A two digit number is 18 less than the sum of the squares of its digits. How many such numbers are there

• Only two possible numbers are :
  47
  and
  67
  
  The equation would be :
  10x + y = (x^2) + (y^2) - 18
  Substitute different values from the options.
• 9 years agoHelpfull: Yes(7) No(5)
• let no be 10x + y
  now 10x + y =x^2 + y^ 2+ 18
  the possible ans are 47 n 67
• 9 years agoHelpfull: Yes(1) No(3)
• ans=2 63 and 82
• 9 years agoHelpfull: Yes(0) No(11)
• Take N = 10a+b.
  Given that, 10a+b+18 = (a+b)2
  for a = 1 to 9, the L.H.S. will be, 28+b, 38+b, 48+b,.....,108+b.
  As LHS is perfect square for the values of b = 1 to 9, only 28+b, 48+b, 58+b, 78+b can be equal to 36, 49, 64, 81 for b = 8, 1, 6, 3 respectively. But only 78+b = 81 for b = 3 So only one such number is possible. I.e, 63
• 9 years agoHelpfull: Yes(0) No(4)