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(D) Given $y+z=5$ and $\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$.From the second equation,$\frac{y+z}{yz} = \frac{5}{6}$.Substituting $y+z=5$,we get $\frac

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$12$

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(D) Given $y+z=5$ and $\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$.From the second equation,$\frac{y+z}{yz} = \frac{5}{6}$.Substituting $y+z=5$,we get $\frac{5}{yz} = \frac{5}{6}$,which implies $yz = 6$.We have $y+z=5$ and $yz=6$. The quadratic equation $t^2 - 5t + 6 = 0$ has roots $t=2$ and $t=3$.Since $y > z$,we have $y=3$ and $z=2$.The number $n = 2^{x} \cdot 3^{3} \cdot 5^{2}$.An odd divisor of $n$ must be of the form $3^{a} \cdot 5^{b}$,where $0 \le a \le 3$ and $0 \le b \le 2$.The number of choices for $a$ is $(3+1) = 4$.The number of choices for $b$ is $(2+1) = 3$.Total number of odd divisors $= 4 	imes 3 = 12$.

$A$ 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 $y^{-1}+z^{-1}=\frac{5}{6}$,with $y > z$. Then the number of odd divisors of $n$,including $1$,is ..... .

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