Sixfold fun in 2015, Part 37

First, we have $\phi(m)$ * $\phi(n)$ = $\phi(m*n)$ if m and n are coprime

Second, we have 12090 = 2 x 3 x 5 x 13 x 31, mean that with every positive interger divisors $d$ of 12090, we have $d$ and $12090/d$ are coprime, in other word $\phi(d)$ * $\phi(12090/d)$ = $\phi(12090)$

Then we have $S=\sum_{d|12090}\frac{1}{\phi(d)}=\sum_{d|12090}\frac{\phi(d)}{\phi(12090)}=\frac{\sum_{d|12090}\phi(d)}{\phi(12090)}$

We can easily see that $\sum_{d|12090}\phi(d) = (\phi(2)+1)*(\phi(3)+1)*(\phi(5)+1)*(\phi(13)+1)*(\phi(31)+1)$

Because RHS when expand will have 32 parts, each part will have a form of $\phi(i)*\phi(j)*\phi(k)...$ , in which i,j,k... are in the set of (2,3,5,13,31), which equal to $\phi(d), d|12090$ , equal to LHS.

If d is a prime then $\phi(d) = d-1$ , So $S=\frac{(\phi(2)+1)*(\phi(3)+1)*(\phi(5)+1)*(\phi(13)+1)*(\phi(31)+1)}{\phi(2)*\phi(3)*\phi(5)*\phi(13)*\phi(31)}=\frac{2*3*5*13*31}{1*2*4*12*30}=\frac{403}{96}$

$f(n)=\sum_{d|n}\frac{1}{\phi(d)}$ is a multiplicative function, with $f(p)=\frac{1}{\phi(1)}+\frac{1}{\phi(p)}=\frac{p}{p-1}$ for primes.

Thus $S=f(12090)=f(2*3*5*13*31)=\frac{2\times 3\times 5\times 13\times 31}{1\times 2\times 4\times 12\times 30}=\frac{403}{96}$ , so that $a=\boxed{96}$