Interpret the factorization

The given prime factorization is $n=2^x3^y5^z$ where $y$ and $z$ satisfy: $y+z=5$ and $y^{-1}+z^{-1}=\frac{5}{6}, \quad y>z$

Here $y^{-1}=\frac{1}{y}$ and $z^{-1}=\frac{1}{z}$.

Use the second condition

We have $\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$ $\frac{y+z}{yz}=\frac{5}{6}$

Since $y+z=5$, substitute: $\frac{5}{yz}=\frac{5}{6}$ $yz=6$

Find $y$ and $z$

Now $y$ and $z$ satisfy $y+z=5, \quad yz=6$

So they are roots of $t^2-5t+6=0$ $(t-2)(t-3)=0$

Hence the pair is $2,3$. Since $y>z$, $y=3, \quad z=2$

Find the number of odd divisors of $n$

Since $n=2^x3^35^2$ an odd divisor cannot contain the factor $2$.

So an odd divisor is of the form $3^a5^b$ where $0\le a\le 3, \quad 0\le b\le 2$

Number of choices for $a$ is $4$, and for $b$ is $3$.

Therefore, total number of odd divisors is $4\times 3=12$

Compare with options

$12$ corresponds to Option C.

Compare with stored correct answer

Stored correct answer is C, which matches our result.