Question

A natural number has prime factorization given by$\begin{array}{rcl}n& =& 2^x{3}^y{5}^z\end{array}$, where $y$and $z$are such that$\begin{array}{rcl}y+z& =& 5,\frac{1}{y}+\frac{1}{z}=\frac{5}{6}andyz=6\end{array}$. Then the number of odd divisors of$n$, including$1$, is

• A. $11$
• B. $16$
• C. $12$
• D. $6$

Answer 1

The correct option is C

$12$

Explanation for the correct option:

Calculate the number of odd divisors of $n$ :

Given $\begin{array}{rcl}n& =& 2^x{3}^y{5}^z,y+z=5,\frac{1}{y}+\frac{1}{z}=\frac{5}{6}andyz=6\end{array}$

Also,

$\begin{array}{rcl}{(y-z)}^2& =& {(y+z)}^2-4yz\\ & =& 25-24\\ & =& \pm 1\end{array}$

$$\begin{gathered} y+z=5......\left(1\right) \\ y-z=\pm 1......\left(2\right) \end{gathered}$$

By solving these equations we get $y=3or2,z=2or3$

To calculate the odd divisors of $n,x$must be$0$.

$n=2^0{3}^3{5}^{2}orn=2^0{3}^2{5}^3$

Total odd divisors must be $(3+1)(2+1)=12$

Hence, the correct option is C.