# A natural number has prime factorization given

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:

1. (1) 11

2. (2) $6 x$

3. (3) 12

4. (4) 6

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 $(3+1)(2+1)=12$