jmc

We know that the rule for divisibility by $3$ is that the digits of the number must add up to a multiple of $3$. So, it's clear that there are no other such two-digit numbers beyond the ones listed in the problem. Every number divisible by $3$ between $100$ and $199$ is in the sequence, so that gets us through the $39$th term of the sequence. Using the rule for divisibility by $3$, it is fairly simple to list out the remaining $11$ terms of the sequence: $201,210,213,216,219,231,261,291,312,315,318$. Thus, the $50$th term is $\boxed{318}$.