Multiplicative Functions and Totient Function

Question

I have two questions.

1. If $f(n)$ is a multiplicative function defined on the positive integers, is

$g(n)=\frac{f(n)}{n}$ multiplicative as well?

I think the answer is yes, but I don't know how to prove it.

2. Evaluate $$\sum_{d\mid2016} \frac{\phi(d)}{d}$$

where $\phi(n)$ is the totient function.

Answer 1

As regards the second question, note that $$g(n):=\sum_{d|n} \frac{\phi(d)}{d}$$ is multiplicative (it is the Dirichlet convolution of two multiplicative functions).

Moreover, $n=2016=2^5\cdot3^2\cdot 7$, and for any prime $p$, $$g(p^k)=\sum_{d|p^k} \frac{\phi(d)}{d}=1+\sum_{j=1}^k\frac{p^j-p^{j-1}}{p^j}=1+k\left(1-\frac{1}{p}\right).$$ Hence $$g(2016)=g(2^5)\cdot g(3^2)\cdot g(7)=\frac{7}{2}\cdot\frac{7}{3}\cdot \frac{13}{7}=\frac{91}{6}.$$

Answer 2

1) Show that for positive integers $m,n$, one has $g(mn) =g (n)g(m) $. Note this is precisely the definition of a function being 'multiplicative.'

2) We have $d$ as any divisor of 2016. Now observe that $2016=2^5×3^2×7$.

Use the fact that the Euler Totient function is multiplicative for coprime numbers to simplify your counting process.