The sum of the digits of a two-digit number X is Y. The sum of the digits of X^2 is Y^2. Find X.

• X = 10,20 or 30
  
  When X = 10
  sum of digits = 1+0 = 1=Y
  X^2 = 100
  sum of digits of 100 = 1+0+0 = 1 = Y^2
  
  When X = 20
  sum of digits = 2+0 = 2=Y
  X^2 = 400
  sum of digits of 400 = 4+0+0 = 4 = Y^2
  
  When X = 30
  sum of digits = 3+0 = 3=Y
  X^2 = 900
  sum of digits of 900 = 9+0+0 = 9 = Y^2
  
• 10 years agoHelpfull: Yes(1) No(1)
• Let X be "ab" ... first digit is a, 2nd digit is b
  
  (a+b) = y -->(1)
  (ab)^2 = y^2 -->(2)
  ab = y
  
  so
  ab= a+b
  
  a+b-ab = 0
  a+b(1-a) = 0
  
  a != 1 ... it ll produce b=0
  So when a=2
  2+b(-1) = 0
  b=2
  
  a = 2, b = 2 , X="ab" = 22
  sub in (1) and (2)
  2+2 = y => y=4
  (22)^2 = 4^2 => 484 = 16
  Given sum of digits of x^2 = Y^2
  x^2 = 484 => 4+8+4 = 16 = y^2
  
  Ans : x=22, y=4
• 10 years agoHelpfull: Yes(1) No(1)