Q. If x,y,z are non negative integers and x=6 then find the number of solutions of 1/x +1/y = 1/z

• 1/x+1/y=1/z
  1/6+1/y=1/z
  (y+6)/6y=1/z
  z=6y/(6+y)
  z=6(y+6-6)/(6+y)
  z=6{1-6/(6+y)}
  z=6-36/(y+6)
  for z to be integer
  y=3 so z=2
  y=6 so z=3
  y=12 so z=4
  y=30 so z=5
  hence,no. of nonnegative integral solutions=4
  (x,y,z)=(6,3,2),(6,6,3),(6,12,4),(6,30,5)
  ans 4
• 10 years agoHelpfull: Yes(61) No(5)
• Rakesh ji your answer is correct but we should also include zero in the solution(i.e y=0 z=0 x=6) because it is given that x,y,z are non negative.
  so there will be 5 solutions.
• 10 years agoHelpfull: Yes(4) No(7)
• 1/6+1/y=1/z
  (y+6)/(6y)=1/z
  z=(6y)/(6+y)
  partition fraction
  z=6-36/(y+6)
  now put the value of y=3,6,12,30
  and get the value of z.....
• 10 years agoHelpfull: Yes(3) No(0)
• solving the equation you will get 6-z=6*z/y
  means the range of z will be from 1 to 6
  from eq y=6*z/6-z calculate y for every z if it is integer than it will be true..
  Hence there will 4 sol that will satisfy the y equal to integer for z = 2,3,4,5
  
  4 solutions
• 7 years agoHelpfull: Yes(2) No(1)
• no of solution are 4
  {(x,y,z)= (6,3,2),(6,6,3),(6,12,4),(6,30,5)}
• 10 years agoHelpfull: Yes(1) No(8)
• it is a condition in which there is 1 eqtn and two variable , hence infinite solutions
• 10 years agoHelpfull: Yes(1) No(3)
• @chaurasia jee when you take y=0,x=0,x=6 then it will not satisfied the eqn:
  1/x=(1/z-1/y).so i think we cant take that set.Is it clear DEEPAK BABU.. :)
• 9 years agoHelpfull: Yes(1) No(0)
• it is infinite solutions.
• 8 years agoHelpfull: Yes(1) No(0)
• @Naveen Pandey...frnd how did you come up with this solution ?Is there any standard approach you used ?
• 10 years agoHelpfull: Yes(0) No(1)
• what r the options.
  
• 10 years agoHelpfull: Yes(0) No(0)
• 3,2
  6,3
  12,4
  30,6
• 10 years agoHelpfull: Yes(0) No(1)
• How it comes z=6-36/(y+6)...cn u explain.........
• 10 years agoHelpfull: Yes(0) No(0)