If n is a natural number and n=p₁ x₁ p₂ x₂ p₃ x₃, where p₁, p₂, p₃ are distinct prime factors, then the… · ABC26GN0412
Subject: General Aptitude · Chapter: Number System · Exam: CDS 2016
If $n$ is a natural number and $n=p_{1}{ }^{x_{1}} p_{2}{ }^{x_{2}} p_{3}{ }^{x_{3}}$, where $p_{1}, p_{2}, p_{3}$ are distinct prime factors, then the number of prime factors for $n$ is
Given $n=p_{1}{ }^{x_{1}} p_{2}{ }^{x_{2}} p_{3}{ }^{x_{3}}$ where $p_{1}, p_{2}, p_{3}$ are distinct prime factors Number of prime factors form $=\left(x_{1} \times x_{2} \times x_{3}\right)=\mathrm{x}_{1} x_{2} x_{3}$ Hence, option (b) is correct
Explanation as extracted from the printed page; notation may be imperfect.