Question

Three positive integers P, Q and Rare in A.P. such that 2Q + 3R = 61.
If Q and Rare prime numbers, then the polynomial whose zeroes are
P, Q and R can be​

Answer 1

Answer:

Step-by-step explanation:

If 2 numbers A and B have LCM as L, where L = a  

x

×b  

y

×c  

z

 ., 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