Question 2252
Sol: By given condition,  we have
 A+B+C = 13...(1)
 C+D+E = 13...(2)
 E+F+G = 13...(3)
 G+H+I = 13...(4)
 Add up (1)+(2)+(3)+(4), we get 
 52=13*4 = A + B + C+ C + D + E+ E + F + G+ G + H + I = C+E+G+(A + B + C+ D +E + F + G + H + I)
    = C+E+G+(1+2+ 3+ 4+ 5+ 6+ 7+ 8+ 9) = C+E+G+(1+9)*9/2 = C+E+G + 45
  
 Hence,C+E+G = 52 - 45 = 7...(5)

 By (5), we have E = 7-C-G <= 7-3 = 4 (since, the smallest C,G are 1 or 2) 
 By (2)-(5)., we get D-G = 6, or D = G+6...(6) 
             Since D,G are between 1 and 9,
             so D= 7, 8,or 9, and we have G=1,2 or 3

 By (3)-(5)., we get F-C = 6, or F = C+6...(7) 
             Since F,C are between 1 and 9,
             so F= 7, 8,or 9, and we have C=1,2 or 3

 By (6) ,(7) & (5):
 Case(i) When G =1, D = G+6 =7, 
         and then C = 2, F = C+6=8, E = 7-C-G= 4
        Or C = 3, F = 9, E = 3 (invalid,since the same C,E value)

 Case(ii) When G =2, D = G+6= 8, 
           and then C = 1, F = C+6=7, E= 7-C-G=4
         Or C = 3, F = 9, E = 2(invalid,since the same C,G value)

 Case(iii) When G =3, D = G+6=9,
        and then C = 1, F = C+6=7, E =7-C-G= 3(invalid,since the same E,G value)
        Or C = 2, F = C+6=8, E=7-C-G = 2(invalid,since the same C,E value)

 In these three possible cases of the values of G, we obtain that the only valid value of E is 4.


Kenny