A natural number has prime factorisation given by n=2^x3^y5^z, where y and z are such that…

Question

A natural number has prime factorisation given by $n=2^x3^y5^z$, where $y$ and $z$ are such that $y+z=5$ and $y^{-1}+z^{-1}=\dfrac56$, $y>z$. Then the number of odd divisors of $n$, including 1, is

Accepted Answer

$y+z=5$, $\dfrac{y+z}{yz}=\dfrac56 \Rightarrow yz=6$.

Solving $y+z=5,\ yz=6$: $y,z$ are roots of $t^2-5t+6=0 \Rightarrow t=2,3$. Since $y>z$: $y=3,z=2$.

For odd divisors, $x$ must contribute nothing, i.e. we only consider $3^y5^z=3^3\times5^2$.

$$\text{Number of odd divisors} = (3+1)(2+1) = 12$$

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