Let $P(x)$ be the product of all linear polynomials $ax+b$, where $a,b\in \{0,\ldots,2016\}$ and $(a,b)\neq (0,0)$. Let $R(x)$ be the remainder when $P(x)$ is divided by $x^5-1$. Determine the remainder when $R(5)$ is divided by $2017$.
1. We start by considering the polynomial \( P(x) \) which is the product of all linear polynomials \( ax + b \) where \( a, b \in \{0, \ldots, 2016\} \) and \((a, b) \neq (0, 0)\). We need to find the remainder \( R(x) \) when \( P(x) \) is divided by \( x^5 - 1 \).

2. Since we are working modulo \( 2017 \) (a prime number), we can use properties of polynomials over finite fields. Note that \( P(x) \) can be expressed as:
   \[
   P(x) = -\prod_{a=1}^{2016} ((ax)^{2016}-1)(ax)
   \]
   This is because for each \( a \), the polynomial \( ax + b \) can be factored into linear terms.

3. Simplifying the expression, we get:
   \[
   P(x) = x^{2016} (x^{2016} - 1)^{2016}
   \]
   Since \( 2016 \equiv 1 \pmod{5} \), we have:
   \[
   P(x) \equiv x (x - 1)^{2016} \pmod{x^5 - 1}
   \]

4. Next, we use the fact that \( (x - 1)^{2016} \equiv \frac{x^{2017} - 1}{x - 1} \pmod{2017} \). This simplifies to:
   \[
   P(x) \equiv x \left( \frac{x^{2017} - 1}{x - 1} \right) \pmod{x^5 - 1}
   \]

5. Since \( x^{2017} \equiv x \pmod{x^5 - 1} \), we have:
   \[
   \frac{x^{2017} - 1}{x - 1} \equiv \frac{x - 1}{x - 1} = 1 \pmod{x^5 - 1}
   \]
   Therefore:
   \[
   P(x) \equiv x \cdot 1 = x \pmod{x^5 - 1}
   \]

6. Now, we need to find \( R(5) \) modulo \( 2017 \). Since \( P(x) \equiv x \pmod{x^5 - 1} \), we have:
   \[
   R(x) = x
   \]
   Thus:
   \[
   R(5) = 5
   \]

7. Finally, we need to find the remainder when \( 5 \) is divided by \( 2017 \). Clearly, \( 5 \) is already less than \( 2017 \), so the remainder is \( 5 \).

The final answer is \( \boxed{5} \).