A+B+C+D=D+E+F+G=G+H+I=17 given A=4.Find value of G and H?

• As,
  A+B+C+D+D+E+F+G+G+H+I=17+17+17=51
  Now,
  A+B+C+D+E+F+G+H+I,there are 9 characters and the value will lie as 0 to 9.so,each character will corrosponds to any one digit from 0 to 9.so,A+B+C+D+E+F+G+H+I=1+2+3+4+5+6+7+8+9=45.
  
  so,D+G=51-45=6
  now,combinations for D and G will be [(1,5),(5,1),(4,2),(2,4),(3,3)]
  But D wil be not equal to G,so (3,3) not possible.And also A=4,so,(4,2) and (2,4),not possible.
  Now possible combinations-
  Case 1.D=1 and G=5
  so, B+C=17-A-D=17-4-1=12
  B+C=12[possible combinations are (3,9) and (9,3)]
  And E+F=17-D-G=17-1-5=11[ combinations are (7,4),(4,7),(6,5),(5,6),(9,2),(2,9),(8,3),(3,8)]
  But A=4,G=5,B OR C=9 OR 3..
  so these combinations are not possible.
  
  so,let us take,
  case 2.D=5 and G=1
  B+C=17-4-5=8 [possible combination(2,6),(6,2)]
  E+F=17-6=11[possible combinations (3,8),(8,3)]
  H+I=16 [possible combination [(9,7),(7,9)]
  so,
  H=9 or 7,D=5 And G=1
  
  
  
  
  
  
  
  
  
  
  
  
  
  
• 10 years agoHelpfull: Yes(14) No(1)
• solution:
  Step 1.
  a + b + c + d = d + e + f + g = g + h + i = 17
  NOW ADD;
  (a + b + c + d) + (d + e + f + g) + (g + h + i) = 17 +17+17
  (a + b + c + d) + (d + e + f + g) + (g + h + i) = 51
  a + b + c + d + e + f + g + h + i + ( d + g ) = 51
  Step 2.
  But 'a' to ' i ' takes values only from 1 to 9
  So,
  a + b + c + d + e + f + g + h + i = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
  
  Step 3.
  From step 1 and step 2
  d + g = 51 - 45 = 6
  possible values of d & g are (1, 5), ( 5,1 ), (2,4) & (4, 2 )
  ( 2, 4 ) & ( 4 , 2 ) are not possible as "a = 4 "
  We have to try with other values (1, 5 ) & ( 5 , 1 )
  Step 4.
  When d = 1 and g = 5
  1. a + b + c + d = 4 + b +c + 1 = 17
  b + c = 12
  only possible values of b & c are ( 9, 3 ) or ( 3, 9 )
  2.d + e + f + g = 1 + e + f + 5 = 17
  e + f = 17 - 6 = 11
  possible combinations for 11 are (2,9) ; (3,8) ; (4,7) & (5,6).
  Out of the above four combinations nothing is possible because b & c takes values 3 & 9, g = 5 (assumed) and a = 4 (given).
  So, d = 1 & g = 5 is not the solution.
  
  Case 2.
  When d = 5 and g = 1
  1. a + b + c + d = 4 + b + c + 5 = 17
  b + c = 17 - 9 = 8
  only possible combination is 2 & 6
  a + b + c + d = 4 + ( 2 + 6 ) + 5 = 17
  2. d + e + f + g = 5 + e + f + 1 = 17
  e + f = 17 - 6 = 11
  The only possible value is 3 & 8
  d + e + f + g = 5 + ( 3 + 8 ) + 1 = 17
  g + h + i = 1 + h + i = 17
  h + i = 17 - 1 = 16
  The only possible value for h & i are 7 & 9
  g + ( h+ i ) = 1 + ( 7 + 9 ) = 17
  So the values of d = 5 & g = 1
  ANS: G=1 & H=7
  
• 10 years agoHelpfull: Yes(5) No(1)
• plzzz explain..?
• 10 years agoHelpfull: Yes(4) No(0)
• actually we cant solve this prob because there are 8 unknowns (from B to I) and we require 8 equ to solve it. use trial and error method to solve it.
• 10 years agoHelpfull: Yes(4) No(2)
• A=4,B=6,C=2,D=5,E=8,F=3,G=1,H=7,I=9
  
  
• 10 years agoHelpfull: Yes(3) No(10)