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Step-by-step Solution
Step 1: Understand the problem
We have a natural number defined by
$ n = 2^x \times 3^y \times 5^z $,
where y and z satisfy:
1) $y + z = 5$
2) $ \frac{1}{y} + \frac{1}{z} = \frac{5}{6}$
and $y > z$. We want to find the total count of odd divisors of $n$. Note that odd divisors occur when $x = 0$, because any power of 2 would make the divisor even.
Step 2: Solve the system of equations for y and z
1) $y + z = 5$
2) $ \frac{1}{y} + \frac{1}{z} = \frac{5}{6}$
From the second equation:
\[
\frac{1}{y} + \frac{1}{z} = \frac{5}{6} \quad \Rightarrow \quad \frac{y+z}{yz} = \frac{5}{6}.
\]
Since $y+z = 5$, substitute this to get:
\[
\frac{5}{yz} = \frac{5}{6} \quad \Rightarrow \quad yz = 6.
\]
We also have $y + z = 5$, so we can find $y$ and $z$ by using:
\[
(y - z)^2 = (y + z)^2 - 4yz = 25 - 4 \times 6 = 25 - 24 = 1.
\]
\[
y - z = 1 \quad (\text{given } y > z).
\]
Solving the system
\[
\begin{cases}
y+z=5 \\
y-z=1
\end{cases}
\]
gives $y = 3$ and $z = 2$.
Step 3: Identify the exponent x for odd divisors
To count odd divisors, we must set $x=0$, because any higher power of 2 would contribute a factor of 2, making the divisor even. Hence, we focus on
\[
n_\text{odd} = 2^0 \times 3^3 \times 5^2.
\]
Step 4: Count the number of odd divisors
A divisor of $3^3 \times 5^2$ can be written as $3^a \times 5^b$, where $a$ can be 0, 1, 2, or 3 and $b$ can be 0, 1, or 2. Thus:
- Possible exponents for 3: 0, 1, 2, 3 (which is 4 choices).
- Possible exponents for 5: 0, 1, 2 (which is 3 choices).
The total number of such odd divisors is
\[
(3 + 1) \times (2 + 1) = 4 \times 3 = 12.
\]
Final Answer
The number of odd divisors of $n$ is 12.
