Let $m,n$ be any two positive integers. Note $\widehat X$ the set of positive divisors of $x$.
$$\widehat X = \{ d : d \mid x\}$$
(do not confuse it with $\hat a = \{x : x \equiv a \mod m\}$)
Assume $(m,n)=1$. How could one prove that
$$\widehat {MN}=\{d:d\mid mn\}$$
is the product $\widehat M \cdot \widehat N$ in the sense that all divisors of $mn$ appear in the product
$$\left(\sum d_i+\sum d_i d_j+\cdots+ d_1 d_2 \cdots d_r \right) \left(\sum e_i+\sum e_i e_j+\cdots+ e_1 e_2 \cdots e_r \right)$$
(clearly $e_1 e_2 \cdots e_r=m$ and $ d_1 d_2 \cdots d_r=n$) where $d_i$ are the divors of $n$ and $e_i$ those of $m$?
This is essential in the proof that if $f$ is multiplicative, then $$F(n)=\sum_{d \mid n}f(d)$$ also is.