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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.

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    How many divisors does $2^{p-1}(2^p-1)$ have? Don't forget, $2^p-1$ must itself be prime. – 2017-02-16
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    See here: https://oeis.org/A061645 – 2017-02-16
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    @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? – 2017-02-16
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    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! – 2017-02-16

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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.

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    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 – 2017-02-24
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    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... – 2017-02-24
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    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. – 2017-02-24
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    Thank you Lulu - much appreciated. – 2017-02-24