A two digit no is 18 less than square of the sum of its digit. How many such no are there..??

• let a two digit number be (10x+y), x=tenth digit & y=unit digit
  
  Given,
  10x+y=(x+y)^2-18
  
  (x+y)^2 should be less than 118 & greater than 28 to be two digit no.
  (x+y)^2 may be 100,81,64,49,36
  
  (x+y)^2 -18 = 10x+y
  check for these
  1. 100-18 = 82 => (82+18)=(8+2)^2
  2. 81 -18 = 63 => (63+18)=(6+3)^2
  3. 64 -18 = 46 => (46+18)=(4+6)^2 [not satisfied]
  4. 49 -18 = 31 => (31+18)=(3+1)^2 [not]
  5. 36 -18 = 18 => (18+18)=(1+8)^2 [not]
  
  Hence, only 2 numbers are possible
  82 & 63
• 10 years agoHelpfull: Yes(44) No(3)
• 1234+6543=11110
• 10 years agoHelpfull: Yes(0) No(6)
• 63 & 82 are the no.s
  2 ans
  
• 10 years agoHelpfull: Yes(0) No(0)
• short trick plz
• 10 years agoHelpfull: Yes(0) No(0)