A natural number has prime factorization given by n = 2 ^ x 3 ^ y 5...

$y + z = 5$

$\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{5}{6}y > z$

$⇒ y = 3, z = 2$

$⇒ n = {2}^{x} . {3}^{3} . {5}^{2} = \left(2 . 2 . 2 \ldots\right)\left(3 . 3 . 3\right)\left(5 . 5\right)$

For calculating the odd divisor of$n = {2}^{x}{3}^{y}{5}^{z},$ $x$ must be $0$.

Hence, Number of odd divisors $= \left(3 + 1\right) \times \left(2 + 1\right) = 4 \times 3 = 12$.