Suppose there is a number 235 such that sum of first two digits is equal to the third digit. How many such 3 digits numbers are possible?

• The last digit can not be 0.
  
  If the last digit is 1, the only possible number is 101.
  (Note that 011 is not a 3-digit number)
  
  If the last digit is 2, the possible numbers are 202 and
  112.
  
  If the last digit is 3, the possible numbers are 303, 213
  and 123.
  
  If the last digit is 4, the possible numbers are 404, 314,
  224 and 134.
  
  If the last digit is 5, the possible numbers are 505, 415,
  325, 235 and 145.
  
  Note the pattern here - If the last digit is 1, there is
  only one number. If the last digit is 2, there are two
  numbers. If the last digit is 3, there are three numbers.
  If the last digit is 4, there are four numbers. If the last
  digit is 5, there are five numbers. And so on.....
  
  Thus, total numbers are
  1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
  
  So Ans is 45 only
• 10 years agoHelpfull: Yes(59) No(1)
• As we know that total digits are
  0 ,1, 2, 3,4,5,6,7,8,9
  if three digit no is xyz
  this means x+y=z must hold
  so we have to select two no out of 10
  
  ans is 10 c 2=45
  
  Ans : 45
• 10 years agoHelpfull: Yes(12) No(2)
• 
  Ans is 45
  
  101 last digit is one
  202,112 last digit is 2
  
  if last digit is one it comes one time
  if last digit is two it comes two time
  so
  1+2+3+4+5+6+7+8+9=45
• 10 years agoHelpfull: Yes(2) No(2)
• As the 3rd digit must be greater than first digit and for 1 there are such 9 digits nd for 2 there are 8 digits nd so on so we can directly calculate the sum of first 9 natural numbers =9(9+1)/2=45.
• 7 years agoHelpfull: Yes(1) No(0)