2.In the following figure:A B C D E F G H I Each of the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 is: a)Represented by a different letter in the figure above. b)Positioned in the figure above so that each of A + B + C,C + D +E,E + F + G, and G + H + I is equal to 13. Which digit does E represent?

• A+B+C=13 => 9+3+1=13
  C+D+E=13 => 1+8+4=13
  E+F+G=13 => 4+7+2=13
  G+H+I=13 => 2+5+6=13
  HENCE E IS 4.
• 13 years agoHelpfull: Yes(15) No(7)
• A+B+C=13=9+3+1=13
  C+D+E=13=1+8+4=13
  E+F+G=13=4+7+2=13
  G+H+I=13=2+5+6=13
  HENCE E IS 4.
  
• 13 years agoHelpfull: Yes(4) No(5)
• A+B+c=13 -->(eq 1)
  c+D+E=13 -->(eq 2)
  E+F+G=13 -->(eq 3)
  G+H+I=13 -->(eq 4)
  
  
  adding all eqns
  we get
  
  (A+b+2c+D+2E+F+2G+H+I )=4*13
  E*G*C(A+B+C+D+E+F+G+H+I)=52
  subsuiting 1,2,3,4,5,6,7,8,9 (bcoz each digit should has atleast one value)
  hence
  
  E*G*C=52-45
  
  E*G*C=7 --> (eq 5)
  
  subtracting 5,2 we get
  
  D-G=5 can be D=5+G ,G=D-5 ( hence we can put all possible values ifG=1,2,3,4 then D=6,7,8,9 )
  
  
  
  
  E*G*C=7 --> (eq 5)
  
  subtracting 5,3 we get
  
  F-C=5 can be F=5+C ,C=F-5 ( hence we can put all possible values ifC=1,2,3,4 then F=6,7,8,9 )
  
  the rqn 5 can be written as E=7-G-C --eqn (6)
  
  assume G=1 C=2 then E=4 (possible values of G or C can be)
  
  apply E,C value in (eqn 2)
  you got D=7
  
  apply E,G value in (eqn 3)then F=8
  
  now we have E=4 ,c=2 , G=1,F=8,D=6 remain values are (9,5,3,6)
  
  consider ( eqn 1) A+B+2=13 then (A and B should be 5 and 6 which yields 5+6+2=13)
  
  similarly H=9 I=3 and G=1 hence (eqn 4) becomes 13
  
  ALL VALUES HAVE FOUND ..!!
  thank you by kannan
  
• 12 years agoHelpfull: Yes(4) No(4)
• A+B+C=13
  C+D+E=13
  E+F+G+I=13
  1+2+3+4+5+6+7+8+9=45
  13+13+13+13=52
  52-45=7
  C+D+G=7-3=4
  HENCE E IS 4
• 10 years agoHelpfull: Yes(1) No(4)