If $k=10q+r$, and $q\geq 1$, then obviously the product has at least $q$ zeroes on the end, which is more zeroes as $10q+r$, if $q\geq 2$. Thus, $k\leq 19$. Furthermore, if $k$ is $11$ to $19$, it will end with a $0$, so it cannot be. Thus, $k\leq 10$. $k=1$: $1$ $k=2$: $3\cdot 4$ $k=3$: No, it has to be even $k=4$: $1\cdot 2\cdot 3\cdot 4$ $k=5$: No, it will be even and multiple of $5$, thus will end in $0$. For same reason, $6$, $7$, $8$, $9$ fail. So, final candidate is $10$. But, obviously it will have two multiples of $5$ and five multiples of $2$, so actually will be multiple of $100$. Thus it is impossible. Thus, $\boxed{1,2,4}$