Sixfold fun in 2015, Part 38

it is a multiplicative function. let $f(d)=\dfrac{\phi(d)}{d}$ then f(1)=1, and f(ab) for coprime a,b $f(ab)=\dfrac{\phi(ab)}{ab}=\dfrac{\phi(a)}{a}\dfrac{\phi(b)}{b}=f(a)f(b)$ the summation at 1 is one. that is also multiplicative. so $\sum_{d|2*3*5*13*31}\dfrac{\phi(d)}{d}=\left(\sum_{d|2}\dfrac{\phi(d)}{d}\right)\left(\sum_{d|3}\dfrac{\phi(d)}{d}\right)\left(\sum_{d|5}\dfrac{\phi(d)}{d}\right)\left(\sum_{d|13}\dfrac{\phi(d)}{d}\right)\left(\sum_{d|31}\dfrac{\phi(d)}{d}\right)$ $=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{2}{3}\right)\left(1+\dfrac{4}{5}\right)\left(1+\dfrac{12}{13}\right)\left(1+\dfrac{30}{31}\right)=\dfrac{13725}{\Large{\boxed{806}}}$

Very nice problem! My approach was quite different from the posted solutions. For the sake of variety, let me add my solution here too.

We note $n:=12090=2\times 3\times 5\times 13\times 31$ .

$S:=\sum_{d\mid n}\frac{\phi(d)}{d}=\sum_{d\mid n}\frac{d\cdot\mathcal{P}(d)}{d}=\sum_{d\mid n}\mathcal{P}(d)$

Here, $\large\mathcal{P}(d)=\prod\limits_i\left(1-\frac 1{q(d,i)}\right)$ where the $q(d,i)$ 's are the prime factors of $d$ , i.e., we have $\{q(d,i)\}\subseteq\{2,3,5,13,31\}=\{p_i\}~\forall~d:d\mid n$

Since multiplicity of every prime factor of $n$ is $1$ , it makes our work simple. We calculate $r_i=\large\left(1-\frac 1{p_i}\right)$ for every element of $\{p_i\}_{i\geq 1}=\{2,3,5,13,31\}$ .

Note that,

$\large S=\sum_{d\mid n}\mathcal{P}(d)=1+\sum_{k=1}^5\sum\limits_{1\leq i_1 < i_2 < \cdots < i_k \leq 5}\prod_{j=1}^k r_{i_j}$

Now, consider the polynomial $f(x)=(x-1)\prod\limits_{i=1}^5\left(x-r_i\right)=x^6+a_5x^5+a_4x^4+\cdots a_1x+a_0~\textrm{(say)}$

By Vieta's formulas and with a bit of inspection, we can conclude that $f(-1)=1+\sum\limits_{k=0}^5(-1)^ka_k=2S$

So, we calculate $f(-1)$ which is,

$\begin{aligned}2S=f(-1)&=\left(-1-1\right)\prod_{i=1}^5\left(-1-r_i\right)\\&=\left(-1-1\right)\left(-1-\frac 12\right)\left(-1-\frac 23\right)\left(-1-\frac 45\right)\left(-1-\frac{12}{13}\right)\left(-1-\frac{30}{31}\right)=\frac{13725}{403}\end{aligned}$

$\implies S=\frac{13725}{806}$

Since $\gcd(13725,806)=1$ , for $aS=\dfrac{13725a}{806}$ to be an integer, we need $a\equiv 0\pmod{806}$ , i.e., $a=806m~\forall~m\in\Bbb Z$ with the smallest positive integer $a$ being $806$ at $m=1$ .

Here is a compressed version of Aareyan's fine solution:

$f(n)=\sum_{d|n}\frac{\phi(d)}{d}$ is multiplicative, with $f(p)=\frac{\phi(p)}{p}+\frac{\phi(1)}{1}=\frac{2p-1}{p}$ for primes $p$ , so that $f(2*3*5*13*31)=f(2)f(3)f(5)f(13)f(31)=\frac{3*5*9*25*61}{2*3*5*13*31}=\frac{13725}{806}$ .

As an interesting aside, note that $f(n)=\frac{1}{n}\sum_{k=1}^{n}\gcd(k,n)$ .