Just as the theorem on multiples of abundant numbers shows that multiples of abundant numbers are also abundant, it is also true that multiples of semiperfect numbers are also semiperfect, and T. Foregger's proof of the abundant number theorem lays bare a simple mechanism that we can also employ for semiperfect numbers.

Given the divisors $d_1, \ldots, d_{k - 1}$ of $n$ (where $k = \tau(n)$ and $\tau(x)$ is the divisor function), sorted in ascending order for our convenience, and with a smart iterator $i$ that somehow knows to skip over those divisors that contribute to $n$ 's abundance, we can show that the divisors of $nm$ (with $m > 0$ ) will include $d_1m, \ldots, d_{k - 1}m$ . With our smart iterator $i$ and thanks to the distributive property of multiplication, it follows that $$\sum_{i = 1}^{k - 1} d_im = nm,$$ our desired result.