Question:
A natural number has prime factorization given by $\mathbf{n}=2^{x} 3^{y} 5^{z}$, where $y$ and $z$ are such that $y+z=5$ and $y^{-1}+z^{-1}=\frac{5}{6}, y>z$. Then the number of odd divisors of $n$, including 1 , is:
Correct Option: , 3
Solution:
$y+z=5 \ldots(1)$
$\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$
$\Rightarrow \frac{y+z}{y z}=\frac{5}{6}$
$\Rightarrow \frac{5}{y z}=\frac{5}{6}$
$\Rightarrow y z=6$
Also $(y-z)^{2}=(y+z)^{2}-4 y z$
$\Rightarrow(y-z)^{2}=(y+z)^{2}-4 y z$
$\Rightarrow(y-z)^{2}=25-4(6)=1$
$\Rightarrow y-z=1$
from (1) and (2), $y=3$ and $z=2$
for calculating odd divisor of $\mathrm{p}=2^{\mathrm{x}} .3^{\mathrm{y}} .5^{z}$
$x$ must be zero
$P=2^{0} \cdot 3^{3} \cdot 5^{2}$
$\therefore$ total odd divisors must be $(3+1)(2+1)=12$