1/x+1/y=1/z where x=12 and y,z positive integer how many possible value of y and z ?
a) 4
b) 7
C) 12
d) infinite

• Correct Ans = b) = 7
  ---------Explanation------------
  1/12 = 1/z - 1/y
  
  y = 12 * z / ( 12 - z )
  
  0 < z < 12
  
  Only for z = 3 ,4 ,6 ,8 , 9 ,10 and 11 , y is intiger
  
  Hence seven solutions
• 8 years agoHelpfull: Yes(25) No(0)
• 1/x+1/y=1/z
  on putting x=12 and solving the eq we get
  y=12z/12-z
  y is +integer hence z should be 1 to 11
  out of which z=3,4,6,8,9,10,11 are possible
  hence required sol be 7
  
• 8 years agoHelpfull: Yes(2) No(0)
• 1/z -1/y = 1=12
  (y-z)/1 =yz/12
  let (y-z)/1 =yz/12=p
  y-z=p
  yz=12p .:y,z are positive integer then p is positive integer number
  y=z+p
  z(Z+p)=12p
  Z^2 +zp - 12p=0
  use
  b^2 -4ac
  p^2 - 4(-12p)*1
  p^2 + 48p
  then p positive integer
  p^2 + 48p >0
  infinite
• 8 years agoHelpfull: Yes(0) No(2)
• jitendra , how can u say p is + integer ..... p can be + real number or a - integer if yz is not divisible by 12 or y
• 8 years agoHelpfull: Yes(0) No(0)
• 1/12 = 1/ z- 1/y = (y-z) /yz => 12( y-z) = yz,
  12 y -12 z =yz
  y = z ( 1+( y /12))
  y = 12z / (12 + z)
  now put the values of z which r multiples of 12 such that if z =12k then
  y = 12 * 12 k / 12 +12k
  y= 12k / k+1 now choose k s.t. k+1 always divides 12put k = 1,2,3 , 5 and 11 we get
  z=12 , y=6 ......1
  z= 24 , y=8.......2
  
  we can say option a) is correct
• 8 years agoHelpfull: Yes(0) No(0)
• awnser should be 7
  on solving we get y=12z/12-z
  y should be +ve and intiger
  so only possible value of z for y=+ve and an integer is 3,4,6,8,9,10,11
  
  so awnser should be option (b) 7
  
• 8 years agoHelpfull: Yes(0) No(0)
• The pattern would be somewhat like this
  
  12(y-z)=yz
  
  4---3 (12)*(4-3)=4*3
  6---4
  12--6
  24--12
  
  
  so ans=4 hence a
• 8 years agoHelpfull: Yes(0) No(2)
• 1/x=1/y+1/z
  1/12=1/y+1/z
  12=y+z
  z=12-y
  so y should be between 0 to 11 to make z positive integer so ans is 12
  
• 8 years agoHelpfull: Yes(0) No(0)