posts: 8 Thanks: 0 The system of equations (64 equations -12 unknown) solution Regards, when programming an application I faced with the following system of equations Code: n1=i1*i5*i9 n2=i1*i5*i10 n3=i1*i5*i11 n4=i1*i5*i12 n5=i1*i6*i9 n6=i1*i6*i10 n7=i1*i6*i11 n8=i1*i6*i12 n9=i1*i7*i9 n10=i1*i7*i10 n11=i1*i7*i11 n12=i1*i7*i12 n13=i1*i8*i9 n14=i1*i8*i10 n15=i1*i8*i11 n16=i1*i8*i12 n17=i2*i5*i9 n18=i2*i5*i10 n19=i2*i5*i11 n20=i2*i5*i12 n21=i2*i6*i9 n22=i2*i6*i10 n23=i2*i6*i11 n24=i2*i6*i12 n25=i2*i7*i9 n26=i2*i7*i10 n27=i2*i7*i11 n28=i2*i7*i12 n29=i2*i8*i9 n30=i2*i8*i10 n31=i2*i8*i11 n32=i2*i8*i12 n33=i3*i5*i9 n34=i3*i5*i10 n35=i3*i5*i11 n36=i3*i5*i12 n37=i3*i6*i9 n38=i3*i6*i10 n39=i3*i6*i11 n40=i3*i6*i12 n41=i3*i7*i9 n42=i3*i7*i10 n43=i3*i7*i11 n44=i3*i7*i12 n45=i3*i8*i9 n46=i3*i8*i10 n47=i3*i8*i11 n48=i3*i8*i12 n49=i4*i5*i9 n50=i4*i5*i10 n51=i4*i5*i11 n52=i4*i5*i12 n53=i4*i6*i9 n54=i4*i6*i10 n55=i4*i6*i11 n56=i4*i6*i12 n57=i4*i7*i9 n58=i4*i7*i10 n59=i4*i7*i11 n60=i4*i7*i12 n61=i4*i8*i9 n62=i4*i8*i10 n63=i4*i8*i11 n64=i4*i8*i12 The system of equations resulting from these matrix, with permutation of all elements Code: i1 i5 i9 i2 i6 i10 i3 i7 i11 i4 i8 i12 In the above equations, n (n1, n2 ..., n64) are familiar values, while the i (i1, i2, ..., i12) are unknown values The whole day i am trying to solve this, but no success. Is it possible to solve this system of equations? if so. How? Thanks in advance
 January 17th, 2013, 11:27 PM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: The system of equations (64 equations -12 unknown) solut If the "*" sign in your system of equations is multiplication, then there is no general method for these since the system is non-linear. Matrices are being used to solve a system of linear equations.
 January 18th, 2013, 02:50 AM #3 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: The system of equations (64 equations -12 unknown) solut Write le logarithm of each equation and replace the unknowns by their logarithms. This will lead to a linear system of equations.
January 18th, 2013, 04:15 AM   #4
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Joined: Mar 2012
From: India, West Bengal

Posts: 3,871
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Math Focus: Number Theory
Re: The system of equations (64 equations -12 unknown) solut

Quote:
 Originally Posted by JJacquelin Write le logarithm of each equation and replace the unknowns by their logarithms. This will lead to a linear system of equations.
Yes, that could work. Nice, JJacquelin !

January 18th, 2013, 04:50 AM   #5
Newbie

Joined: Jan 2013

Posts: 8
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Re: The system of equations (64 equations -12 unknown) solut

Quote:
 Originally Posted by JJacquelin Write le logarithm of each equation and replace the unknowns by their logarithms. This will lead to a linear system of equations.
thanks very much, i was forgot to use logarithms

now i have next system

Code:
n1=log(i1)+log(i5)+log(i9)
n2=log(i1)+log(i5)+log(i10)
n3=log(i1)+log(i5)+log(i11)
n4=log(i1)+log(i5)+log(i12)
n5=log(i1)+log(i6)+log(i9)
n6=log(i1)+log(i6)+log(i10)
n7=log(i1)+log(i6)+log(i11)
n8=log(i1)+log(i6)+log(i12)
n9=log(i1)+log(i7)+log(i9)
n10=log(i1)+log(i7)+log(i10)
n11=log(i1)+log(i7)+log(i11)
n12=log(i1)+log(i7)+log(i12)
n13=log(i1)+log(i8)+log(i9)
n14=log(i1)+log(i8)+log(i10)
n15=log(i1)+log(i8)+log(i11)
n16=log(i1)+log(i8)+log(i12)
n17=log(i2)+log(i5)+log(i9)
n18=log(i2)+log(i5)+log(i10)
n19=log(i2)+log(i5)+log(i11)
n20=log(i2)+log(i5)+log(i12)
n21=log(i2)+log(i6)+log(i9)
n22=log(i2)+log(i6)+log(i10)
n23=log(i2)+log(i6)+log(i11)
n24=log(i2)+log(i6)+log(i12)
n25=log(i2)+log(i7)+log(i9)
n26=log(i2)+log(i7)+log(i10)
n27=log(i2)+log(i7)+log(i11)
n28=log(i2)+log(i7)+log(i12)
n29=log(i2)+log(i8)+log(i9)
n30=log(i2)+log(i8)+log(i10)
n31=log(i2)+log(i8)+log(i11)
n32=log(i2)+log(i8)+log(i12)
n33=log(i3)+log(i5)+log(i9)
n34=log(i3)+log(i5)+log(i10)
n35=log(i3)+log(i5)+log(i11)
n36=log(i3)+log(i5)+log(i12)
n37=log(i3)+log(i6)+log(i9)
n38=log(i3)+log(i6)+log(i10)
n39=log(i3)+log(i6)+log(i11)
n40=log(i3)+log(i6)+log(i12)
n41=log(i3)+log(i7)+log(i9)
n42=log(i3)+log(i7)+log(i10)
n43=log(i3)+log(i7)+log(i11)
n44=log(i3)+log(i7)+log(i12)
n45=log(i3)+log(i8)+log(i9)
n46=log(i3)+log(i8)+log(i10)
n47=log(i3)+log(i8)+log(i11)
n48=log(i3)+log(i8)+log(i12)
n49=log(i4)+log(i5)+log(i9)
n50=log(i4)+log(i5)+log(i10)
n51=log(i4)+log(i5)+log(i11)
n52=log(i4)+log(i5)+log(i12)
n53=log(i4)+log(i6)+log(i9)
n54=log(i4)+log(i6)+log(i10)
n55=log(i4)+log(i6)+log(i11)
n56=log(i4)+log(i6)+log(i12)
n57=log(i4)+log(i7)+log(i9)
n58=log(i4)+log(i7)+log(i10)
n59=log(i4)+log(i7)+log(i11)
n60=log(i4)+log(i7)+log(i12)
n61=log(i4)+log(i8)+log(i9)
n62=log(i4)+log(i8)+log(i10)
n63=log(i4)+log(i8)+log(i11)
n64=log(i4)+log(i8)+log(i12)
but I have still not solve this system

 January 18th, 2013, 09:20 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,938 Thanks: 984 Re: The system of equations (64 equations -12 unknown) solut Perhaps this will answer your question: a+c+e = 14 a+c+f = 11 a+d+e = 13 a+d+f = 10 b+c+e = 13 b+c+f = 10 b+d+e = 12 b+d+f = 9 8 equations, 6 unknowns...can you solve?
 January 18th, 2013, 10:56 PM #7 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: The system of equations (64 equations -12 unknown) solut Hi ZeusTheMunja ! You have not yet made the change of unknown : x1 = log(i1) x2 = log(i2) etc. Then, you have to check if it is possible that the system of 64 linear equations could have a solution (x1, x2, ..., x12), since they are too many equations compared to the number of unknowns.
January 19th, 2013, 08:19 AM   #8
Math Team

Joined: Oct 2011

Posts: 13,938
Thanks: 984

Re: The system of equations (64 equations -12 unknown) solut

Quote:
 Originally Posted by JJacquelin You have not yet made the change of unknown : x1 = log(i1) x2 = log(i2)
Can't see how that will help...this equation system that I gave as shorter example (but similar):
a+c+e = 14
a+c+f = 11
a+d+e = 13
a+d+f = 10
b+c+e = 13
b+c+f = 10
b+d+e = 12
b+d+f = 9

...has 27 solutions keeping the variables > 0 ; (a,b,c,d,e,f):
1: 2,1,3,2,9,6
2: 2,1,4,3,8,5
...
26: 7,6,3,2,4,1
27: 8,7,2,1,4,1

 January 20th, 2013, 01:48 AM #9 Newbie   Joined: Jan 2013 Posts: 8 Thanks: 0 Re: The system of equations (64 equations -12 unknown) solut Denis, that is exactly what i want. All possible solutions. Can you tell me how you solve it. In my case all variables are > 0
 January 20th, 2013, 09:11 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,938 Thanks: 984 Re: The system of equations (64 equations -12 unknown) solut What you are asking is not clear...are you simply writing a program to result in obtaining 64 multiplications each of 3 integers, given 12 integers? The 12 integers would be in an array, and you need the results of multiplying all combinations of: [element 1 to 4] * [element 5 to 8] * [element 9 to 12]? This would be 64 solutions: 4 * 4 * 4 = 64. If that's all you're after, then this simple program will handle: .................................................. .................................................. ..................... Open array N(64), I(12) : array N receives the solutions, array I contains the 12 integers k = 0 : this variable will increase by 1 at each calculation, in order to place result in array N loop a from 1 to 4 loop b from 5 to 8 loop c from 9 10 12 k = k + 1 N(k) = I(a) * I(b) + I(c) print N(k) , I(a) , I(b) , I(c) : if you want results printed as you go along endloops .................................................. .................................................. .................................... As example, if I(2) = 7, I(7) = 2 and I(10) = 4, and if a = 2, b = 7 and c = 10 then this will be result: N(k) = 7 * 2 * 4 = 56 If that's not what you're after, then post problem again CLEARLY, and give an actual example.

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