smc

Working with the second part of the problem first, we know that $3n$ has $30$ divisors. We try to find the various possible prime factorizations of $3n$ by splitting $30$ into various products of $1,2$ or $3$ integers. $30\to p^{29}$ $2\cdot 15\to p{q}^{14}$ $3\cdot 10\to p^2{q}^9$ $5\cdot 6\to p^4{q}^5$ $2\cdot 3\cdot 5\to p{q}^2{r}^4$ The variables $p,q,r$ are different prime factors, and one of them must be $3$. We now try to count the factors of $2n$, to see which prime factorization is correct and has $28$ factors. In the first case, $p=3$ is the only possibility. This gives $2n=2\cdot p^{28}$, which has $2\cdot 29$ factors, which is way too many. In the second case, $p=3$ gives $2n=2q^{14}$. If $q=2$, then there are $16$ factors, while if $q\ne 2$, there are $2\cdot 15=30$ factors. In the second case, $q=3$ gives $2n=2p{3}^{13}$. If $p=2$, then there are $3\cdot 13$ factors, while if $p\ne 2$, there are $2\cdot 2\cdot 13$ factors. In the third case, $p=3$ gives $2n=2\cdot 3\cdot q^9$. If $q=2$, then there are $11\cdot 2=22$ factors, while if $q\ne 2$, there are $2\cdot 2\cdot 10$ factors. In the third case, $q=3$ gives $2n=2\cdot p^2\cdot 3^8$. If $p=2$, then there are $4\cdot 9$ factors, while if $p\ne 2$, there are $2\cdot 3\cdot 9$ factors. In the fourth case, $p=3$ gives $2n=2\cdot 3^3\cdot q^5$. If $q=2$, then there are $7\cdot 4=28$ factors. This is the factorization we want. Thus, $3n=3^4\cdot 2^5$, which has $5\cdot 6=30$ factors, and $2n=3^3\cdot 2^6$, which has $4\cdot 7=28$ factors. In this case, $6n=3^4\cdot 2^6$, which has $5\cdot 7=35$ factors, and the answer is $\boxed{35}$