More problems in 2016 part 7

all $\mu,\tau,\lambda$ are multiplicative functions, implying the summations are as well.

we use multiplicity and the fact that $\sum_{d|n}\lambda(d)=\begin{cases} 1&&\text{if n is a perfect square}\\0&&\text{otherwise}\end{cases}$ . the first part becomes zero. and the remaining $f(p^k)=\sum_{d|p^k}(\mu(p^k)\tau(p^k))=1-2+3-4+...+(-1)^k(k+1)=\begin{cases} \dfrac{k+2}{2} &&\text{if k is even}\\\dfrac{-k-1}{2}&&\text{if k is odd}\end{cases}$ so $f(41)=-1,f(83)=-1,f(97)=-1$ since f is multiplicative: $f(330091)=f(41)f(83)f(97)=-1*-1*-1=-1$ . so the final answe is $0+(-1)=\boxed{-1}$