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

• 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
  (a + b + c + d) + (d + e + f + g) + (g + h + i) = 17 x 3
  (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
  
• 10 years agoHelpfull: Yes(15) No(4)
• wrong ques ...full ques is all the alphabets having range from 1 to 9 and distinct value
  
• 10 years agoHelpfull: Yes(9) No(0)
• a+b+c+d+d+e+f+g+g+h+i= 51,
  a+b+c+d+e+f+g+h+i=45,
  => d + g = 6. d ≠ g => d, g, ≠ 3. a = 4 => d, g ≠ 4, 2.
  => a, d = 1, 5; or 5, 1.
  Now h+i = 16 or 12; h+i = 16 => h, i = 7, 9; h+i = 12
  => h, i = 3, 9 => h or i = 9;
  Also e + f = 11 => e + f = 2+9 or 3+8; But e, f ≠ 9
  => e, f = 3, 8.
  b+c+d = 13, d, g = 1, 5. b, c = 2, 3, 6, 7.
  h, i = 9 and either 3 or 7.
  => b, c = 2, 6 => d = 5 and g = 1.
• 11 years agoHelpfull: Yes(8) No(23)
• A+B+C+D=17
  4+2+6+5=17
  D=5
  
  D+E+F+G=17
  5+3+8+1=17
  G=1
  
  G+H+I=17
  1+9+7=17
  G=1,H=9.
• 11 years agoHelpfull: Yes(6) No(27)
• can anyone explain clearly please?
• 6 years agoHelpfull: Yes(0) No(0)