Re:Playing with numbers

*To*: mathgroup at smc.vnet.net*Subject*: [mg50223] Re:[mg50135] Playing with numbers*From*: "Fred Simons" <f.h.simons at tue.nl>*Date*: Sat, 21 Aug 2004 03:04:29 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

The question is to solve with Mathematica the set of equations (1) H - F = I (2) F x G = C (3) E / B = F (4) A + B = H (5) C - D = A where every variable from A to I stands for a number from 1 to 9 and each number appears only once. Here is a solution where Mathematica does the computations that we otherwise had to do by hand. Since the symbols represent different integers between 1 and 9, we conclude from the equations i+f=h, a+b=h that h has to be at least 5. From the equations c=f*g, e=f*b we conclude that f is between 2 and 4, that g is at least 2, that e is at least 2, that c is at least 6 and that e is at least 6. Given values for f, h, c and e, we can compute the other variables. The result has to be a permutation of the numbers 1, ..., 9. In[21]:= solution = {a, b, c, d, e, f, g, h, i} /. Solve[{h - f == i, f*g == c, e/b == f, a + b == h, c - d == a}, {a, b, d, g, i}][[1]] Out[21]= {-((e - f*h)/f), e/f, c, c + e/f - h, e, f, c/f, h, -f + h} In[23]:= Timing[Do[If[Length[Union[solution]] == 9, Print[solution]], {f, 2, 4}, {c, Ceiling[5/f]*f, 9, f}, {e, Ceiling[5/f]*f, 9, f}, {h, 5, 9}]] From In[23]:= {5,4,6,1,8,2,3,9,7} Out[23]= {0. Second,Null} So indeed there is only one solution. Fred Simons Eindhoven University of Technology

**Follow-Ups**:**Re: Re:Playing with numbers***From:*DrBob <drbob@bigfoot.com>