Multiple of 6 is Semiperfect

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Theorem

Let $n \in \Z_{>0}$ be a multiple of $6$.

Then $n$ is semiperfect.


Proof

Let $n = 6 k$.

Then:

$n = 2 \times 3 k$

and so $3 k$ is a factor of $n$.

$n = 3 \times 2 k$

and so $2 k$ is a factor of $n$.

$n = 6 \times k$

and so $k$ is a factor of $n$.

But:

$n = k + 2 k + 3 k$

and so is the sum of a subset of its factors

Hence the result by definition of semiperfect.

$\blacksquare$