IF
A+B+C=13
C+D+E=13
E+F+G=13
G+H+I=13
Then the value of E ?

• can any one explain how have you chosen the values of a,b,c...... ?
• 9 years agoHelpfull: Yes(14) No(0)
• A + B + C | 9 + 3 + 1 = 13 = 6 + 5 + 2
  C + D + E | 1 + 8 + 4 = 13 = 2 + 7 + 4
  E + F + G | 4 + 7 + 2 = 13 = 4 + 8 + 1
  G + H + I | 2 + 5 + 6 = 13 = 1 + 3 + 9
  Value of E = 4
• 9 years agoHelpfull: Yes(13) No(5)
• I got from one site.I just share in detail. i learn you also learn how it solved.
  
  A + B + C | 9 + 3 + 1 = 13 = 6 + 5 + 2
  C + D + E | 1 + 8 + 4 = 13 = 2 + 7 + 4
  E + F + G | 4 + 7 + 2 = 13 = 4 + 8 + 1
  G + H + I | 2 + 5 + 6 = 13 = 1 + 3 + 9
  Value of E = 4
  
  13*4 = 52
  1+2+3+....+9 = 45
  
  Then C+D+E must be equal to 7
  or they must from 1 , 2 or 4.
  
  1 and 2 for E cannot satisfy as because 1 and 2 cannot
  come together in C+D+E
  hence 1 and 4 and 2 and 4 should be together.
  
  so you get two solutions and putting them any of the either
  way you get E = 4. and everynumber gets satisfied with a
  unique value.
  
• 9 years agoHelpfull: Yes(6) No(6)
• please explain how are u choosing the values
• 9 years agoHelpfull: Yes(5) No(0)
• 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
• 9 years agoHelpfull: Yes(4) No(3)
• A + B + C | 9 + 3 + 1 = 13 = 6 + 5 + 2
  C + D + E | 1 + 8 + 4 = 13 = 2 + 7 + 4
  E + F + G | 4 + 7 + 2 = 13 = 4 + 8 + 1
  G + H + I | 2 + 5 + 6 = 13 = 1 + 3 + 9
  Value of E = 4
• 9 years agoHelpfull: Yes(3) No(1)
• ans is 4,
  Because choose values randomly,
  for example
  A+B+C=13(A=4,B=5,C=4)
  C+D+E=13(C=4,D=5,E=4)
  E+F+G=13(E=4,F=5,G=4)
  G+H+I=13(G=4,H=5,I=4)
• 9 years agoHelpfull: Yes(2) No(6)
• 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
• 9 years agoHelpfull: Yes(1) No(3)
• e=1
  (a=0,b=4,c=9)=13
  (c=9,d=3,e=1)=13
  (e=1,f=7,g=5)=13
  (g=5,h=6,i=2)=13
• 9 years agoHelpfull: Yes(1) No(0)
• how u r choose a,b,c,d,.... values
• 9 years agoHelpfull: Yes(0) No(0)
• 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
• 9 years agoHelpfull: Yes(0) No(10)
• this can solve using heat & trail method
• 9 years agoHelpfull: Yes(0) No(0)
• A + B + C | 9 + 3 + 1 = 13 = 6 + 5 + 2
  C + D + E | 1 + 8 + 4 = 13 = 2 + 7 + 4
  E + F + G | 4 + 7 + 2 = 13 = 4 + 8 + 1
  G + H + I | 2 + 5 + 6 = 13 = 1 + 3 + 9
  Value of E = 4
  its all about hit and trial just keep in mind that we dont have to put two big no. together oderwise we would get stuck as sum 13 could not be possible with left over digit..!!
• 9 years agoHelpfull: Yes(0) No(3)
• 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
  
  Then consider
  A B C
  C TO C is diagonally connected so put C = 1
  C D E
  As we have to find out E keep E as it is
  E F G
  G TO G is diagonally connected so put G = 2
  G H I
  
  from eq (5)
  So E = 7 - C - G = 7 - 1 - 2 = 4
• 1 year agoHelpfull: Yes(0) No(0)