Number of real roots of $\frac{a_1}{a_1-x}+\frac{a_2}{a_2-x}+...+\frac{a_n}{a_n-x}=2016$ for $0Does it have exactly $n$ roots? Would replacing the R.H.S. of the equation with any other real number change the outcome? I can show that the equation has no complex roots. But how to find the number of roots?
 A: Let $f(x)$ be the left side.  Note that $f(x) \to 0$ as $x \to \pm \infty$, $+\infty$ as $x \to a_i-$ and $-\infty$ as $x \to a_i+$.  So there will be at least one root in each interval $(-\infty, a_1)$, $(a_1, a_2)$, ..., $(a_{n-1}, a_n)$.  Since the equation is equivalent to a polynomial equation with degree $n$, there are at most $n$ roots.  Therefore there are exactly $n$ real roots.  
EDIT: There is nothing special about $2016$: any other positive real number would work (for a negative number, you would replace $(-\infty, a_1)$ by $(a_n, \infty)$ and still have $n$ real roots).  For $0$, on the other hand, there would only be $n-1$ roots.
A: The equation is nothing but a polynomial of degree $n$ in disguise (except at $x=a_i$). Hence, it will have $n$ roots. You can show that the roots are interlaced by consecutive $a_i$'s and hence will have $n$ real roots. We have
$$p(x) = 2016 \prod_{k=1}^n (a_k-x) - \sum_{i=1}^n \left(a_i \prod_{\overset{k=1}{k \neq i}}^n (a_k-x) \right)$$
We have
$$p(a_1) < 0, p(a_2)>0, p(a_3)<0, \cdots,$$
This ensures there are $n-1$ real roots that are interlaced. Hence, the $n^{th}$ root must also be real.
