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Let $n$ be a natural number. Denote by $\omega(n)$ the number of prime divisors of $n$, and denote by $\Omega(n)$ the number of prime-power divisors of $n$.

If $n=p_1^{e_1}\cdots p_r^{e_r}$ is the factorization of $n$, then $\omega(n)=r$ and $\Omega(n)=e_1+\cdots + e_r$.

Which of the following statements are true? Select ALL that apply.


If $n$ is a square (i.e. $n=m^2$), then $\omega(n)=\omega(\sqrt{n})$.


If $n$ is square-free (i.e. $n$ is not divisible by a square of any prime number), then $\Omega(n)=\omega(n)$.


If $n$ is a square, then $\Omega(n)=2\omega(n)$.


If $n$ is not a square, then $\Omega(n)\neq 2\omega(n)$.

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