How many two digit numbers are there which when substracted from the number formed by reversing it's digits as well as when added to the number formed by reversing its digits, result in a perfect square.

• The answer is only 1 i,e the number is 56 (or say) 65.
  Explanation:
  (10x + y) - (10y + x) = k^2 < This is the first condition >
  => 9(x-y) = k^2 ______equation(1)
  
  (10x + y) + (10y +x) = z^2 < This is the second condition>
  =>11(x+y) = z^2
  Here 11 is nt a perfect square,so (x+y) must be 11 to make LHS a perfect square.
  so, the probables of (x+y ) are____ x y
  2 9 not possible
  3 8 not possible
  4 7 not possible
  5 6 possible
  
  now from equation 1 also we can say that... 9 is a perfect square ,so (x-y) must have to be 1 or 4..but this doesnt follow the given conditions.
  
  so,the answer is 56 (or) 65.
• 9 years agoHelpfull: Yes(13) No(0)
• Ans is 5
  nos are 38,47,65,83,92
  look those no.sum=11
  38
  + 83
  121 perfect square
  so all pair unit + tens place = 11.
  and further square not possible .
• 9 years agoHelpfull: Yes(0) No(9)