Find the two digit no.which exceeds by 12 the sum of squares of digits and by 16 the doubled product of its digits.?

• Ans= 64
  
  Let the digits of a two-digit number be x and y and the number be 10x+y, then
  According to first condition: 10x+y= (x^2+y^2)+ 12 ----(i) and
  According to second condition: 10x+y= 2xy+16 ----(ii)
  Comparing (i) & (ii), (x^2+y^2)+ 12 = 2xy+16
  => x^2+y^2-2xy =4
  => (x-y)^2 = 4 or x-y= 2
  Now checking for the combinations of x,y with difference of 2, and satisfying 1st and 2nd condition, we have x=6, y=4 , where 64= 6^2+4^2+12 and 64=2*6*4+16
• 7 years agoHelpfull: Yes(4) No(0)
• yes,it is 64
• 7 years agoHelpfull: Yes(0) No(0)