Multiple of 6 is Semiperfect

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.

[[Category:6]]