What is the maximum possible value of $5-|6x-80|$ over all integers $x$?

$\textbf{(A) }{-}1\qquad\textbf{(B) }0\qquad\textbf{(C) }1\qquad\textbf{(D) }3\qquad\textbf{(E) }5$
1. We start with the expression \(5 - |6x - 80|\). To maximize this expression, we need to minimize the absolute value term \(|6x - 80|\).
2. The absolute value \(|6x - 80|\) is minimized when \(6x - 80\) is as close to 0 as possible. This occurs when \(6x = 80\).
3. Solving for \(x\), we get:
   \[
   x = \frac{80}{6} = \frac{40}{3} \approx 13.3333
   \]
   Since \(x\) must be an integer, we consider the closest integers to \(\frac{40}{3}\), which are \(x = 13\) and \(x = 14\).
4. Calculate \(|6x - 80|\) for \(x = 13\):
   \[
   6 \cdot 13 - 80 = 78 - 80 = -2 \quad \Rightarrow \quad |6 \cdot 13 - 80| = | -2 | = 2
   \]
5. Calculate \(|6x - 80|\) for \(x = 14\):
   \[
   6 \cdot 14 - 80 = 84 - 80 = 4 \quad \Rightarrow \quad |6 \cdot 14 - 80| = | 4 | = 4
   \]
6. Since \(|6 \cdot 13 - 80| = 2\) is smaller than \(|6 \cdot 14 - 80| = 4\), the minimum value of \(|6x - 80|\) is 2, which occurs at \(x = 13\).
7. Substitute this value back into the original expression:
   \[
   5 - |6 \cdot 13 - 80| = 5 - 2 = 3
   \]

The final answer is \(\boxed{3}\).