The answer is 0.

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We can apply a fundamental result of the mobius function . Specifically, if $n > 1$ then $\sum_{d|n} \mu(d) = \boxed{0}.$

The proof is as follows. Let $n = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdot \ldots \cdot p_r^{\alpha_r}$ ( $\alpha_i \ge 1$ for all i =1,...,r). If $d|n$ , $d = p_1^{\beta_1} \cdot p_2^{\beta_2} \cdot \ldots \cdot p_r^{\beta_r}$ with $0 \leq \beta_i \leq \alpha_i$ for i = 1,...,r. If $\beta_i \ge 2$ for some i = 1, ... , r , then $\mu(d) = 0$ . Therefore, $\large \sum_{d|n} \mu(d) = \sum_{(\beta_1, \ldots , \beta_r)/ \beta_i = 0 \quad \text{or} \quad 1} \mu(p_1^{\beta_1} \cdot \ldots \cdot p_r^{\beta_r}) =$ $\large = 1 - \binom{r}{1} + \binom{r}{2} - \ldots + (-1)^{r} \binom {r}{r} = (1 - 1)^{r} = 0.$