One more Ramanujan Sum

Via Moebius' inversion formula we can write a sum:

$c_{2015}(n) = \sum_{d|(2015,n)}μ(2015/d) \times d$

μ denotes the Moebius function.

This means we only have to consider the dividers of 2015 which are 1, 5, 13, 31, 65, 155, 403 and 2015. For coprime values Ramanujan's sum is -1. The others' sums will be determined by the numbers above.
The minimum value is obtained when n=403: $c_{2015}(403) = -1 +13 +31 - 403 = -360$

Remark: For all numbers greater than 2015 we can separate a multiplier $e^{2\pi i}$ (equals 1) from any $n^{th}$ power root and the result will not be influenced.

The solution I had in mind is analogous to Mr Wendler's.

Since the Ramanujan sum $c_m(n)$ is multiplicative in $m$ , we have $c_{2015}(n)=c_5(n)c_{13}(n)c_{31}(n)$ . For prime numbers $p$ we know that $c_p(n)=p-1$ if $p|n$ and $c_p(n)=-1$ otherwise. Thus there are eight possible values for $c_{2015}(n)$ , the minimal of which is $c_{2015}(n)=(-1)\times 12 \times 30=\boxed{-360}$ , attained when $n$ is divisible by $13\times 31=403$ but not by 5.