2015 · Q35
If n is a natural number and n=p^{x₁₁p^{x₂₂p^{x₃₃ where p₁,p₂,p₃ are distinct prime factors, then the number of prime factors for n is
Elementary MathematicsNumber System
- Ax₁+x₂+x₃
- Bx₁x₂x₃
- C(x₁+1)(x₂+1)(x₃+1)
- DNone of the above
Explanation
For n = p₁^x¹p₂^x²p₃^x³, each prime's exponent can be chosen in (xᵢ+1) ways, giving (x₁+1)(x₂+1)(x₃+1) divisors in all (option C; this is the divisor-count formula).