The correct option is
C 225
If 2 numbers A and B have LCM as L, where
L = ax×by×cz ., then, number of ordered pairs of numbers having the above LCM will be:
(2x+1)×(2y+1)×(2z+1)
Let us consider r: we need the power r in either p or q to be at least 2.
If the power of r in p is 2, then in q it should be 0 or 1 or 2 →3 cases
If the power of r in q is 2, then in q it should be 0 or 1 or 2 →3 cases
But (2, 2) has been counted twice. Thus, there are 3+3−1=5 cases
Total of five cases for the exponents of r such that we have the given LCM which is 2×2+1
Similarly, the others follow.
Here, x=2,y=4,z=2.
Therefore, the answer = (2×2+1)⋅(2×4+1)⋅(2×2+1)=225
Hence, (C) is correct.