The number of distinct prime factors in (216) $^{3 / 5}$ (2500) $^{2 / 5}$ (300) $^{1 / 5}$ is

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None of these

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Solution

The correct option is D
None of these

$\begin{matrix} {(216)}^{3 / 5} . {(2500)}^{2 / 5} {(300)}^{1 / 5} = {(3^{3} {.2}^{3})}^{3 / 5} {(5^{2} {.10}^{2})}^{2 / 5} {(3 \times 10^{2})}^{1 / 5} = 3^{9 / 5} . 2^{9 / 5} {(\frac{10^{4}}{2^{2}})}^{2 / 5} . 3^{1 / 5} \times 10^{2 / 5} = 3^{(\frac{9}{5} + \frac{1}{5})} . 2^{(\frac{9}{5} - \frac{4}{5})} . 10^{(\frac{8}{5} + \frac{2}{5})} = 3^{2} . 2^{1} . 1 0^{2} = 3^{2} . 2^{1} . 2^{2} . 5^{2} = 2^{3} . 3^{2} . 5^{2} ∴ N u m b e r o f p r i m e f a c t o r s = 3 . \end{matrix}$

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The number of distinct prime factors in (216) $^{3 / 5}$ (2500) $^{2 / 5}$ (300) $^{1 / 5}$ is

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$(216)^{\frac{3}{5}} \times (2500)^{\frac{2}{5}} \times (300)^{\frac{1}{5}}$ as a product of its primes is

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