Is there an even perfect number with exactly 22 divisors? I know that even perfect numbers have the form $n=(2^{p-1})\cdot(2^p-1)$ but don't really know where to go from here.

  • 2
    $\begingroup$ How many divisors does $2^{p-1}(2^p-1)$ have? Don't forget, $2^p-1$ must itself be prime. $\endgroup$ – lulu Feb 16 '17 at 14:08
  • $\begingroup$ See here: oeis.org/A061645 $\endgroup$ – Rohan Feb 16 '17 at 14:10
  • $\begingroup$ @lulu The divisors are $1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1$, and $n$. So that means that one of these has to equal 22? Which never can happen? $\endgroup$ – harry55 Feb 16 '17 at 14:14
  • $\begingroup$ No...we're after the count, not the actual divisors. In general, if $n=\prod p_i^{a_i}$ then the number of divisors of $n$ is $d(n)=\prod (a_i+1)$. That formula is easy to apply in this case! $\endgroup$ – lulu Feb 16 '17 at 14:22

In general, if the prime factorization of $n$ is $n=\prod p_i^{a_i}$ then the number of divisors of $n$ is $d(n)=\prod (a_i+1)$. That is especially easy to apply in the case of an even perfect number. After all, such a number only has two prime factors ($2,2^p-1$) so $$d(2^{p-1}(2^p-1))=p\times 2=2p$$

You are asking for this to be $22$ so we are lead to consider $p=11$. This will work iff $2^{11}-1$ is a Mersenne prime, so we have to check that. Alas $$2^{11}-1=23\times 89$$

As this was the only case which might have worked out, we conclude that no example exists.

  • $\begingroup$ Lulu, I want to understand this part "d(2^{p-1}(2^p-1))" How does it equal 2p? Thank you for the answer btw $\endgroup$ – user975 Feb 24 '17 at 14:17
  • $\begingroup$ Just apply the formula. We are using Euler's characterization of the even perfect numbers...they have the form $n=2^{p-1}M_p$ where $M_p=2^p-1$ is a Mersenne prime. Thus we already have the prime factorization of $n$...it has two prime factors, $2$ and $M_p$. The exponent of $2$ is $p-1$ and the exponent of $M_p$ is $1$ so... $\endgroup$ – lulu Feb 24 '17 at 14:21
  • $\begingroup$ Alternatively, just list the divisors. they are of the form $1,2,2^2,\cdots, 2^{p-1}$ and $M_p,2M_p,2^2M_p,\cdots, 2^{p-1}M_p$ and we can see that there are $2p$ of them. $\endgroup$ – lulu Feb 24 '17 at 14:22
  • $\begingroup$ Thank you Lulu - much appreciated. $\endgroup$ – user975 Feb 24 '17 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.