If the remainder is $2013$ when a polynomial with coefficients from the set $\{0,1,2,3,4,5\}$ is divided by $x-6$, what is the least possible value of the coefficient of $x$ in this polynomial?

$ 
\textbf{(A)}\ 5
\qquad\textbf{(B)}\ 4
\qquad\textbf{(C)}\ 3
\qquad\textbf{(D)}\ 2
\qquad\textbf{(E)}\ 1
$
To solve this problem, we need to use the Remainder Theorem, which states that the remainder of the division of a polynomial \( P(x) \) by \( x - a \) is \( P(a) \). Here, we are given that the remainder is 2013 when the polynomial is divided by \( x - 6 \). Therefore, we have:

\[ P(6) = 2013 \]

We need to find the polynomial \( P(x) \) with coefficients from the set \(\{0, 1, 2, 3, 4, 5\}\) such that the coefficient of \( x \) is minimized.

1. **Express the polynomial in general form:**
   Let \( P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \), where \( a_i \in \{0, 1, 2, 3, 4, 5\} \).

2. **Evaluate the polynomial at \( x = 6 \):**
   \[ P(6) = a_n \cdot 6^n + a_{n-1} \cdot 6^{n-1} + \cdots + a_1 \cdot 6 + a_0 = 2013 \]

3. **Simplify the problem:**
   We need to find the coefficients \( a_i \) such that the sum equals 2013. We aim to minimize \( a_1 \), the coefficient of \( x \).

4. **Convert 2013 to base 6:**
   To find the coefficients, we convert 2013 to base 6. This will help us determine the coefficients \( a_i \).

   \[
   2013 \div 6 = 335 \quad \text{remainder} \quad 3 \quad (a_0 = 3)
   \]
   \[
   335 \div 6 = 55 \quad \text{remainder} \quad 5 \quad (a_1 = 5)
   \]
   \[
   55 \div 6 = 9 \quad \text{remainder} \quad 1 \quad (a_2 = 1)
   \]
   \[
   9 \div 6 = 1 \quad \text{remainder} \quad 3 \quad (a_3 = 3)
   \]
   \[
   1 \div 6 = 0 \quad \text{remainder} \quad 1 \quad (a_4 = 1)
   \]

   Therefore, the base-6 representation of 2013 is \( 11353_6 \).

5. **Construct the polynomial:**
   Using the base-6 representation, the polynomial is:
   \[
   P(x) = 1 \cdot x^4 + 1 \cdot x^3 + 3 \cdot x^2 + 5 \cdot x + 3
   \]

6. **Identify the coefficient of \( x \):**
   The coefficient of \( x \) in the polynomial \( P(x) \) is 5.

Thus, the least possible value of the coefficient of \( x \) in this polynomial is 5.

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