Let a,b,c be three distinct positive numbers. 
Consider  the quadratic polynomial $P (x) =\frac{c(x - a)(x - b)}{(c -a)(c -b)}+\frac{a(x - b)(x - c)}{(a - b)(a - c)}+\frac{b(x -c)(x - a)}{(b - c)(b - a)}+ 1$.
The value of $P (2017)$ is

(A): $2015$ (B): $2016$ (C): $2017$ (D): $2018$ (E): None of the above.
1. Consider the given polynomial:
   \[
   P(x) = \frac{c(x - a)(x - b)}{(c - a)(c - b)} + \frac{a(x - b)(x - c)}{(a - b)(a - c)} + \frac{b(x - c)(x - a)}{(b - c)(b - a)} + 1
   \]

2. We need to show that the expression:
   \[
   Q(x) = \frac{c(x - a)(x - b)}{(c - a)(c - b)} + \frac{a(x - b)(x - c)}{(a - b)(a - c)} + \frac{b(x - c)(x - a)}{(b - c)(b - a)}
   \]
   is equal to zero for all \( x \).

3. Notice that \( Q(x) \) is a quadratic polynomial in \( x \). Let's denote it as:
   \[
   Q(x) = A x^2 + B x + C
   \]

4. To determine the coefficients \( A \), \( B \), and \( C \), we evaluate \( Q(x) \) at \( x = a \), \( x = b \), and \( x = c \):
   \[
   Q(a) = \frac{c(a - a)(a - b)}{(c - a)(c - b)} + \frac{a(a - b)(a - c)}{(a - b)(a - c)} + \frac{b(a - c)(a - a)}{(b - c)(b - a)} = 0 + 1 + 0 = 1
   \]
   \[
   Q(b) = \frac{c(b - a)(b - b)}{(c - a)(c - b)} + \frac{a(b - b)(b - c)}{(a - b)(a - c)} + \frac{b(b - c)(b - a)}{(b - c)(b - a)} = 0 + 0 + 1 = 1
   \]
   \[
   Q(c) = \frac{c(c - a)(c - b)}{(c - a)(c - b)} + \frac{a(c - b)(c - c)}{(a - b)(a - c)} + \frac{b(c - c)(c - a)}{(b - c)(b - a)} = 1 + 0 + 0 = 1
   \]

5. Since \( Q(a) = Q(b) = Q(c) = 1 \), and \( Q(x) \) is a quadratic polynomial, the only way for \( Q(x) \) to be equal to 1 at three distinct points is if \( Q(x) \) is identically equal to 1. Therefore:
   \[
   Q(x) = 1 \quad \text{for all } x
   \]

6. Substituting \( Q(x) \) back into the original polynomial \( P(x) \):
   \[
   P(x) = Q(x) + 1 = 1 + 1 = 2
   \]

7. Therefore, \( P(x) = 2 \) for all \( x \), including \( x = 2017 \).

The final answer is \(\boxed{2}\)