Let $x$ be an integer, and let $p_{x}$ be the smallest prime factor of $x$. Then, if $L(a)=x$, we note that we must have $a=p x$ for some prime $p \leq p_{x}$. (Otherwise, if $p>p_{x}$, then $\frac{p x}{p_{x}}>x$. If $p$ is composite, then $k x>x$ for some factor $k$ of $x$.)

So we have:
- $L(a)=2,4,8,10,16,20,40 \Rightarrow 1$ value for $a$
- $L(a)=5 \Rightarrow 3$ values for $a$

Hence, we note that, since $m$ and $n$ are composite, we cannot have $L(m)=1$ or $L(n)=1$, so the possible pairs $(L(m), L(n))$ are $(2,40),(4,20),(5,16),(8,10)$ and vice-versa.

We add the number of choices for each pair, and double since $m$ and $n$ are interchangeable, to get $2(1 \times 1+1 \times 1+3 \times 1+1 \times 1)=12$ possible ordered pairs $(m, n)$.