An old man distributed all the coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 36 times the difference between the two numbers. How many coins did the old man have?

• Given a^2-b^2=36(a-b)
  Solving this we get a+b=36
  Thus, the old man had 36 coins
• 12 years agoHelpfull: Yes(2) No(1)
• 36 coins
  
  If coins are divided as x and y, then
  x^2-y^2 = 36(x-y)
  (x+y)*(x-y)=36*(x-y)
  x+y=36
• 12 years agoHelpfull: Yes(2) No(1)
• Let x and y
  be the number of gold coins the two sons received. Since we are given that the differencebetween the squares of the two numbers is 36 times the difference between the two numbers, we have the equation
  
  
  (x)2-(y)2= 36(x–y)(x–y)(x+y)
  = 36(x–y)
  by the Difference of Squares formula
  a2–b2=(a–b)(a+b)
  
  x+y = 36
  by canceling (x–y) from both sidesHence,
  the total number of gold coins the old man had is
  
  x+ y which equals 36.
  
  
• 12 years agoHelpfull: Yes(1) No(3)