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Question

If r,s and t are prime numbers and p,q are positive integers such that the LCM of p,q is r2t4s2 then the number of ordered pair (p,q) is

A
254
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B
252
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C
225
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D
224
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Solution

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+31=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.

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