Here is another question that I got stuck with. Please see attachment. I typed the question in " Microsoft word" . If you can't load the attachment , please do let me know.
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Did you consider using the Mobius inversion formula?
If $\displaystyle g(n)=\sum_{d|n} f(d)$is multiplicative then,
$\displaystyle f(n)=\sum_{d|n} \mu (n/d) \cdot g(d) $
Now, $\displaystyle g(n)$ is multiplicative function
And $\displaystyle \mu(n)$ (the Mobius mu function) is also multiplicative.
Thus,
$\displaystyle g(n)\cdot \mu(n)$ is also multiplicative.
Thus,
$\displaystyle f(n)$ is multiplicative (because of the summation theorem).