Is there an even perfect number with exactly 22 divisors?

Question

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.

Answer 1

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.