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.
$\blacksquare$