Least Number with Three Given Fractions

Proof
Let $a, b, c$ be the given aliquot parts.

Let $d, e, f$ be the numbers called by the same name as the aliquot parts $a, b, c$.

From, let:
 * $g = \lcm \set {d, e, f}$

So $g$ has aliquot parts called by the same name as $d, e, f$.

Therefore $g$ has the aliquot parts $a, b, c$.

Suppose there exists $h \in \N: h < g$ which has the aliquot parts $a, b, c$.

By, $h$ will be measured by numbers called by the same name as the aliquot parts $a, b, c$.

Therefore $h$ is measured by $d, e, f$.

But $h < g$ which is impossible.

Therefore there is no number less than $g$ which has the aliquot parts $a, b, c$.