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If n is a natural number and n = $${p_1}^{{x_1}}$$ $${p_2}^{{x_2}}$$ $${p_3}^{{x_3}}$$ where p1, p2, p3 are distinct prime factors, then the number of prime factors for n is :
Answer & Solution
Correct Answer:
Option
B
Given n = $${p_1}^{{x_1}}{p_2}^{{x_2}}{p_3}^{{x_3}}$$
Where p1, p2, p3 are distinct prime factors
Number of prime factors form :
= (x1 × x2 × x3)
= x1 x2 x3
Hence option (B) is correct
Where p1, p2, p3 are distinct prime factors
Number of prime factors form :
= (x1 × x2 × x3)
= x1 x2 x3
Hence option (B) is correct
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