Let p, q, r are prime numbers and $α, β, γ$ are positive integers such that LCM of $α, β, γ$ is $p^{3} q^{2} r$ and greatest common divisor of $α, β, γ$ is $p q r$ . If the number of possible triplets $(α, β, γ)$ will be K, then number of divisors of K which are of the form 3n, $n \in N$ are

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Detailed Solution

$α = p^{m_{1}} q^{n_{1}} r, β = p^{m_{2}} q^{n_{2}} r, γ = p^{m_{3}} q^{n_{3}} r$

$\min {m_{1}, m_{2}, m_{3}} = 1, \max {m_{1}, m_{2}, m_{3}} = 3$ Gives number of possible $m_{1}, m_{2}, m_{3}$ are 12. also $\min {n_{1}, n_{2}, n_{3}} = 1, \max {n_{1}, n_{2}, n_{3}} = 2$ Gives number of possible $n_{1}, n_{2}, n_{3}$ are 6 $∴$ total values are 72