A four digit number numbered from 0000 to 9999 is said to be lucky, if the sum of the first two digits is equal to the sum of the last two digits. Find how many such numbers are possible ?

• 670 such numbers are possible.
  
  Minimum Sum for first two and last two digits = 0 (0000)
  Maximum Sum for first two and last two digits = 18(9999)
  
  Sum of first two = 0 --> 1 Case (0000)
  Sum of first two = 1 --> 2 Cases (10. 01)
  .....
  .....
  Sum of first two = 9 --> 10 Cases(09,18,27,36,54,45,63,72,18,90)
  Sum of first two = 10 --> 9 Cases(19,28,37,46,55,64,73,82,91)
  Sum of first two = 11 => 8 Cases (38,47,56,65,74,83,92,29)
  .....
  .....
  Sum of first two = 18 => 1 Case (99)
  
  Thus, Total will be = 2(1^2 + 2^2 + .... + 9^2) + 10^2 = 670 such numbers,
• 12 years agoHelpfull: Yes(2) No(2)
• 670 such numbers are possible.
  
  Sum of the first two digit = Sum of the last two digit
  when
  Sum of first two digits = 0... 1 possible Case
  Sum of first two digits = 1 .. 2 Cases possible
  .....
  .....
  Sum of first two = 10 .. 9 Cases
  Sum of first two = 11 .. 8 Cases
  .....
  
  Sum of first two digits = 18 .... 1 Case
  
  Thus, Total will be = 2(1^2 + 2^2 + .... + 9^2) + 10^2 = 670
• 12 years agoHelpfull: Yes(2) No(1)