A + B + C +D = D + E + F + G = G + H + I =17.
IF A = 4 WHAT ARE THE VALUES OF D AND G. EACH
LETTER TAKEN ONLY ONE OF THE DIGIT FROM 1 TO 9.

• Step 1.
  
  a + b + c + d = d + e + f + g = g + h + i = 17
  
  Means, ( a + b + c + d )+ (d +e + f + g )+ (g +h + i ) = 17 +17+17 = 17 x 3 = 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 = sum of the numbers from 1 to 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 beacuse
  
  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
• 10 years agoHelpfull: Yes(11) No(0)
• D=5 g=1 or g=5 d=1
• 10 years agoHelpfull: Yes(0) No(0)
• By trial and error,
  a+b+c+d=4+2+6+5=17
  d+e+f+g=5+8+3+1=17
  g+h+i= 1+9+7=17
  
  therefore, d=5 and g=1
• 10 years agoHelpfull: Yes(0) No(0)