s,t = 2 and v = 8. Positive divisors greater than 1 = {2,4,8}.
Therefore, answer would be three (B)

Tough one. Nice question. Tricky.

Note :

if we add 1 to the exponents of each primes and multiply them we get the total number of factors including 1 and the number itself.

Total factor : (3+1) (3+1) = 16.

Remember that 16 is the factor of \(v^2\). So, v has 4 factors for sure as v's factors was squared.

So, v has 4 factors.

Now , it's stated in the question that we need to find out the number of divisors of v that are greater than 1. In our factor calculation we consider 1 too. So it's time to deduct now.

Total number of factors greater than 1 = 4-1 = 3.

The correct answer is B.

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Given: V is an integer => \(\sqrt{S^3 * T^3}\) => must be an integer

but given S and T are prime numbers, then \(\sqrt{S^3 * T^3}\) cannot be an integer, unless and until both the primes are same.

for V to be an integer, S = T , \(V^2 = S^3 * S^3 = S^6\); \(\sqrt{ V^2 }\) = \(\sqrt{(S^3)^2}\)
=> \(V = S^3\)
=> number of factors of V = (3 + 1) = 4 (rules)
=> Number of factors of V greater than 1 = (4 - 1) = 3

#Reference GMATCLUB.

https://youtu.be/aUTz3b5kgfY
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