jmc

Call Jan's number $J$. $12=2^2\cdot 3$ and $15=3\cdot 5$, so $J$ has at least two factors of 2, one factor of 3, and one factor of 5 in its prime factorization. If $J$ has exactly two factors of 2, then the prime factorization of $J$ is of the form $2^2\cdot 3^a\cdot 5^b\cdots$. Counting the number of positive factors of this yields $(2+1)(a+1)(b+1)\cdots =3k$, where $k$ is some integer. But we know $J$ has 16 factors, and since 16 is not divisible by 3, $16\ne 3k$ for any integer $k$. So $J$ cannot have exactly two factors of 2, so it must have at least 3. This means that $J$ is divisible by $2^3\cdot 3\cdot 5=120$. But 120 already has $(3+1)(1+1)(1+1)=16$ factors, so $J$ must be $\boxed{120}$ (or else $J$ would have more than 16 factors).