Rational Solver
+1 vote
For any integer $n \geq 1$, let

$d(n)=$ number of positive divisors of $n$

$v(n)=$ number of distinct prime divisors of $n$

$\omega(n)=$ number of prime divisors of $n$ counted with multiplicity

[for example: If $p$ is prime, then $\left.d(p)=2, v(p)=v\left(p^{2}\right)=1, \omega\left(p^{2}\right)=2\right]$

1. if $n \geq 1000$ and $\omega(n) \geq 2$, then $d(n)>\log n$

2. there exists $n$ such that $d(n)>3 \sqrt{n}$

3. for every $n, 2^{v(n)} \leq d(n) \leq 2^{\omega(n)}$

4. if $\omega(n)=\omega(\mathrm{m})$, then $d(n)=d(m)$
in Abstract Algebra by Expert (2.4k points)

Please log in or register to answer this question.

Welcome to Rational Solver, where you can ask questions and receive answers from other members of the community.
53 questions
12 answers
1 comment
1,801 users