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January 17th, 2013, 07:35 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 Code: i1 i5 i9 i2 i6 i10 i3 i7 i11 i4 i8 i12 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, 10:27 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 nonlinear. Matrices are being used to solve a system of linear equations. 
January 18th, 2013, 01:50 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, 03:15 AM  #4  
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 Quote:
 
January 18th, 2013, 03:50 AM  #5  
Newbie Joined: Jan 2013 Posts: 8 Thanks: 0  Re: The system of equations (64 equations 12 unknown) solut Quote:
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)  
January 18th, 2013, 08:20 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,771 Thanks: 862  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, 09:56 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, 07:19 AM  #8  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,771 Thanks: 862  Re: The system of equations (64 equations 12 unknown) solut Quote:
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, 12:48 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, 08:11 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,771 Thanks: 862  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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