# A natural number has prime factorization given by $n = 2^x3^y5^z$, 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 divisons of n, including 1, is: Option: 1 6x Option: 2 6 Option: 3 11 Option: 4 12

Given that

$\\y+z=5 \\ \frac{1}{y}+\frac{1}{z}=\frac{5}{6} \\y>z\\\Rightarrow y=3,\;z=2$

for calculating odd divisor of n = 2x3y5z

x must be zero

$\\\Rightarrow \mathrm{n}=2^{\mathrm{0}} .3^{3} \cdot 5^{2}\\\therefore \text { total odd divisors must be }(3+1)(2+1)=12$

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