A natural number has prime factorization given by $$n = 2^x 3^y 5^z$$, where $$y$$ and $$z$$ are such that $$y + z = 5$$ and $$y^{-1} + z^{-1} = \frac{5}{6}$$, $$y > z$$. Then the number of odd divisors of $$n$$, including 1, is:

We are given $$n = 2^x 3^y 5^z$$ where $$y + z = 5$$ and $$y^{-1} + z^{-1} = \dfrac{5}{6}$$, with $$y > z$$.

From the second equation, $$\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{y + z}{yz} = \dfrac{5}{yz} = \dfrac{5}{6}$$, which gives $$yz = 6$$.

So $$y$$ and $$z$$ are roots of $$t^2 - 5t + 6 = 0$$, i.e., $$(t - 3)(t - 2) = 0$$. Since $$y > z$$, we get $$y = 3$$ and $$z = 2$$.

Thus $$n = 2^x \cdot 3^3 \cdot 5^2$$. The odd divisors of $$n$$ are the divisors of $$3^3 \cdot 5^2$$ (we exclude all factors of 2). The number of such divisors is $$(3 + 1)(2 + 1) = 12$$.

The answer is $$12$$.