jmc

We have $$729_{16}=7\cdot 16^2+2\cdot 16+9.$$ We could actually convert this to base 10 and divide by $15$, but we can also use a clever trick here, writing $16$ as $15+1$ and $16^2$ as $15\cdot 17+1$. Then $$\begin{array}{rl}729_{16}& =7(15\cdot 17+1)+2(15+1)+9\\ & =15(7\cdot 17+2)+(7+2+9)\\ & =(\text{a multiple of 15})+18,\end{array}$$ so when we divide by $15$, the remainder is the same as the remainder left by $18$. This is $\boxed{3}$.

(This trick may remind you of the justification for the test of divisibility by $9$ in base $10$. That isn't a coincidence!)