[superprismatic]

Solve the following set of 10

nonlinear equations in 10 positive

integer variables A,B,C,D,E,F,G,H,I,J:

B(CE+A)+AI = 33

AB(CGH+1)+AI = 99

(FH+1)AI+BCE = 1413

H(AFI+J+ABCG) = 1464

J(DFI+H)+AB = 320

(A+E)I+ACDF+HJ = 180

H(AFI+J)+EI+D(H+ACF) = 1569

BC(AGH+E)+I(DFJ+E)+HJ = 413

AC(DF+BGH) = 204

H(ABCG+J)+BCE = 84

[/code]

Here, as is usual, multiplication is

assumed when terms are concatenated.

Note also that the left hand sides of

both the 2[sup]nd[/sup] and 3[sup]rd[/sup] equations

contain a 1 (I don't want anyone to

confuse these 1s with Is).  This

system has a unique solution in the

positive integers.


If need be, I will post hints every

several days to get people started

on the way to solve this one.

[Guest]

  Reveal hidden contents

Can't do it right now, but for anyone with enough time (or a program to do it) use Gaussian Elimination to find the unknowns

[Guest]

  Reveal hidden contents

A=3, B=2, C=1, D=4, E=3, F=11, G=2, H=6, I=7, J=1

[superprismatic]

  On 11/30/2009 at 11:55 PM, Andrew1258 said:

  Reveal hidden contents

A=3, B=2, C=1, D=4, E=3, F=11, G=2, H=6, I=7, J=1

Will you tell us how you arrived at the solution? Please do so.

[Guest]

  Reveal hidden contents

From 2, A is a multiple of 3, or 9, or 11, 33

(3)-(1) gives

AFHI - AB = 1380 which means A couldn't be a multiple of 9 or 11,

A = 1 or 3

(4) - (10) gives

AFHI - BCE = 1380

thus we have : A = CE

(8) - (10) gives

I(DJF + E) = 329 = 7X47

I could 1, 7 ,47 or 329.But could (1)it's less than 33, so

I = 1 or 7

If A=1, I=1, from (1) we have

B= 16

If I = 1, I = 7, from (1) we have

B= 13

But (2)-(1) gives

AB(CGH-1) = 66. Neither B=13 nor B=16 fits

So A=3, and if I = 1, B will be 6 which is ruled out.

So I = 7, giving B = 2

The others are similarly obtained.

Seemed to have a simpler method but6 forgot.