The number of positive integers $$k$$ such that the constant term in the binomial expansion of $$\left(2x^3 + \frac{3}{x^k}\right)^{12}, x \neq 0$$ is $$2^8 \cdot l$$, where $$l$$ is an odd integer, is ______

In the expansion of $$\left(2x^3 + \frac{3}{x^k}\right)^{12}$$, the general term is:

$$T_{r+1} = \binom{12}{r} (2x^3)^{12-r} \left(\frac{3}{x^k}\right)^r = \binom{12}{r} 2^{12-r} \cdot 3^r \cdot x^{3(12-r) - kr}$$

For the constant term, the power of $$x$$ must be zero:

$$36 - 3r - kr = 0 \implies r(3 + k) = 36 \implies r = \frac{36}{3 + k}$$

Since $$0 \leq r \leq 12$$ and $$r$$ must be a positive integer, $$(3 + k)$$ must divide 36 and $$\frac{36}{3+k} \leq 12$$, meaning $$3 + k \geq 3$$, i.e., $$k \geq 0$$. Since $$k$$ is a positive integer, $$k \geq 1$$.

Divisors of 36 that are at least 4: $$4, 6, 9, 12, 18, 36$$.

So $$k \in \{1, 3, 6, 9, 15, 33\}$$ with $$r \in \{9, 6, 4, 3, 2, 1\}$$ respectively.

The constant term is $$\binom{12}{r} \cdot 2^{12-r} \cdot 3^r$$.

We need this to equal $$2^8 \cdot l$$ where $$l$$ is odd. This means the highest power of 2 dividing the constant term must be exactly 8.

The power of 2 in the constant term is $$(12 - r) + v_2\binom{12}{r})$$, where $$v_2$$ denotes the 2-adic valuation.

For each case:

$$k = 1, r = 9$$: $$v_2(2^3) + v_2\binom{12}{9} = 3 + v_2(220) = 3 + 2 = 5$$. Not 8.

$$k = 3, r = 6$$: $$v_2(2^6) + v_2\binom{12}{6} = 6 + v_2(924) = 6 + 2 = 8$$. Yes!

$$k = 6, r = 4$$: $$v_2(2^8) + v_2\binom{12}{4} = 8 + v_2(495) = 8 + 0 = 8$$. Yes!

$$k = 9, r = 3$$: $$v_2(2^9) + v_2\binom{12}{3} = 9 + v_2(220) = 9 + 2 = 11$$. Not 8.

$$k = 15, r = 2$$: $$v_2(2^{10}) + v_2\binom{12}{2} = 10 + v_2(66) = 10 + 1 = 11$$. Not 8.

$$k = 33, r = 1$$: $$v_2(2^{11}) + v_2\binom{12}{1} = 11 + v_2(12) = 11 + 2 = 13$$. Not 8.

The values of $$k$$ that work are $$k = 3$$ and $$k = 6$$.

Hence the answer is $$\boxed{2}$$.