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Question

The number of solutions (x, y, z) to the equation x - y - z = 25, where x, y, and z are positive integers such that x40,y12, and z12 is


  1. 87

  2. 105

  3. 99

  4. 101


Solution

The correct option is C

99


xyz=25 where x, y, z are positive integers such that x40,y12,z12,  y+z=x251
Now,  for LHS to have feasible solutions under the given conditions, RHS = (x - 25) can take values from a minimum 2 to the maximum of 15.
i.e. 2(y+z)15
Now, let's consider y+z15
y+z+k=152, where k is any variable.
Putting y=(y+1),z=(z+1) in equation 2), we get
y+1+z+1+k=15y+z+k=13
Number of solutions for the above equation =(13+31)C(31)=15C2=15×142=105
But the above number of solutions contains solutions with y12,z12. Those have to removed.
Number of invalid solutions = 2 + 4 = 6 ( 2 solutions for RHS = 13 and 4 solutions for RHS = 12)
Hence, final number of possible solutions = 105 - 6 = 99.
 

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