Let for the $$9^{th}$$ term in the binomial expansion of $$(3 + 6x)^n$$, in the increasing powers of $$6x$$, to be the greatest for $$x = \frac{3}{2}$$, the least value of $$n$$ is $$n_0$$. If $$k$$ is the ratio of the coefficient of $$x^6$$ to the coefficient of $$x^3$$, then $$k + n_0$$ is equal to

We need to find the least value $$n_0$$ such that the 9th term in the binomial expansion of $$(3 + 6x)^n$$ (in increasing powers of $$6x$$) is greatest when $$x = \frac{3}{2}$$, and then compute $$k + n_0$$ where $$k$$ is the ratio of the coefficient of $$x^6$$ to the coefficient of $$x^3$$.

Since the $$(r+1)$$-th term in the expansion of $$(3 + 6x)^n$$ is given by $$T_{r+1} = \binom{n}{r} 3^{n-r} (6x)^r,$$ substituting $$x = \frac{3}{2}$$ yields $$T_{r+1} = \binom{n}{r} 3^{n-r} \cdot 6^r \cdot \left(\frac{3}{2}\right)^r = \binom{n}{r} 3^{n-r} \cdot \frac{6^r \cdot 3^r}{2^r} = \binom{n}{r} 3^n \cdot \frac{6^r}{2^r} = \binom{n}{r} 3^n \cdot 3^r = \binom{n}{r} 3^{n+r}.$$

From this, the ratio of consecutive terms is $$\frac{T_{r+1}}{T_r} = \frac{\binom{n}{r} 3^{n+r}}{\binom{n}{r-1} 3^{n+r-1}} = \frac{n-r+1}{r} \cdot 3.$$ For the 9th term (corresponding to $$r = 8$$) to be the greatest, we require both $$T_9 \ge T_8$$ and $$T_9 \ge T_{10}$$.

From $$\frac{T_9}{T_8} \ge 1 \quad(\text{ratio at } r = 8),$$ we have $$\frac{3(n - 7)}{8} \ge 1 \implies n \ge \frac{29}{3} \approx 9.67 \implies n \ge 10.$$

Similarly, from $$\frac{T_{10}}{T_9} \le 1 \quad(\text{ratio at } r = 9),$$ we get $$\frac{3(n - 8)}{9} \le 1 \implies n - 8 \le 3 \implies n \le 11.$$ Hence $$10 \le n \le 11$$ and the least value satisfying these inequalities is $$n_0 = 10$$.

Next, to determine $$k$$, the ratio of the coefficient of $$x^6$$ to that of $$x^3$$ in $$(3 + 6x)^{10},$$ note that the coefficient of $$x^r$$ is $$\binom{10}{r} 3^{10-r} \cdot 6^r.$$ Therefore, $$\text{Coefficient of }x^6 = \binom{10}{6} \cdot 3^4 \cdot 6^6,$$ $$\text{Coefficient of }x^3 = \binom{10}{3} \cdot 3^7 \cdot 6^3,$$ and hence $$k = \frac{\binom{10}{6} \cdot 3^4 \cdot 6^6}{\binom{10}{3} \cdot 3^7 \cdot 6^3} = \frac{\binom{10}{6}}{\binom{10}{3}} \cdot \frac{6^3}{3^3} = \frac{210}{120} \cdot \frac{216}{27} = \frac{7}{4} \cdot 8 = 14.$$

Finally, combining these results gives $$k + n_0 = 14 + 10 = 24,$$ so the answer is $$\boxed{24}$$.