More problems in 2016 part 9 Grand Total

We know that $\prod_{d|n}\phi(d)=(\phi(n))^{\tau(n)/2}$

for a square-free $n$ ; to prove this formula, just group the divisors in $\tau(n)/2$ pairs whose product is $n$ . For the given $n$ , with 4 prime divisors, we have $\tau(n)/2=2^4/2=8$ , so that the first term in our sum is $-8$ .

A straightforward computation shows that the second summand is $\phi(2^4+18*24*42*44)=389120$ , for a grand total of $\boxed{389112}$