Rational Solver
+1 vote
5 views
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) | 5 views

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.
49 questions
8 answers
1 comment
1,546 users