Knowing that every even perfect number is of the form $$n=2^{k}p=2^k(2^{k+1}-1)$$makes it easy to list all its factors, and this enables us to prove some interesting results. Since $p$ is a prime, the factors of $n=2^{k}p$ must either be a power of $2$ (up to $2^k$) or a product of a power of $2$ and $p$. Listing these factors in a table we get:

For example, the factors of $n=28=2^4\times 7$ are:

We claimed that the product of the factors of an even perfect number $n=2^{k}p$ is always equal to $n^{k+1}$. For $n=24$ this is clearly true, since $$1\times 2\times 4\times 7\times 14\times 28=21952={28}^3,$$ as required. For a general proof, note that the product of the factors in the first column of the table above is $$1\times 2\times 2^2...\times 2^k=2^{1+2+...+k}.$$ As we already noted in the main article, the sum of the first $k$ integers is $$1+2+..+k=k(k+1)/2,$$ so (1) is equal to $$2^{1+2+...+k}=2^{k(k+1)/2}.$$ Note that every factor in the second column of the table above is $p$ times the corresponding factor in the first table. Since there are $k+1$ entries in each column, the product of all the factors in the second column is therefore $$p^{k+1}{2}^{k(k+1)/2}.$$ The product of all the factors is $$2^{k(k+1)/2}{p}^k{2}^{k(k+1)/2}=p^{k+1}{2}^{k(k+1)}=(2^{k}p{)}^{k+1}=n^{k+1}.$$ This proves our result.

A similar result involves adding up the reciprocals of all the factors of an even perfect number. Taking our example of $n=28$ again we see that $$\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}=2.$$ In fact the sum is equal to $2$ for every perfect number! As one might expect, the proof proceeds in the same way as above, replacing addition with multiplication. We leave it to you as an exercise.

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